JEE Heat Transfer Questions

$$C = \frac{5}{9}(F - 32)$$

$$C = \frac{5}{9}F - \frac{160}{9}$$

• Slope: The slope ($$\frac{5}{9}$$) is positive, which indicates that the temperature in Celsius increases linearly as the Fahrenheit temperature increases.
• Vertical Intercept ($$C$$-axis): When $$F = 0$$, $$C = -\frac{160}{9} \approx -17.8$$. This means the graph must intersect the vertical axis below the origin.
• Horizontal Intercept ($$F$$-axis): When $$C = 0$$ (the freezing point of water), the equation gives $$F = 32$$. Therefore, the graph must intersect the horizontal axis at a positive value.
  

Among the provided figures, only graph (D) correctly illustrates this correctly.

First convert the given dimensions to SI units:
length $$l = 9\text{ cm} = 0.09\text{ m}$$, width $$d = 4\text{ cm} = 0.04\text{ m}$$.

The initial area of the rectangular sheet is
$$A_0 = l \times d = 0.09 \times 0.04 = 0.0036\ \text{m}^2$$ $$-(1)$$

Heat supplied to the sheet is $$Q = 8.1 \times 10^{2}\ \text{J}$$.
Using the relation $$Q = m\,C_v\,\Delta T$$, the rise in temperature is
$$\Delta T = \frac{Q}{m\,C_v} = \frac{8.1 \times 10^{2}}{0.1 \times 900} = 9\ \text{K}$$ $$-(2)$$

For an isotropic solid, the coefficient of superficial (area) expansion is
$$\beta = 2\alpha$$ $$-(3)$$
where $$\alpha = 3.1 \times 10^{-5}\ \text{K}^{-1}$$ is the linear expansion coefficient.

From $$(3)$$,
$$\beta = 2 \times 3.1 \times 10^{-5} = 6.2 \times 10^{-5}\ \text{K}^{-1}$$ $$-(4)$$

The fractional change in area due to heating through $$\Delta T$$ is $$\beta\,\Delta T$$, so the absolute change in area is
$$\Delta A = A_0 \,\beta \,\Delta T$$ $$-(5)$$

Substituting from $$(1)$$, $$(2)$$ and $$(4)$$ into $$(5)$$:
$$\Delta A = 0.0036 \times 6.2 \times 10^{-5} \times 9$$

Compute step by step:
$$6.2 \times 10^{-5} \times 9 = 5.58 \times 10^{-4}$$;
$$0.0036 \times 5.58 \times 10^{-4} = 2.01 \times 10^{-6}\ \text{m}^2$$.

Rounding to two significant figures,
$$\Delta A \approx 2.0 \times 10^{-6}\ \text{m}^2$$.

Hence, the change in area of the rectangular sheet is $$2.0 \times 10^{-6}\ \text{m}^2$$.
Option A is correct.

We need to find the total heat required to convert ice at $$-10°C$$ to steam at $$110°C$$.

Given data:

Mass $$m = 10^{-3}$$ kg, specific heat of ice $$c_{ice} = 2100 \text{ J kg}^{-1}\text{K}^{-1}$$, specific heat of water $$c_w = 4180 \text{ J kg}^{-1}\text{K}^{-1}$$, specific heat of steam $$c_s = 1920 \text{ J kg}^{-1}\text{K}^{-1}$$, latent heat of ice $$L_f = 3.35 \times 10^5 \text{ J kg}^{-1}$$, latent heat of steam $$L_v = 2.25 \times 10^6 \text{ J kg}^{-1}$$.

$$Q_1 = m \cdot c_{ice} \cdot \Delta T = 10^{-3} \times 2100 \times 10 = 21 \text{ J}$$

$$Q_2 = m \cdot L_f = 10^{-3} \times 3.35 \times 10^5 = 335 \text{ J}$$

$$Q_3 = m \cdot c_w \cdot \Delta T = 10^{-3} \times 4180 \times 100 = 418 \text{ J}$$

$$Q_4 = m \cdot L_v = 10^{-3} \times 2.25 \times 10^6 = 2250 \text{ J}$$

$$Q_5 = m \cdot c_s \cdot \Delta T = 10^{-3} \times 1920 \times 10 = 19.2 \text{ J}$$

$$Q = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \text{ J}$$

$$Q \approx 3043 \text{ J}$$

The correct answer is Option 1: 3043 J.

We need to find the mass of the bullet that melts completely after penetrating the wooden block. Since the total heat required has two components, the first part is the heat to raise the temperature from 300 K to the melting point at 600 K, which is given by $$Q_1 = mc\Delta T$$, and the second part is the heat to melt the bullet at 600 K, given by $$Q_2 = mL$$.

Given that $$c = 125 \text{ J kg}^{-1}\text{K}^{-1}$$, $$L = 2.5 \times 10^{4} \text{ J kg}^{-1}$$, and $$\Delta T = 600 - 300 = 300 \text{ K}$$, the total heat supplied is $$Q = 625 \text{ J}$$.

Substituting these values into the expression for total heat, we have $$Q = mc\Delta T + mL$$, which becomes $$625 = m\bigl(125 \times 300 + 2.5 \times 10^{4}\bigr)$$. Therefore, $$625 = m(37500 + 25000)$$, leading to $$625 = m \times 62500$$. This gives $$m = \frac{625}{62500} = 0.01 \text{ kg} = 10 \text{ grams}$$.

Therefore, the correct answer is Option 1: 10 grams.

Newton’s law of cooling: the rate of fall of temperature of a body is proportional to the temperature difference between the body and the surroundings.

Write it as $$\frac{dT}{dt} = -k\,(T - T_s)$$, where $$T_s$$ is the surrounding temperature and $$k$$ is a positive constant.

Integrating, we obtain the exponential form
$$T - T_s = (T_0 - T_s)\,e^{-kt}$$ $$-(1)$$, where $$T_0$$ is the initial temperature at $$t = 0$$.

Given: $$T_s = 16^{\circ}C$$ and $$T_0 = 40^{\circ}C$$.
Initial temperature difference: $$\Delta T_0 = 40 - 16 = 24^{\circ}C$$.

After $$4$$ min the body cools to $$24^{\circ}C$$.
Temperature difference then: $$\Delta T_4 = 24 - 16 = 8^{\circ}C$$.

Insert these values into $$(1)$$ to determine $$k$$:
$$8 = 24\,e^{-k \times 4}$$

$$e^{-4k} = \frac{8}{24} = \frac{1}{3}$$

Taking natural logarithm on both sides:
$$-4k = \ln\!\left(\frac{1}{3}\right) = -\ln 3$$

Thus $$k = \frac{\ln 3}{4}$$.

We want the temperature after another 4 min, i.e. at $$t = 8$$ min.
From $$(1)$$:

$$T_8 - 16 = 24\,e^{-k \times 8}$$

Substitute $$k = \frac{\ln 3}{4}$$:
$$T_8 - 16 = 24\,e^{-2\ln 3} = 24\,(e^{\ln 3})^{-2} = 24 \times \frac{1}{3^2} = \frac{24}{9} = \frac{8}{3}$$

Hence $$T_8 = 16 + \frac{8}{3} = \frac{48}{3} + \frac{8}{3} = \frac{56}{3}\,^{\circ}C$$.

Therefore the temperature of the body after the next 4 minutes is $$\frac{56}{3}^{\circ}C$$. 

This corresponds to Option D.

For each physical quantity, first recall its definition, then write the factors on which it depends, and finally write the corresponding SI unit.

Case A - Heat capacity of a body
Definition: Heat required to raise the temperature of the entire body by $$1\;{\rm K}$$.
Heat $$Q$$ needed is proportional only to temperature change $$\Delta T$$: $$Q = C\,\Delta T$$ where $$C$$ is heat capacity.
Hence unit of $$C$$ is Joule per Kelvin, $$\rm J\,K^{-1}$$. Match: (A) $$\rightarrow$$ (II).

Case B - Specific heat capacity
Definition: Heat required to raise temperature of unit mass of a substance by $$1\;{\rm K}$$.
Formula: $$Q = m\,c\,\Delta T$$. Here $$c$$ relates heat to both mass and temperature difference, so its unit is Joule per kilogram per Kelvin, $$\rm J\,kg^{-1}\,K^{-1}$$. Match: (B) $$\rightarrow$$ (III).

Case C - Latent heat
Definition: Heat required to change the phase of unit mass of a substance without temperature change.
Formula: $$Q = m\,L$$, where $$L$$ is latent heat.
Only mass appears in the relation, so the unit is Joule per kilogram, $$\rm J\,kg^{-1}$$. Match: (C) $$\rightarrow$$ (I).

Case D - Thermal conductivity
Definition (Fourier’s law for one-dimensional steady conduction):$$\displaystyle \frac{dQ}{dt} = k\,A\,\frac{\Delta T}{\Delta x}$$.
Rearrange for $$k$$: $$k = \frac{1}{A}\,\frac{dQ}{dt}\,\frac{\Delta x}{\Delta T}$$.
Units:
$$\frac{dQ}{dt}$$ has unit $$\rm J\,s^{-1}$$ (power), $$A$$ is $$\rm m^{2}$$, $$\Delta x$$ is $$\rm m$$, $$\Delta T$$ is $$\rm K$$.
Thus $$k$$ has unit $$\dfrac{\rm J\,s^{-1}}{\rm m^{2}}\times \rm m \times \rm K^{-1}=J\,m^{-1}\,K^{-1}\,s^{-1}$$. Match: (D) $$\rightarrow$$ (IV).

Combining all matches:
(A)-(II), (B)-(III), (C)-(I), (D)-(IV).

The option with this sequence is Option D.

For steady one-dimensional conduction through a rod, the heat current is
$$Q=\frac{K\,A}{L}\,\left(T_{\text{hot}}-T_{\text{cold}}\right)$$

When two rods are joined in series, the same heat current $$Q$$ flows through each of them in the steady state. Let $$T$$ be the interface temperature (at the junction of rods A and B).

Given ratios
$$\frac{L_A}{L_B}=\frac12 \;\Rightarrow\; L_A=\frac{L_B}{2}$$
$$\frac{r_A}{r_B}=2 \;\Rightarrow\; r_A=2r_B$$
$$\frac{K_A}{K_B}=\frac12 \;\Rightarrow\; K_A=\frac{K_B}{2}$$

Cross-sectional areas: $$A=\pi r^{2}$$, so
$$A_A=\pi r_A^{2}=\pi\,(2r_B)^{2}=4\pi r_B^{2}=4A_B$$
thus $$\frac{A_A}{A_B}=4$$.

Heat current through rod A:
$$Q=\frac{K_AA_A}{L_A}\,(400-T)$$

Heat current through rod B:
$$Q=\frac{K_BA_B}{L_B}\,(T-200)$$

Equate the two heat currents:

$$\frac{K_AA_A}{L_A}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Substitute the ratio values:
$$K_A=\frac{K_B}{2},\;A_A=4A_B,\;L_A=\frac{L_B}{2}$$

$$\frac{\left(\tfrac{K_B}{2}\right)\,(4A_B)}{\tfrac{L_B}{2}}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Simplify the fraction on the left:
$$\frac{(\tfrac{K_B}{2})\,(4A_B)}{\tfrac{L_B}{2}}=\frac{2K_BA_B}{\tfrac{L_B}{2}}=\frac{2K_BA_B\cdot2}{L_B}=\frac{4K_BA_B}{L_B}$$

Hence
$$\frac{4K_BA_B}{L_B}\,(400-T)=\frac{K_BA_B}{L_B}\,(T-200)$$

Cancel the common factor $$\frac{K_BA_B}{L_B}$$ from both sides:

$$4(400-T)=T-200$$

Solve for $$T$$:
$$1600-4T=T-200$$
$$1600+200=5T$$
$$1800=5T$$
$$T=360\,\text{K}$$

The temperature at the interface of the two rods is $$\mathbf{360\;K}$$.

Treat it as two conductors $$k_1,k_2$$​ in parallel, connected in series with conductor $$k_3$$

Let area of conductors 1 and 2 be A.

Given for conductor 3:

$$A_3=2A$$

All lengths are same l.

Thermal resistance:

$$R=\frac{l}{kA}$$

For conductor 1:

$$R_1=\frac{l}{60A}$$

For conductor 2:

$$R_2=\frac{l}{120A}$$

These are in parallel:

$$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{60A}{l}+\frac{120A}{l}=l60A​+l120A​$$

$$=\frac{180A}{l}$$

So

$$R_p=\frac{l}{180A}$$

Equivalent conductivity of left section:

$$k_{eq}=180$$

Now conductor 3:

$$R_3=\frac{l}{135(2A)}=\frac{l}{270A}$$

Suppose junction temperature is θ\thetaθ.

Heat flow through left section:

$$Q_1=\frac{(100-\theta)}{R_p}$$

$$=(100-\theta)\frac{180A}{l}$$

Heat flow through right section:

$$Q_2=\frac{\theta}{R_3}=\theta\frac{270A}{l}$$

Steady state:

$$Q_1=Q_2$$

$$180(100-\theta)=270\theta$$

$$18000=450\theta$$

$$\theta=40^{\circ}C$$

The relationship between temperature changes on Celsius and Fahrenheit scales is given by $$ \Delta F = \frac{9}{5} \Delta C $$. If the change in Celsius is $$\Delta C = 40°C$$, substituting this into the equation yields

$$ \Delta F = \frac{9}{5} \times 40 = 72°F $$. Therefore, the correct answer is Option 3: 72°F.

The mass of the sample is $$m = 1 \text{ kg}$$ (given).

For a temperature-dependent specific heat capacity $$C = kT$$, the small amount of heat $$dQ$$ needed to raise the temperature by $$dT$$ is
$$dQ = m\,C\,dT = m\,(kT)\,dT = mkT\,dT.$$

Total heat supplied when the temperature rises from $$T_1$$ to $$T_2$$ is obtained by integration:
$$Q = \int_{T_1}^{T_2} mkT\,dT = mk \int_{T_1}^{T_2} T\,dT.$$

Evaluating the integral:
$$\int_{T_1}^{T_2} T\,dT = \left[\frac{T^{2}}{2}\right]_{T_1}^{T_2} = \frac{T_2^{2} - T_1^{2}}{2}.$$

Hence
$$Q = mk\,\frac{T_2^{2} - T_1^{2}}{2}.$$

The temperature limits must be in kelvin. Using the relation $$T(\text{K}) = T(^{\circ}\text{C}) + 273$$:
$$T_1 = -73^{\circ}\text{C} + 273 = 200 \text{ K},$$
$$T_2 = 27^{\circ}\text{C} + 273 = 300 \text{ K}.$$

Substituting $$m = 1 \text{ kg}$$, $$T_1 = 200 \text{ K}$$, $$T_2 = 300 \text{ K}$$:
$$Q = 1 \times k \times \frac{300^{2} - 200^{2}}{2}.$$

Calculate the bracketed term:
$$300^{2} - 200^{2} = 90000 - 40000 = 50000.$$

Therefore,
$$Q = k \times \frac{50000}{2} = 25000\,k.$$

The heat required is given in the problem as $$nk$$, so comparing,
$$n = 25000.$$

Final answer: 25000

For ice at −10°C heated to steam at 100°C, the qualitative heating curve should have:

1. Rising temperature from −10°C to 0°C (ice warming)
2. Horizontal plateau at 0°C (melting, temperature constant)
3. Rising temperature from 0°C to 100°C (water warming)
4. Horizontal plateau at 100°C (boiling, temperature constant)

A body cools from 60°C to 40°C in 6 minutes. Temperature of surroundings is 10°C. Find the temperature after the next 6 minutes.

According to Newton's Law of Cooling: $$T(t) - T_s = (T_0 - T_s)e^{-kt}$$ where $$T_s = 10°$$C is the surrounding temperature.

At $$t=6$$ min, $$T=40°$$C, $$T_0=60°$$C, so $$40-10=(60-10)e^{-6k}$$, giving $$30=50e^{-6k}$$ and hence $$e^{-6k}=\frac{3}{5}$$.

At $$t=12$$ min from the start, the temperature satisfies $$T-10=50e^{-12k}=50(e^{-6k})^2=50\left(\frac{3}{5}\right)^2=18$$, so $$T=28°\text{C}$$.

Verification using the interval method: Alternatively, treating the second interval separately with initial temperature 40°C: $$T-10=(40-10)e^{-6k}=30\cdot\frac{3}{5}=18$$, hence $$T=28°\text{C}$$ ✓

Answer: 28°C

We need to find $$d$$ where the change in diameter is $$d \times 10^{-3}$$ cm when temperature changes from 27°C to 177°C.

Given data.

Initial diameter $$D_0 = 5$$ cm, $$\alpha = 1.6 \times 10^{-5}$$ /°C, $$\Delta T = 177 - 27 = 150$$°C.

Apply linear expansion.

When a metal sheet with a hole is heated, the hole expands as if it were filled with the same metal.

$$\Delta D = D_0 \times \alpha \times \Delta T = 5 \times 1.6 \times 10^{-5} \times 150$$

$$= 5 \times 2.4 \times 10^{-3} = 12 \times 10^{-3}$$ cm

Conclusion.

$$d = 12$$.

Therefore, the answer is 12.

We need to convert $$-95°$$X to the Fahrenheit scale, using Celsius as an intermediate.

On the X scale, the freezing point is $$-15°$$X and the boiling point is $$65°$$X, giving a range of $$65 - (-15) = 80$$ divisions. The conversion formula from X to Celsius is:

$$\frac{T_X - (-15)}{80} = \frac{T_C - 0}{100}$$

Substituting $$T_X = -95$$:

$$\frac{-95 + 15}{80} = \frac{T_C}{100}$$

$$\frac{-80}{80} = \frac{T_C}{100}$$

$$T_C = -100°\text{C}$$

Now, converting to Fahrenheit:

$$T_F = \frac{9}{5}T_C + 32 = \frac{9}{5}(-100) + 32 = -180 + 32 = -148°\text{F}$$

So, the answer is $$-148°$$F.

A body cools from $$60°\text{C}$$ to $$40°\text{C}$$ in $$7$$ minutes and the surrounding temperature remains at $$10°\text{C}$$, so we wish to determine its temperature after the next $$7$$ minutes.

According to Newton’s Law of Cooling in exponential form, the temperature at time $$t$$ satisfies the relation $$T(t) - T_s = (T_0 - T_s)\,e^{-kt}$$ where $$T_s = 10°\text{C}$$ is the ambient temperature.

By setting $$t = 7$$ minutes and using the initial drop from $$60°\text{C}$$ to $$40°\text{C}$$, we substitute into the formula to obtain $$40 - 10 = (60 - 10)\,e^{-7k}$$, which simplifies to $$30 = 50\cdot e^{-7k}$$ and hence $$e^{-7k} = \dfrac{3}{5}$$.

For the subsequent $$7$$ minutes, the temperature $$T$$ starting from $$40°\text{C}$$ satisfies $$T - 10 = (40 - 10)\,e^{-7k}$$. Substituting the value $$e^{-7k} = \dfrac{3}{5}$$ yields $$T - 10 = 30 \times \dfrac{3}{5} = 18$$, so $$T = 18 + 10 = 28°\text{C}$$.

Therefore, the temperature after the next $$7$$ minutes is 28°$$\text{C}$$.

Use the general relation for any linear temperature scales:

$$\frac{\text{Reading}-\text{Lower fixed point}}{\text{Upper fixed point}-\text{Lower fixed point}}=\text{constant}$$

For scale P:

• Lower fixed point = 30
• Upper fixed point = 180

For scale Q:

• Lower fixed point = 0
• Upper fixed point = 100

So,

$$\frac{t_P-30}{180-30}=\frac{t_Q-0}{100-0}$$

$$\frac{t_P-30}{150}=\frac{t_Q}{100}$$

To find the value of $$t_2$$, we apply Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the surrounding temperature.

1. Newton's Law of Cooling Formula

The temperature difference $$\Delta T$$ at any time $$t$$ is given by:

$$\Delta T = \Delta T_0 e^{-\lambda t}$$

Where:

• $$\Delta T = T - T_0$$ (Temperature difference at time $$t$$)
• $$\Delta T_0 = T_i - T_0$$ (Initial temperature difference at $$t = 0$$)
• $$\lambda$$ = Cooling constant
• Initial state ($$t = 0$$): $$\Delta T_0 = 80^\circ\text{C} - 20^\circ\text{C} = 60^\circ\text{C}$$ (assuming $$T_0 = 20^\circ\text{C}$$ based on the $$60$$ in the solution).
• At $$t = 6 \text{ min}$$: The graph shows $$\Delta T = 40^\circ\text{C}$$.
• At $$t = t_2$$: The graph shows $$\Delta T = 20^\circ\text{C}$$.

2. Step-by-Step Calculation

Based on the experimental data provided:

$$40 = 60e^{-\lambda(6)}$$

$$\frac{40}{60} = e^{-6\lambda} \implies \frac{2}{3} = e^{-6\lambda}$$

Taking natural log on both sides:

$$6\lambda = \ln\left(\frac{3}{2}\right) = \ln(1.5)$$

$$20 = 60e^{-\lambda t_2}$$

$$\frac{20}{60} = e^{-\lambda t_2} \implies \frac{1}{3} = e^{-\lambda t_2}$$

Taking natural log on both sides:

$$t_2 \lambda = \ln(3)$$

3. Solving for $$t_2$$

Divide the two logarithmic equations to eliminate $$\lambda$$:

$$\frac{t_2 \lambda}{6 \lambda} = \frac{\ln(3)}{\ln(1.5)}$$

$$\frac{t_2}{6} = \frac{\ln(3)}{\ln(1.5)}$$

Using the values $$\ln(3) \approx 1.0986$$ and $$\ln(1.5) \approx 0.4055$$:

$$t_2 = 6 \times \frac{1.0986}{0.4055} \approx 6 \times 2.709$$

$$t_2 \approx 16.25 \text{ min}$$

Final Result:

Rounding to the nearest integer as suggested by the solution:

$$\boxed{t_2 \approx 16}$$

Given:

Mass of ice: $$m_{ice} = 120 \text{ g} = 0.12 \text{ kg}$$

Temperature of ice: $$T_{ice} = 0°C$$

Mass of water: $$m_w = 300 \text{ g} = 0.3 \text{ kg}$$

Temperature of water: $$T_w = 25°C$$

Specific heat capacity of water: $$c = 4200 \text{ J kg}^{-1} \text{ K}^{-1}$$

Latent heat of ice: $$L = 3.5 \times 10^5 \text{ J kg}^{-1}$$

The water cools from $$25°C$$ to $$0°C$$. The heat released by the water is:

$$Q = m_w \cdot c \cdot \Delta T = 0.3 \times 4200 \times 25 = 31500 \text{ J}$$

This heat is used to melt the ice. The mass of ice that melts:

$$Q = m \cdot L$$

$$m = \dfrac{Q}{L} = \dfrac{31500}{3.5 \times 10^5} = 0.09 \text{ kg} = 90 \text{ g}$$

Therefore, the value of $$x$$ is $$\boxed{90}$$.

We need to find the rate of combustion of fuel in a geyser that heats water.

Since the flow rate of water is $$2.0$$ kg/min (equivalent to $$2000$$ g/min), the temperature rise is $$70°$$C $$- 30°$$C = $$40°$$C, the heat of combustion is $$8 \times 10^3$$ J/g, and the specific heat of water is $$4.2$$ J g$$^{-1}$$ °C$$^{-1}$$, we first calculate the heat required per minute.

Using $$Q = m \times c \times \Delta T$$ gives $$Q = 2000 \times 4.2 \times 40 = 336000$$ J/min.

Now, if $$x$$ g of fuel is burnt per minute, then $$x \times 8 \times 10^3 = 336000$$. Substituting and solving yields $$x = \frac{336000}{8000} = 42$$ g/min.

The rate of combustion of fuel is 42 g min$$^{-1}$$.

For the bimetallic strip to maintain a constant length difference (20 cm), both metals must expand by the same amount when the temperature changes.

The change in length for each material is:

$$\Delta L = L \alpha \Delta T$$

For the difference to remain constant, the expansions must be equal:

$$\Delta L_{\text{brass}} = \Delta L_{\text{iron}}$$

$$L_{\text{brass}} \times \alpha_{\text{brass}} \times \Delta T = L_{\text{iron}} \times \alpha_{\text{iron}} \times \Delta T$$

Cancelling $$\Delta T$$ from both sides:

$$L_{\text{brass}} \times \alpha_{\text{brass}} = L_{\text{iron}} \times \alpha_{\text{iron}}$$

$$40 \times 1.8 \times 10^{-5} = L_{\text{iron}} \times 1.2 \times 10^{-5}$$

$$L_{\text{iron}} = \dfrac{40 \times 1.8}{1.2} = \dfrac{72}{1.2} = 60 \text{ cm}$$

We can verify: The difference $$L_{\text{iron}} - L_{\text{brass}} = 60 - 40 = 20 \text{ cm}$$, which is the required scale length.

Therefore, the length of iron is $$\boxed{60}$$ cm.

The total surface area of the cube is $$24$$ m$$^2$$.

A cube has 6 faces, so the area of each face is:

$$\frac{24}{6} = 4 \text{ m}^2$$

The side length of the cube is:

$$a = \sqrt{4} = 2 \text{ m}$$

The original volume of the cube is:

$$V = a^3 = 2^3 = 8 \text{ m}^3$$

The coefficient of volume expansion is:

$$\gamma = 3\alpha = 3 \times 5.0 \times 10^{-4} = 1.5 \times 10^{-3} \text{ °C}^{-1}$$

The increase in volume is:

$$\Delta V = V \gamma \Delta T = 8 \times 1.5 \times 10^{-3} \times 10$$

$$\Delta V = 0.12 \text{ m}^3$$

Converting to cm$$^3$$ ($$1 \text{ m}^3 = 10^6 \text{ cm}^3$$):

$$\Delta V = 0.12 \times 10^6 = 1.2 \times 10^5 \text{ cm}^3$$

Hence, the correct answer is Option B.

The mass of the nail is $$100$$ g, the mass of the hammer is $$1.5$$ kg, the velocity of the hammer is $$60$$ m/s, and the specific heat of iron is $$0.42$$ J g$$^{-1}$$ $$°$$C$$^{-1}$$. The kinetic energy of the hammer is calculated as $$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.5 \times (60)^2 = \frac{1}{2} \times 1.5 \times 3600 = 2700 \text{ J}$$. One fourth of this energy is transferred to the nail, so $$Q = \frac{1}{4} \times 2700 = 675 \text{ J}$$.

Applying $$Q = ms\Delta T$$, where $$m$$ is the mass of the nail, $$s$$ is the specific heat capacity, and $$\Delta T$$ is the temperature rise, gives $$675 = 100 \times 0.42 \times \Delta T$$ and hence $$675 = 42 \times \Delta T$$. Solving for $$\Delta T$$ yields $$\Delta T = \frac{675}{42} = 16.07°\text{C}$$. Therefore, the rise in temperature of the nail is $$16.07°$$C. The correct answer is Option A.

An ice cube of dimensions $$60 \text{ cm} \times 50 \text{ cm} \times 20 \text{ cm}$$ is in an insulated box (wall thickness $$1 \text{ cm}$$) at $$0°C$$, and the room temperature is $$40°C$$. We need to find the rate of melting.

First, the surface area is calculated as $$A = 2(60 \times 50 + 50 \times 20 + 60 \times 20) \text{ cm}^2$$. This gives $$A = 2(3000 + 1000 + 1200) = 2 \times 5200 = 10400 \text{ cm}^2$$ and hence $$A = 10400 \times 10^{-4} \text{ m}^2 = 1.04 \text{ m}^2$$.

Since heat is conducted through the box walls, we use the formula $$\dfrac{dQ}{dt} = \dfrac{kA\Delta T}{d}$$ where $$k = 0.05 \text{ W m}^{-1} °C^{-1}$$, $$A = 1.04 \text{ m}^2$$, $$\Delta T = 40°C$$, and $$d = 1 \text{ cm} = 0.01 \text{ m}$$. Substituting these values yields $$\dfrac{dQ}{dt} = \dfrac{0.05 \times 1.04 \times 40}{0.01} = \dfrac{2.08}{0.01} = 208 \text{ W}$$.

Next, the rate of melting is given by $$\dfrac{dm}{dt} = \dfrac{1}{L} \cdot \dfrac{dQ}{dt}$$ where $$L = 3.4 \times 10^5 \text{ J kg}^{-1}$$. Therefore, $$\dfrac{dm}{dt} = \dfrac{208}{3.4 \times 10^5} = \dfrac{208}{340000} \approx 6.12 \times 10^{-4} \text{ kg s}^{-1}$$ and hence $$\dfrac{dm}{dt} \approx 61.2 \times 10^{-5} \text{ kg s}^{-1} \approx 61 \times 10^{-5} \text{ kg s}^{-1}$$.

The correct answer is Option B: $$61 \times 10^{-5} \text{ kg s}^{-1}$$.

A gold ring of diameter 6.230 cm needs to be heated to fit on a wooden bangle of diameter 6.241 cm. Room temperature is 27°C. $$\alpha_L = 1.4 \times 10^{-5}$$ K$$^{-1}$$.

$$L = L_0(1 + \alpha_L \Delta T)$$

where $$L_0 = 6.230$$ cm is the initial diameter and $$L = 6.241$$ cm is the required diameter.

$$6.241 = 6.230(1 + 1.4 \times 10^{-5} \times \Delta T)$$

$$\frac{6.241}{6.230} = 1 + 1.4 \times 10^{-5} \times \Delta T$$

$$\frac{6.241 - 6.230}{6.230} = 1.4 \times 10^{-5} \times \Delta T$$

$$\frac{0.011}{6.230} = 1.4 \times 10^{-5} \times \Delta T$$

$$1.765 \times 10^{-3} = 1.4 \times 10^{-5} \times \Delta T$$

$$\Delta T = \frac{1.765 \times 10^{-3}}{1.4 \times 10^{-5}} = \frac{1765}{14} \approx 125.7 \text{ °C}$$

$$T = 27 + 125.7 = 152.7 \text{ °C}$$

Hence, the correct answer is Option D: 152.7°C.

We need to find the steady-state junction temperature T between a steel rod and a copper rod connected in series, with the steel end at 0°C and the copper end at 50°C.

Since the system is in steady state, the rate of heat flow through both rods is equal, so $$\frac{K_1 A_1 (T - 0)}{L_1} = \frac{K_2 A_2 (50 - T)}{L_2}$$ where subscript 1 refers to steel and subscript 2 refers to copper.

We are given that $$\frac{K_2}{K_1} = 9$$, $$\frac{A_1}{A_2} = 2$$, and $$\frac{L_1}{L_2} = 2$$.

Substituting these ratios into the heat balance equation gives $$\frac{K_1 A_1}{L_1} \cdot T = \frac{K_2 A_2}{L_2} \cdot (50 - T)\,.$$

From the above, the ratio of the coefficients is $$\frac{K_1 A_1 / L_1}{K_2 A_2 / L_2} = \frac{K_1}{K_2} \cdot \frac{A_1}{A_2} \cdot \frac{L_2}{L_1} = \frac{1}{9} \times 2 \times \frac{1}{2} = \frac{1}{9}\,.$$

Substituting this result into the previous relation yields $$\frac{1}{9} \cdot T = (50 - T)\,, $$ which can be written as $$T = 9(50 - T) = 450 - 9T\,, $$ leading to $$10T = 450$$ and hence $$T = 45°C\,. $$

Therefore, the correct answer is Option C: $$45°C$$.

A lead bullet penetrates a solid object and melts. 40% of its kinetic energy is used to heat it.

Initial temperature $$T_i = 127°$$C, Melting point $$T_m = 327°$$C

Latent heat of fusion $$L_f = 2.5 \times 10^4$$ J/kg

Specific heat capacity $$c = 125$$ J/(kg·K)

Calculate the heat required to melt the bullet.

Temperature rise: $$\Delta T = 327 - 127 = 200$$ K

Heat to raise temperature to melting point:

$$Q_1 = mc\Delta T = m \times 125 \times 200 = 25000m$$ J

Heat to melt:

$$Q_2 = mL_f = m \times 2.5 \times 10^4 = 25000m$$ J

Total heat required:

$$Q = Q_1 + Q_2 = 25000m + 25000m = 50000m$$ J

Relate to kinetic energy.

40% of kinetic energy equals the heat required:

$$0.4 \times \frac{1}{2}mv^2 = 50000m$$

$$0.2v^2 = 50000$$

$$v^2 = 250000$$

$$v = 500$$ m/s

The initial speed of the bullet is $$500$$ m/s.

The correct answer is Option B.

We need to check which statements (A, B, C, D) are correct.

Statement A: When the temperature difference is doubled, rate of heat loss becomes twice.

By Newton's law of cooling, the rate of heat loss is proportional to the temperature difference between the liquid and surroundings:

$$\frac{dQ}{dt} \propto \Delta T$$

If $$\Delta T$$ is doubled, the rate of heat loss doubles. Statement A is correct.

Statement B: Two bodies P and Q at 10°C and 20°C have thermal radiation ratio 1:1.15.

By Stefan's law, power radiated $$\propto T^4$$ (in Kelvin).

$$T_P = 10 + 273 = 283 \text{ K}$$, $$T_Q = 20 + 273 = 293 \text{ K}$$

$$\frac{E_P}{E_Q} = \left(\frac{283}{293}\right)^4 = \left(\frac{283}{293}\right)^4$$

$$\frac{283}{293} \approx 0.9659$$

$$(0.9659)^4 \approx 0.871$$

So $$E_P : E_Q \approx 0.871 : 1 \approx 1 : 1.15$$. Statement B is correct.

Statement C: Carnot engine between 100 K and 400 K has efficiency 75%.

Carnot efficiency: $$\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 - \frac{100}{400} = 1 - 0.25 = 0.75 = 75\%$$

Statement C is correct.

Statement D: When temperature difference is quadrupled, rate of heat loss becomes twice.

By Newton's law of cooling, rate $$\propto \Delta T$$. If $$\Delta T$$ is quadrupled, rate becomes 4 times (not twice).

Statement D is incorrect.

Statements A, B, and C are correct.

The correct answer is Option A: A, B, C only.

We have three points whose temperatures are fixed: $$A \;(200^{\circ}\text{C}),\; B \;(100^{\circ}\text{C}),\; D \;(125^{\circ}\text{C})$$. An identical rod $$AB$$ has total thermal resistance $$10\;\text{K W}^{-1}$$, and rod $$CD$$ also has thermal resistance $$10\;\text{K W}^{-1}$$. Because the connection to rod $$CD$$ is made at the mid-point $$C$$ of $$AB$$, rod $$AB$$ is effectively split into two equal halves: $$AC$$ and $$CB$$.

The resistance of a portion is directly proportional to its length. Since each half is half the full length, each half has half the resistance: $$R_{AC}=R_{CB}=\dfrac{10}{2}=5\;\text{K W}^{-1}.$$

Let the steady-state temperature of the junction $$C$$ be $$T\;({}^{\circ}\text{C}).$$ For any rod the heat current $$Q$$ is given by the basic conduction relation $$Q=\dfrac{\Delta T}{R},$$ where $$\Delta T$$ is the temperature difference across the rod and $$R$$ is its thermal resistance.

The heat current from $$A$$ to $$C$$ is therefore $$Q_{AC}=\dfrac{200-T}{5}.$$

The heat current from $$C$$ to $$B$$ is $$Q_{CB}=\dfrac{T-100}{5}.$$

The heat current from $$C$$ to $$D$$ is $$Q_{CD}=\dfrac{T-125}{10}.$$

At junction $$C$$ the system is in steady state, so the heat arriving at $$C$$ from $$A$$ must equal the heat leaving $$C$$ towards $$B$$ and $$D$$: $$Q_{AC}=Q_{CB}+Q_{CD}.$$ Substituting the expressions, we get $$\dfrac{200-T}{5}=\dfrac{T-100}{5}+\dfrac{T-125}{10}.$$

To clear denominators, multiply the entire equation by $$10$$: $$2(200-T)=2(T-100)+(T-125).$$

Simplifying each side, we obtain $$400-2T=2T-200+T-125.$$ Combining like terms on the right, $$400-2T=3T-325.$$

Now move all the $$T$$ terms to one side and the constants to the other: $$400+325=3T+2T.$$ $$725=5T.$$ So, $$T=\dfrac{725}{5}=145^{\circ}\text{C}.$$

The required heat current in rod $$CD$$ is $$Q_{CD}=\dfrac{T-125}{10}=\dfrac{145-125}{10}=\dfrac{20}{10}=2\;\text{W}.$$

Hence, the heat current in $$CD$$ is $$2\;\text{W}$$, so the value of $$P$$ is $$2$$.

So, the answer is $$2$$.

We use Newton's Law of Cooling. For JEE Main, we use the standard approximate form for finite time intervals:

$$\dfrac{T_1 - T_2}{\Delta t} = k \left( \dfrac{T_1 + T_2}{2} - T_s \right)$$

where $$T_1$$ and $$T_2$$ are the initial and final temperatures in the interval, $$\Delta t$$ is the time interval, $$T_s$$ is the surrounding (room) temperature, and $$k$$ is a constant.

Given: Room temperature $$T_s = 25°C$$.

First interval (0 to 5 min): Body cools from $$75°C$$ to $$65°C$$.

$$\dfrac{75 - 65}{5} = k \left( \dfrac{75 + 65}{2} - 25 \right)$$

$$\dfrac{10}{5} = k \left( 70 - 25 \right)$$

$$2 = 45k$$

$$k = \dfrac{2}{45}$$

Second interval (5 to 10 min): Body cools from $$65°C$$ to $$T°C$$.

$$\dfrac{65 - T}{5} = \dfrac{2}{45} \left( \dfrac{65 + T}{2} - 25 \right)$$

$$\dfrac{65 - T}{5} = \dfrac{2}{45} \times \dfrac{65 + T - 50}{2}$$

$$\dfrac{65 - T}{5} = \dfrac{2}{45} \times \dfrac{15 + T}{2}$$

$$\dfrac{65 - T}{5} = \dfrac{15 + T}{45}$$

Cross-multiplying:

$$45(65 - T) = 5(15 + T)$$

$$2925 - 45T = 75 + 5T$$

$$2925 - 75 = 45T + 5T$$

$$2850 = 50T$$

$$T = \dfrac{2850}{50} = 57$$

The temperature of the body at the end of the next 5 minutes is $$57°C$$.

According to Newton’s Law of Cooling, the rate at which the temperature $$T$$ of a body changes is proportional to the excess temperature over that of the surroundings. If $$T_s$$ is the constant temperature of the surroundings and $$k$$ is the proportionality constant, the law is written as

Integrating between an initial temperature $$T_1$$ and a final temperature $$T_2$$ over a time interval $$t$$ gives the well-known working formula

We first use the given cooling from 61 °C to 59 °C in 4 min to determine the constant $$k$$. Here

Substituting in the formula, we have

Solving for $$k$$,

Now we need the time required for the body to cool from 51 °C to 49 °C. For this second stage we put

Using the same surroundings temperature $$T_s = 30^\circ\text{C}$$ and the same $$k$$, the time $$t'$$ for this cooling is

We substitute the value of $$k$$ obtained earlier:

Evaluating the logarithms (natural logarithms):

So

The calculated time is practically 6 minutes.

Thermal stress in a material arises when the material is constrained and prevented from expanding or contracting due to temperature changes. The stress is given by $$\sigma = Y \alpha \Delta T$$, where $$Y$$ is Young's modulus, $$\alpha$$ is the coefficient of linear expansion, and $$\Delta T$$ is the temperature change.

When a rod is lying freely and is heated, it is free to expand without any constraint. Since there is no external force or support preventing the expansion, no thermal stress develops in the rod. Therefore, Assertion A is true.

Reason R states that on heating, the length of the rod increases. This is indeed true for most materials, as they undergo thermal expansion when heated.

However, R is not the correct explanation of A. The fact that the rod's length increases on heating is not why thermal stress is absent. Thermal stress is absent because the rod is free to expand. If the rod were fixed at both ends, it would still try to expand on heating, but the constraint would cause thermal stress. The key factor is the freedom to expand, not the expansion itself.

Therefore, both A and R are true but R is NOT the correct explanation of A, which corresponds to Option (3).

We know that the heat supplied per unit time is the power. If a body of mass $$m$$ is heated at a constant rate (power) $$P$$, then during a small interval of time $$dt$$ the heat given is $$dQ = P\,dt$$.

The calorimetric relation for a temperature rise says $$dQ = m\,c\,d\theta,$$ where $$c$$ is the specific heat capacity and $$d\theta$$ is the corresponding rise in temperature.

Substituting $$dQ = P\,dt$$ in the calorimetric relation, we obtain

$$P\,dt = m\,c\,d\theta.$$

Rearranging, the rate of rise of temperature becomes

$$\frac{d\theta}{dt} = \frac{P}{m\,c}.$$

The quantity $$\dfrac{d\theta}{dt}$$ is the slope of the temperature-time graph. Hence the slope $$S$$ of that graph is

$$S = \frac{P}{m\,c}.$$

For the two metal blocks A and B the mass $$m$$ and the heating power $$P$$ are identical, therefore

$$S \propto \frac{1}{c}\qquad\Longrightarrow\qquad c \propto \frac{1}{S}.$$

Thus the ratio of their specific heat capacities is the inverse ratio of the slopes:

$$\frac{c_A}{c_B} = \frac{S_B}{S_A}.$$

We now read the slopes directly from the straight-line portions of the given graph. Choosing the first $$4\ \text{min}$$ interval (any convenient interval gives the same result):

For body A the temperature rises from $$20^{\circ}\text{C}$$ to $$60^{\circ}\text{C},$$ so

$$\Delta\theta_A = 60 - 20 = 40^{\circ}\text{C}, \qquad \Delta t = 4\ \text{min},$$

and hence

$$S_A = \frac{\Delta\theta_A}{\Delta t} = \frac{40}{4} = 10^{\circ}\text{C}\,\text{min}^{-1}.$$

During the same time the temperature of body B rises from $$20^{\circ}\text{C}$$ to $$35^{\circ}\text{C},$$ giving

$$\Delta\theta_B = 35 - 20 = 15^{\circ}\text{C},$$

so that

$$S_B = \frac{\Delta\theta_B}{\Delta t} = \frac{15}{4} = 3.75^{\circ}\text{C}\,\text{min}^{-1}.$$

Taking their ratio,

$$\frac{S_B}{S_A} = \frac{15/4}{40/4} = \frac{15}{40} = \frac{3}{8}.$$

Substituting this into $$\dfrac{c_A}{c_B} = \dfrac{S_B}{S_A}$$ gives

$$\frac{c_A}{c_B} = \frac{3}{8}.$$

Hence, the correct answer is Option B.

For two identical wires (same length $$l$$ and cross-sectional area $$A$$) connected in series, the total thermal resistance is the sum of individual resistances. The thermal resistance of each wire is $$R_i = \frac{l}{K_i A}$$, so the total resistance is $$R = \frac{l}{K_1 A} + \frac{l}{K_2 A}$$.

The effective conductivity $$K_{\text{eff}}$$ of the combination (total length $$2l$$, same area $$A$$) satisfies $$R = \frac{2l}{K_{\text{eff}} A}$$. Therefore $$\frac{2l}{K_{\text{eff}} A} = \frac{l}{K_1 A} + \frac{l}{K_2 A}$$.

Simplifying: $$\frac{2}{K_{\text{eff}}} = \frac{1}{K_1} + \frac{1}{K_2} = \frac{K_1 + K_2}{K_1 K_2}$$, which gives $$K_{\text{eff}} = \frac{2K_1 K_2}{K_1 + K_2}$$.

The correct answer is option 1: $$\frac{2K_1 K_2}{K_1 + K_2}$$.

We have a cubic metal box with each side of length $$a$$ at room temperature $$T$$. The coefficient of linear expansion of the metal sheet is $$\alpha$$. The temperature is increased by $$\Delta T$$.

The initial volume of the cube is $$V_0 = a^3$$.

When the temperature increases by $$\Delta T$$, each side of the cube expands. The new length of each side is $$a' = a(1 + \alpha \Delta T)$$.

The new volume is $$V = (a')^3 = a^3(1 + \alpha \Delta T)^3$$.

Since $$\alpha \Delta T$$ is very small (the problem states "small temperature $$\Delta T$$"), we can use the binomial approximation $$(1 + x)^3 \approx 1 + 3x$$ for small $$x$$.

So $$V \approx a^3(1 + 3\alpha \Delta T) = a^3 + 3a^3\alpha \Delta T$$.

The increase in volume is $$\Delta V = V - V_0 = 3a^3\alpha \Delta T$$.

This result is consistent with the general formula for volume expansion: $$\Delta V = V_0 \gamma \Delta T$$, where the coefficient of volume expansion $$\gamma = 3\alpha$$ for isotropic materials.

Hence, the correct answer is Option B.

Let the masses of the liquids be equal and equal to $$m$$. Denote their specific heats by $$c_x,\;c_y,\;c_z$$ and their initial temperatures by $$T_x = 10^\circ\text{C},\;T_y = 20^\circ\text{C},\;T_z = 30^\circ\text{C}$$ respectively.

Whenever two liquids are mixed adiabatically, the heat lost by the hotter liquid equals the heat gained by the colder one. Mathematically we write

$$\text{Heat lost} = \text{Heat gained}.$$

For equal masses this becomes

$$m c_{\text{hot}}\,(T_{\text{hot}}-T_f)=m c_{\text{cold}}\,(T_f-T_{\text{cold}}),$$

where $$T_f$$ is the final (equilibrium) temperature of the mixture.

First we mix liquids $$x$$ and $$y$$. Their final temperature is given to be $$16^\circ\text{C}$$. Here $$x$$ is colder and $$y$$ is hotter, so

$$m c_x\,(16-10)=m c_y\,(20-16).$$

Cancelling the common mass $$m$$ and substituting the numerical differences, we have

$$c_x\,(6)=c_y\,(4).$$

Dividing both sides by 2 gives

$$3c_x=2c_y,$$

which rearranges to

$$c_y=\frac{3}{2}\,c_x.$$

Hence the specific heat of liquid $$y$$ is $$1.5$$ times that of liquid $$x$$.

Next we mix liquids $$y$$ and $$z$$. Their final temperature is given to be $$26^\circ\text{C}$$. Again applying the heat‐balance relation, with $$y$$ colder and $$z$$ hotter, we get

$$m c_y\,(26-20)=m c_z\,(30-26).$$

After cancelling the mass $$m$$ and inserting the numerical differences, this becomes

$$c_y\,(6)=c_z\,(4).$$

So we have

$$6c_y=4c_z \quad\Longrightarrow\quad c_z=\frac{6}{4}\,c_y=\frac{3}{2}\,c_y.$$

We already found $$c_y=\frac{3}{2}\,c_x$$, so substituting this value we get

$$c_z=\frac{3}{2}\left(\frac{3}{2}c_x\right)=\frac{9}{4}\,c_x=2.25\,c_x.$$

Finally we mix liquids $$x$$ and $$z$$. Let the required equilibrium temperature be $$T_f$$. Liquid $$x$$ is colder and liquid $$z$$ is hotter, hence

$$m c_x\,(T_f-10)=m c_z\,(30-T_f).$$

Cancelling the common factor $$m$$ and substituting $$c_z=2.25\,c_x$$ gives

$$c_x\,(T_f-10)=2.25\,c_x\,(30-T_f).$$

Since $$c_x\neq 0$$, it can be divided out, leaving

$$T_f-10=2.25\,(30-T_f).$$

Expanding the right side, we have

$$T_f-10=2.25\times30-2.25\,T_f.$$

Calculating the product, $$2.25\times30=67.5$$, so

$$T_f-10 = 67.5 - 2.25T_f.$$

Now collect the $$T_f$$ terms on the left and the constants on the right:

$$T_f + 2.25T_f = 67.5 + 10.$$

This simplifies to

$$3.25T_f = 77.5.$$

Dividing both sides by $$3.25$$, we obtain

$$T_f = \frac{77.5}{3.25}.$$

Carrying out the division gives

$$T_f = 23.846\ldots^\circ\text{C} \approx 23.84^\circ\text{C}.$$

Hence, the correct answer is Option D.

We begin with the well-known Fourier law of heat conduction in its differential form. For any medium, the instantaneous heat current $$dQ/dt$$ (which we shall denote simply by $$Q$$ for convenience) is related to the temperature gradient by

$$Q = -K A\,\frac{d\theta}{dr},$$

where $$K$$ is the thermal conductivity of the material, $$A$$ is the area normal to the direction of flow and $$\dfrac{d\theta}{dr}$$ is the spatial rate of fall of temperature. Because we are dealing with a spherical geometry, every point at a distance $$r$$ from the common centre lies on a sphere of area

$$A = 4\pi r^{2}.$$

So, for purely radial (i.e. spherically symmetric) flow we can write

$$Q = -K(4\pi r^{2})\,\frac{d\theta}{dr} \;=\; -\,4\pi K r^{2}\,\frac{d\theta}{dr}.$$

Under steady-state conditions the same quantity of heat per unit time must cross every spherical surface, hence $$Q$$ is independent of $$r$$. We now isolate the derivative:

$$\frac{d\theta}{dr} \;=\; -\,\frac{Q}{4\pi K}\,\frac{1}{r^{2}}.$$

Next, we integrate this relation from the inner radius $$r_1$$ (at temperature $$\theta_1$$) to the outer radius $$r_2$$ (at temperature $$\theta_2$$). Writing the integral explicitly, we have

$$\int_{\theta_1}^{\theta_2} d\theta \;=\; -\,\frac{Q}{4\pi K}\,\int_{r_1}^{r_2} \frac{dr}{r^{2}}.$$

The left-hand side integrates to

$$\theta_2 - \theta_1.$$

For the right-hand side, we use the elementary integral $$\displaystyle\int r^{-2}dr = -\,\dfrac{1}{r}$$, giving

$$-\,\frac{Q}{4\pi K}\;\Bigl[-\frac{1}{r}\Bigr]_{r_1}^{r_2} \;=\; -\,\frac{Q}{4\pi K}\;\Bigl(-\frac{1}{r_2}+\frac{1}{r_1}\Bigr) \;=\; -\,\frac{Q}{4\pi K}\;\Bigl(\frac{1}{r_1}-\frac{1}{r_2}\Bigr).$$

Equating both results, we obtain

$$\theta_2 - \theta_1 \;=\; -\,\frac{Q}{4\pi K}\;\Bigl(\frac{1}{r_1}-\frac{1}{r_2}\Bigr).$$

Since $$r_1 < r_2$$, the bracketed term $$\Bigl(\dfrac{1}{r_1}-\dfrac{1}{r_2}\Bigr)$$ is positive. For clarity we take the magnitude of the heat current (the direction being implicitly from the hotter to the colder surface) and remove the minus sign:

$$Q \;=\; 4\pi K\,(\theta_2 - \theta_1)\,\frac{1}{\dfrac{1}{r_1}-\dfrac{1}{r_2}}.$$

Combining the fractions in the denominator,

$$\frac{1}{r_1}-\frac{1}{r_2} = \frac{r_2 - r_1}{r_1 r_2}.$$

Substituting this back, we obtain

$$Q = 4\pi K\,(\theta_2 - \theta_1)\, \frac{r_1 r_2}{\,r_2 - r_1\,}.$$

Thus the rate at which heat flows radially through the material separating the two concentric spherical shells is

$$Q = \frac{4\pi K\,r_1 r_2\,(\theta_2 - \theta_1)}{\,r_2 - r_1\,}.$$

This expression exactly matches Option D.

Hence, the correct answer is Option D.

We first determine how much ice is present inside the pipe. The pipe is 1 m long and its cross-sectional area is 1 cm$$^{2}$$. Converting the area to square metres, we write $$1\text{ cm}^{2}=1\times(10^{-2}\text{ m})^{2}=1\times10^{-4}\text{ m}^{2}.$$ Hence the volume is

$$V=\text{area}\times\text{length}=1\times10^{-4}\text{ m}^{2}\times1\text{ m}=1\times10^{-4}\text{ m}^{3}.$$

Taking the density of ice as $$\rho=10^{3}\text{ kg m}^{-3},$$ the mass contained in this volume is

$$m=\rho V=10^{3}\text{ kg m}^{-3}\times1\times10^{-4}\text{ m}^{3}=0.1\text{ kg}.$$

The ice is initially at $$-10^\circ\text{C}$$. Before melting, it must be raised to $$0^\circ\text{C}$$. The heat needed for this warming is, by the specific-heat formula $$Q=mc\Delta T,$$ where the specific heat of ice is $$c=2\times10^{3}\text{ J kg}^{-1}{}^\circ\text{C}^{-1}$$ and the temperature change is $$\Delta T=10^\circ\text{C}.$$ Thus

$$Q_1=m\,c\,\Delta T=0.1\text{ kg}\times2\times10^{3}\text{ J kg}^{-1}{}^\circ\text{C}^{-1}\times10^\circ\text{C}=0.1\times20\,000\text{ J}=2\,000\text{ J}.$$

Next, the ice at $$0^\circ\text{C}$$ must melt. Using the latent-heat formula $$Q=mL,$$ with latent heat of fusion $$L=3.33\times10^{5}\text{ J kg}^{-1},$$ we get

$$Q_2=mL=0.1\text{ kg}\times3.33\times10^{5}\text{ J kg}^{-1}=33\,300\text{ J}.$$

The total heat required is the sum of these two amounts:

$$Q_{\text{total}}=Q_1+Q_2=2\,000\text{ J}+33\,300\text{ J}=35\,300\text{ J}.$$

This heat is supplied electrically. The power dissipated in a resistor is given by $$P=I^{2}R.$$ Here the current is $$I=0.5\text{ A}$$ and the resistance is $$R=4\text{ k}\Omega=4\,000\ \Omega.$$ Therefore,

$$P=(0.5\text{ A})^{2}\times4\,000\ \Omega=0.25\times4\,000\text{ W}=1\,000\text{ W}.$$

Since power is energy per unit time, $$P=\dfrac{Q}{t},$$ so the time required is

$$t=\dfrac{Q_{\text{total}}}{P}=\dfrac{35\,300\text{ J}}{1\,000\text{ J s}^{-1}}=35.3\text{ s}.$$

Hence, the correct answer is Option 3.

We are told that the solid metal cube expands differently along the three mutually perpendicular directions. Let the coefficients of linear expansion along these directions be denoted by $$\alpha_x,\;\alpha_y$$ and $$\alpha_z$$.

From the statement of the problem we have

$$\alpha_x = 5 \times 10^{-5}\;{}^\circ\!{\rm C}^{-1},$$

$$\alpha_y = 5 \times 10^{-6}\;{}^\circ\!{\rm C}^{-1},$$

$$\alpha_z = 5 \times 10^{-6}\;{}^\circ\!{\rm C}^{-1}.$$

The basic relation connecting the linear and volume expansion coefficients for small temperature changes is:

$$\beta = \alpha_x + \alpha_y + \alpha_z,$$

where $$\beta$$ is the coefficient of volume expansion. This formula simply states that the fractional increase in volume equals the sum of the fractional increases in length along the three perpendicular axes.

Substituting the given numerical values, we get

$$\beta = 5 \times 10^{-5} + 5 \times 10^{-6} + 5 \times 10^{-6}.$$

To combine these terms easily, we convert every term to the same power of ten. Note that

$$5 \times 10^{-5} = 50 \times 10^{-6}.$$

Hence

$$\beta = 50 \times 10^{-6} + 5 \times 10^{-6} + 5 \times 10^{-6}.$$

Adding the coefficients:

$$\beta = (50 + 5 + 5) \times 10^{-6} = 60 \times 10^{-6}\;{}^\circ\!{\rm C}^{-1}.$$

In the problem the volume expansion coefficient is written as $$C \times 10^{-6}\;{}^\circ\!{\rm C}^{-1}$$, so on comparison

$$C = 60.$$

So, the answer is $$60$$.

We begin by noting that the container is perfectly insulated, so the total heat released by the hot substance (steam) must equal the total heat absorbed by the cold substance (ice). In symbols, we write $$Q_{\text{released}} = Q_{\text{absorbed}}.$$

Let $$M$$ grams be the mass of steam at $$100^\circ\text{C}$$ that is mixed with $$200\ \text{g}$$ of ice at $$0^\circ\text{C}$$. Eventually the whole system reaches $$40^\circ\text{C}$$ as liquid water.

We treat the heat changes in two separate stages for each substance.

1. Heat released by the steam

(a) First, the steam condenses to water at the same temperature $$100^\circ\text{C}$$. The formula for heat involved in a phase change is $$Q = mL,$$ where $$m$$ is the mass and $$L$$ is the latent heat. For condensation we use the latent heat of vaporization of water, $$L_v = 540\ \text{cal g}^{-1}.$$ Hence the heat released during condensation is $$Q_1 = M \times 540.$$

(b) Next, the resulting water cools from $$100^\circ\text{C}$$ down to the final temperature $$40^\circ\text{C}$$. The formula for heat exchange during a temperature change is $$Q = mc\Delta T,$$ where $$c$$ is the specific heat capacity. For liquid water, $$c = 1\ \text{cal g}^{-1\ ^\circ\!C}$$. The temperature drop is $$\Delta T = 100 - 40 = 60^\circ\text{C}.$$ Thus the heat released during cooling is $$Q_2 = M \times 1 \times 60 = 60M.$$

Total heat released by the steam is therefore $$Q_{\text{released}} = Q_1 + Q_2 = 540M + 60M = 600M.$$

2. Heat absorbed by the ice

(a) First, the ice melts at $$0^\circ\text{C}$$ to become water at the same temperature. Using $$Q = mL$$ with the latent heat of fusion $$L_f = 80\ \text{cal g}^{-1}$$, we have $$Q_3 = 200 \times 80 = 16\,000.$$

(b) Next, the melt-water warms from $$0^\circ\text{C}$$ to $$40^\circ\text{C}$$. Using $$Q = mc\Delta T$$ with $$\Delta T = 40^\circ\text{C}$$, we get $$Q_4 = 200 \times 1 \times 40 = 8\,000.$$

Total heat absorbed by the ice and cold water is $$Q_{\text{absorbed}} = Q_3 + Q_4 = 16\,000 + 8\,000 = 24\,000.$$

3. Applying energy conservation

Since no heat is lost to the surroundings, we equate the two totals: $$Q_{\text{released}} = Q_{\text{absorbed}},$$ $$600M = 24\,000.$$

Solving for $$M$$ gives $$M = \frac{24\,000}{600} = 40.$$

Thus, $$40\ \text{g}$$ of steam are required.

So, the answer is $$40$$.

We begin by recalling the principle of calorimetry for ideal mixtures. When different masses (or, for a liquid of uniform density, different volumes) of water at temperatures $$T_1,\,T_2,\,T_3$$ are mixed and no heat is lost to the surroundings, the final equilibrium temperature is the weighted average:

$$T_{\text{final}}=\frac{m_1T_1+m_2T_2+m_3T_3}{m_1+m_2+m_3},$$

where $$m_1,m_2,m_3$$ are the respective masses or volumes. Here each litre of water is taken to have the same mass, so we may write all equations directly in litres.

Let the unknown temperatures of the three containers be

$$T_1\;(\text{for }C_1),\qquad T_2\;(\text{for }C_2),\qquad T_3\;(\text{for }C_3).$$

Now we translate every piece of experimental data into an equation.

First mixture: 1 L from $$C_1$$ and 2 L from $$C_2$$ are mixed, giving a final temperature of $$60^\circ\text{C}$$. Substituting in the formula,

$$\frac{1\cdot T_1+2\cdot T_2}{1+2}=60.$$

Multiplying both sides by 3, we obtain

$$T_1+2T_2=180 \quad -(1)$$

Second mixture: 1 L from $$C_2$$ and 2 L from $$C_3$$ are mixed, giving $$30^\circ\text{C}$$. Thus,

$$\frac{1\cdot T_2+2\cdot T_3}{1+2}=30,$$

which simplifies, after multiplying by 3, to

$$T_2+2T_3=90 \quad -(2)$$

Third mixture: 2 L from $$C_1$$ and 1 L from $$C_3$$ are mixed, again giving $$60^\circ\text{C}$$. Hence,

$$\frac{2\cdot T_1+1\cdot T_3}{2+1}=60,$$

and multiplying through by 3 yields

$$2T_1+T_3=180 \quad -(3)$$

We now have a system of three linear equations in the three unknowns $$T_1,T_2,T_3$$:

$$\begin{aligned} T_1+2T_2 &= 180\qquad&(1)\\ T_2+2T_3 &= 90\qquad&(2)\\ 2T_1+T_3 &= 180\qquad&(3) \end{aligned}$$

From equation (1) we solve for $$T_1$$ in terms of $$T_2$$:

$$T_1=180-2T_2. \quad -(4)$$

Substituting (4) in equation (3):

$$2(180-2T_2)+T_3=180.$$

Expanding the left side gives

$$360-4T_2+T_3=180.$$

Isolating $$T_3$$ we find

$$T_3=180+4T_2-360=4T_2-180. \quad -(5)$$

Now substitute (5) into equation (2):

$$T_2+2(4T_2-180)=90.$$

Simplifying the left-hand side,

$$T_2+8T_2-360=90,$$

which combines to

$$9T_2-360=90.$$

Adding 360 to both sides gives

$$9T_2=450,$$

and dividing by 9 yields

$$T_2=50^\circ\text{C}.$$

Back-substitution: Using $$T_2=50$$ in equation (4),

$$T_1=180-2(50)=180-100=80^\circ\text{C}.$$

Using $$T_2=50$$ in equation (5),

$$T_3=4(50)-180=200-180=20^\circ\text{C}.$$

So the individual container temperatures are

$$T_1=80^\circ\text{C},\qquad T_2=50^\circ\text{C},\qquad T_3=20^\circ\text{C}.$$

Finally, we turn to the required mixture of 1 L from each container. The total volume is $$1+1+1=3$$ litres, so

$$\theta=\frac{1\cdot T_1+1\cdot T_2+1\cdot T_3}{3} =\frac{80+50+20}{3} =\frac{150}{3} =50^\circ\text{C}.$$

So, the answer is $$50$$.

We have a bakelite beaker whose capacity at the reference temperature $$30^{\circ}\text{C}$$ is

$$V_{b0}=500\ \text{cc}.$$

It is partly filled with mercury. Let the volume of mercury taken at the same reference temperature be

$$V_m\ \text{cc}.$$

Therefore the empty (unfilled) space inside the beaker at $$30^{\circ}\text{C}$$ is

$$V_{\text{empty,0}} \;=\; V_{b0}-V_m.$$

The condition given in the statement is that this empty space remains the same when the temperature changes. Mathematically, for an infinitesimal rise $$dT$$ in temperature, the change in empty space must be zero:

$$dV_{\text{empty}} \;=\; 0.$$

But

$$dV_{\text{empty}} \;=\; dV_{\text{beaker}} \;-\; dV_{\text{mercury}}.$$

For any substance the change in volume caused by a small rise $$dT$$ is given by the volume-expansion formula

$$dV \;=\; V_0\,\gamma\,dT,$$

where $$\gamma$$ is the coefficient of volume expansion and $$V_0$$ is the initial volume at the reference temperature.

Applying this formula to the bakelite beaker, we get

$$dV_{\text{beaker}} \;=\; V_{b0}\,\gamma_{\text{beaker}}\,dT.$$

Similarly, applying it to the mercury, we have

$$dV_{\text{mercury}} \;=\; V_m\,\gamma_{\text{mercury}}\,dT.$$

Now we impose the zero-change condition:

$$0 \;=\; dV_{\text{empty}} \;=\; V_{b0}\,\gamma_{\text{beaker}}\,dT \;-\; V_m\,\gamma_{\text{mercury}}\,dT.$$

The factor $$dT$$ is common to both terms, so it can be cancelled. Rearranging, we obtain

$$V_m\,\gamma_{\text{mercury}} \;=\; V_{b0}\,\gamma_{\text{beaker}}.$$

Hence the required initial volume of mercury is

$$V_m \;=\; \frac{V_{b0}\,\gamma_{\text{beaker}}}{\gamma_{\text{mercury}}}.$$

We are already given

$$\gamma_{\text{beaker}} = 6 \times 10^{-6}\,^{\circ}\text{C}^{-1}.$$

The standard value for the coefficient of volume expansion of mercury around ordinary temperatures is approximately

$$\gamma_{\text{mercury}} \approx 1.5 \times 10^{-4}\,^{\circ}\text{C}^{-1}.$$

Substituting the numerical data, we get

$$V_m = \frac{500\ \text{cc}\;\bigl(6 \times 10^{-6}\bigr)}{1.5 \times 10^{-4}}.$$

We simplify step by step. First evaluate the ratio of the coefficients:

$$\frac{6 \times 10^{-6}}{1.5 \times 10^{-4}} = \frac{6}{1.5}\times 10^{-6+4} = 4 \times 10^{-2} = 0.04.$$

Now multiply by the beaker volume:

$$V_m = 500 \times 0.04 = 20.$$

Thus

$$V_m \approx 20\ \text{cc}.$$

Hence, the correct answer is Option D (20 cc).

Let us begin by recalling the definition of the temperature coefficient of linear expansion. If a rod of initial length $$L$$ is heated through a small temperature rise $$\Delta T$$, its increase in length $$\Delta L$$ is given by

$$\Delta L \;=\; \alpha\,L\,\Delta T,$$

where $$\alpha$$ is called the (linear) temperature coefficient of expansion for that material.

Now we have two different wires. Their individual data are

Length of first wire: $$L_1\,,\qquad$$ coefficient: $$\alpha_1$$,

Length of second wire: $$L_2\,,\qquad$$ coefficient: $$\alpha_2$$.

The two wires are joined end-to-end (in series), so the overall initial length of the composite wire is

$$L_{\text{total}} \;=\; L_1 + L_2.$$

Suppose the temperature of the whole system rises by the same small amount $$\Delta T$$. Each segment expands according to its own coefficient:

For the first wire we have

$$\Delta L_1 \;=\; \alpha_1\,L_1\,\Delta T.$$

For the second wire we have

$$\Delta L_2 \;=\; \alpha_2\,L_2\,\Delta T.$$

Because the segments are in series, their elongations simply add. Therefore the total elongation of the composite wire is

$$\Delta L_{\text{total}} \;=\; \Delta L_1 + \Delta L_2 \;=\; \alpha_1 L_1 \Delta T \;+\; \alpha_2 L_2 \Delta T.$$

Factorising out the common $$\Delta T$$ we obtain

$$\Delta L_{\text{total}} \;=\; \left( \alpha_1 L_1 + \alpha_2 L_2 \right)\Delta T.$$

Next, we define an effective (or equivalent) coefficient of linear expansion, say $$\alpha_{\text{eff}}$$, for the whole composite wire by using the same basic formula applied to the total length:

$$\Delta L_{\text{total}} \;=\; \alpha_{\text{eff}}\;L_{\text{total}}\;\Delta T \;=\; \alpha_{\text{eff}}\,(L_1 + L_2)\,\Delta T.$$

We now have two expressions for the same quantity $$\Delta L_{\text{total}}$$. Equating them gives

$$\alpha_{\text{eff}}\,(L_1 + L_2)\,\Delta T \;=\; \left( \alpha_1 L_1 + \alpha_2 L_2 \right)\Delta T.$$

Because $$\Delta T$$ is non-zero, it can be cancelled from both sides, leaving

$$\alpha_{\text{eff}}\,(L_1 + L_2) \;=\; \alpha_1 L_1 + \alpha_2 L_2.$$

Finally, solving for $$\alpha_{\text{eff}}$$ we divide by $$L_1 + L_2$$:

$$\alpha_{\text{eff}} \;=\; \frac{\alpha_1 L_1 + \alpha_2 L_2}{L_1 + L_2}.$$

This result matches Option A.

Hence, the correct answer is Option A.

We begin with Newton’s law of cooling, which says that the rate of fall of temperature of a body is directly proportional to the difference between its temperature and the temperature of the surroundings. In equation form, the law is written as

$$\frac{dT}{dt} = -k\,\bigl(T - T_{\text{env}}\bigr),$$

where $$T$$ is the instantaneous temperature of the body, $$T_{\text{env}}$$ is the constant atmospheric (environment) temperature, and $$k$$ is a positive proportionality constant that depends on the nature of the body and its surroundings.

Integrating this differential equation gives the standard exponential cooling formula

$$T - T_{\text{env}} = \bigl(T_0 - T_{\text{env}}\bigr)\,e^{-kt},$$

in which $$T_0$$ is the initial temperature at time $$t = 0$$.

We are told that the metallic sphere has an initial temperature $$T_1 = 50^\circ\text{C}$$ and cools to $$T_2 = 40^\circ\text{C}$$ in $$t_1 = 300\ \text{s}$$ (that is, 5 minutes). The atmospheric temperature is

$$T_{\text{env}} = 20^\circ\text{C}.$$

Substituting these values into the exponential form for the first cooling interval, we write

$$T_2 - T_{\text{env}} = \bigl(T_1 - T_{\text{env}}\bigr)\,e^{-k t_1}.$$

So, using the numbers,

$$40 - 20 = (50 - 20)\,e^{-k \times 300}.$$

Simplifying the numerical differences gives

$$20 = 30\,e^{-300k}.$$

To isolate the exponential factor, we divide both sides by 30:

$$\frac{20}{30} = e^{-300k}.$$

So,

$$\frac{2}{3} = e^{-300k}.$$

Now we take the natural logarithm on both sides to solve for $$k$$:

$$\ln\!\bigl(\tfrac{2}{3}\bigr) = -300k.$$

Hence,

$$k = -\frac{1}{300}\,\ln\!\bigl(\tfrac{2}{3}\bigr) = \frac{1}{300}\,\ln\!\bigl(\tfrac{3}{2}\bigr).$$

With the value of $$k$$ fixed, we now want the temperature $$T_3$$ of the sphere after a further time of another 5 minutes, that is, after a second interval of $$t_2 = 300\ \text{s}$$. The total time elapsed from the initial instant will then be $$t_{\text{total}} = 600\ \text{s}$$.

Using the exponential law again for the whole 600-second span, we have

$$T_3 - T_{\text{env}} = \bigl(T_1 - T_{\text{env}}\bigr)\,e^{-k t_{\text{total}}}.$$

That is,

$$T_3 - 20 = 30 \, e^{-k \times 600}.$$

Because $$e^{-k \times 600} = \bigl(e^{-k \times 300}\bigr)^{2},$$ and from earlier we already found $$e^{-k \times 300} = \tfrac{2}{3},$$ we can write

$$e^{-k \times 600} = \left(\tfrac{2}{3}\right)^{2} = \tfrac{4}{9}.$$

Substituting this value back gives

$$T_3 - 20 = 30 \times \frac{4}{9}.$$

Carrying out the multiplication,

$$T_3 - 20 = \frac{120}{9} = 13.\overline{3}\;^\circ\text{C}.$$

Finally, adding 20°C to both sides yields

$$T_3 = 20 + 13.\overline{3} \approx 33.3^\circ\text{C}.$$

The temperature after the next five minutes is therefore very close to $$33^\circ\text{C}$$, matching option B.

Hence, the correct answer is Option B.

We start by noting that the only significant energy exchange is between the warm water and the ice; the container and surroundings are assumed perfectly insulating. Because the ice is already at its melting point (0 °C) and the water finally cools to 0 °C, no part of the ice-water mixture ever rises above 0 °C. Hence the warm water will simply give up some of its thermal energy, cooling from 25 °C down to 0 °C, and that energy will go exclusively into melting a certain mass of ice.

First we calculate the heat lost by the water as it cools. The specific heat capacity formula is stated as

$$Q = m\,c\,\Delta T,$$

where $$m$$ is the mass of the substance, $$c$$ its specific heat capacity, and $$\Delta T$$ the change in temperature.

The warm water has

$$m_w = 200 \text{ g} = 0.200 \text{ kg},$$

$$c_w = 4200 \text{ J\,kg}^{-1}\text{K}^{-1},$$

$$\Delta T_w = 25^\circ\text{C} - 0^\circ\text{C} = 25 \text{ K}.$$

Substituting these values, we obtain

$$Q_{\text{lost}} = (0.200)\,(4200)\,(25) \text{ J}.$$

Now we multiply step by step:

$$4200 \times 25 = 105,000,$$

$$0.200 \times 105,000 = 21,000.$$

So,

$$Q_{\text{lost}} = 21,000 \text{ J}.$$

This 21,000 J of energy is absorbed by a portion of the ice, melting it while itself staying at 0 °C. For melting, the relevant formula is the latent heat relation

$$Q = m L,$$

where $$m$$ is the mass melted and $$L$$ is the latent heat of fusion of ice.

Given

$$L = 3.4 \times 10^{5} \text{ J\,kg}^{-1},$$

we set

$$Q_{\text{lost}} = Q_{\text{gained}},$$

$$21,000 = m\,L.$$

Solving for $$m$$ we have

$$m = \frac{21,000}{3.4 \times 10^{5}} \text{ kg}.$$

Carrying out the division step by step, we write

$$\frac{21,000}{340,000} = 0.0617647\ldots \text{ kg}.$$

To convert this to grams we multiply by 1000:

$$0.0617647\ldots \text{ kg} \times 1000 = 61.7647\ldots \text{ g}.$$

Rounded sensibly to three significant figures, the mass of ice that melts is approximately

$$\boxed{61.7 \text{ g}}.$$

Hence, the correct answer is Option A.

We have three rods, each of identical length $$L$$ and identical cross-sectional area $$A$$. Their thermal conductivities are $$K_1, K_2$$ and $$K_3$$, respectively. They are joined end to end. The left-most end is kept at $$100^\circ\text{C}$$, the right-most end is kept at $$0^\circ\text{C}$$. In steady state the temperatures at the two junctions are given to be $$70^\circ\text{C}$$ and $$20^\circ\text{C}$$, as shown below:

$$100^\circ\text{C}\;\; |\,\text{Rod }1\,| \;70^\circ\text{C}\;\; |\,\text{Rod }2\,| \;20^\circ\text{C}\;\; |\,\text{Rod }3\,| \;0^\circ\text{C}$$

Because no heat is lost through the lateral surfaces, the same heat current $$Q$$ flows through every section of the composite rod in steady state.

Fourier’s law of heat conduction states first:

$$Q \;=\; \dfrac{K\,A\,\Delta T}{L}$$

where $$K$$ is the thermal conductivity of that section, $$A$$ is the cross-sectional area, $$\Delta T$$ is the temperature difference across the length $$L$$ of that section.

Applying this formula to each rod separately and remembering that $$A$$ and $$L$$ are the same for all rods, we write the heat current for every rod explicitly.

For Rod 1 (from $$100^\circ\text{C}$$ to $$70^\circ\text{C}$$):

$$Q \;=\; \dfrac{K_1\,A\,(100 - 70)}{L} \;=\; \dfrac{K_1\,A\,(30)}{L}$$

For Rod 2 (from $$70^\circ\text{C}$$ to $$20^\circ\text{C}$$):

$$Q \;=\; \dfrac{K_2\,A\,(70 - 20)}{L} \;=\; \dfrac{K_2\,A\,(50)}{L}$$

For Rod 3 (from $$20^\circ\text{C}$$ to $$0^\circ\text{C}$$):

$$Q \;=\; \dfrac{K_3\,A\,(20 - 0)}{L} \;=\; \dfrac{K_3\,A\,(20)}{L}$$

Because the heat current is the same everywhere, we equate these three expressions:

$$\dfrac{K_1\,A\,(30)}{L} \;=\; \dfrac{K_2\,A\,(50)}{L} \;=\; \dfrac{K_3\,A\,(20)}{L}$$

The common factors $$A$$ and $$L$$ cancel from all three terms, giving the simpler equality

$$K_1\,(30) \;=\; K_2\,(50) \;=\; K_3\,(20)$$

We now obtain the required ratios by pairwise comparison. First, comparing the first two terms:

$$K_1\,(30) \;=\; K_2\,(50)$$

Dividing both sides by $$30\,K_2$$:

$$\dfrac{K_1}{K_2} \;=\; \dfrac{50}{30} \;=\; \dfrac{5}{3}$$

Next, comparing the second and third terms:

$$K_2\,(50) \;=\; K_3\,(20)$$

Dividing both sides by $$50\,K_3$$:

$$\dfrac{K_2}{K_3} \;=\; \dfrac{20}{50} \;=\; \dfrac{2}{5}$$

From the above two ratios, we can also find $$K_1 : K_3$$ logically. We already have

$$\dfrac{K_1}{K_3} \;=\; \left(\dfrac{K_1}{K_2}\right)\!\left(\dfrac{K_2}{K_3}\right) \;=\; \left(\dfrac{5}{3}\right)\!\left(\dfrac{2}{5}\right) \;=\; \dfrac{2}{3}$$

Therefore, writing all the ratios in the simplest integer form:

$$K_1 : K_3 \;=\; 2 : 3, \qquad K_2 : K_3 \;=\; 2 : 5$$

These two relations match exactly with Option A of the given choices. No other option is consistent with both ratios simultaneously.

Hence, the correct answer is Option A.

We have a calorimeter whose water equivalent is 20 g. This means it behaves exactly as if it contained 20 g of water: its thermal capacity is $$20 \text{ cal }{}^\circ\text{C}^{-1}$$. Inside it are $$180 \text{ g}$$ of actual water. The initial temperature of everything inside the calorimeter is $$25^\circ\text{C}$$.

Now, let $$m$$ grams of dry steam at $$100^\circ\text{C}$$ be passed into the calorimeter. After all the steam has condensed and equilibrium is reached, the common temperature of the mixture is $$31^\circ\text{C}$$. We must find the numerical value of $$m$$.

Principle used: Heat lost = Heat gained. No heat is lost to the surroundings.

Heat gained is by the existing water plus the calorimeter itself. Their combined mass (water equivalent) is

The temperature rise of this 200 g is from $$25^\circ\text{C}$$ to $$31^\circ\text{C}$$, i.e. $$\Delta T = 31 - 25 = 6^\circ\text{C}$$. Using the specific heat of water $$c = 1 \text{ cal g}^{-1}{}^\circ\text{C}^{-1}$$,

Heat lost is by the incoming steam. Two separate processes occur:

(i) Condensation: Each gram of steam releases its latent heat $$L = 540 \text{ cal g}^{-1}$$ when it turns into water at $$100^\circ\text{C}$$. Thus the heat released in condensation is $$m \times 540 \text{ cal}$$.

(ii) Cooling: The condensed water, whose mass is still $$m$$ grams, then cools from $$100^\circ\text{C}$$ to the final $$31^\circ\text{C}$$. The temperature drop is $$100 - 31 = 69^\circ\text{C}$$, so the heat released in cooling is

Hence the total heat lost by the steam is

Applying conservation of energy,

Solving for $$m$$, we divide both sides by $$609$$:

Carrying out the division,

The options given are 2 g, 4 g, 3.2 g, and 2.6 g. The value we obtained, $$1.972 \text{ g}$$, is closest to $$2 \text{ g}$$.

Hence, the correct answer is Option A.

We start with the bullet mass $$m = 5\;\text{gram}$$ and its speed $$v = 210\;\text{m s}^{-1}$$. For uniformity we change the speed to centimetre-second (cgs) units because the energy conversion factor given at the end is in ergs:

$$210\;\text{m s}^{-1} = 210 \times 100\;\text{cm s}^{-1} = 2.10 \times 10^{4}\;\text{cm s}^{-1}.$$

The kinetic-energy formula is first stated:

$$\text{K.E.} = \dfrac{1}{2} m v^{2}.$$

Substituting the values (keeping the mass in grams and the speed in centimetres per second so that the answer comes out in ergs):

$$ \text{K.E.} \;=\; \dfrac{1}{2}\,(5\;\text{g})\,(2.10 \times 10^{4}\;\text{cm s}^{-1})^{2} \;=\; \dfrac{1}{2}\times 5 \times (2.10)^{2} \times 10^{8} \;=\; 2.5 \times 4.41 \times 10^{8} \;=\; 1.1025 \times 10^{9}\;\text{ergs}. $$

The question tells us that one half of this kinetic energy goes into heating the wooden target. Therefore the remaining half heats the bullet itself. So the heat energy actually absorbed by the bullet is

$$ Q = \dfrac{1}{2}\,\text{K.E.} = \dfrac{1}{2}\times 1.1025 \times 10^{9}\;\text{ergs} = 0.55125 \times 10^{9}\;\text{ergs}. $$

We convert this energy into calories because the specific heat is given in calories per gram per degree Celsius. The conversion factor stated in the question is $$1\;\text{calorie} = 4.2 \times 10^{7}\;\text{ergs}$$, therefore

$$ Q = \dfrac{0.55125 \times 10^{9}\;\text{ergs}}{4.2 \times 10^{7}\;\text{ergs calorie}^{-1}} = \dfrac{0.55125}{4.2}\times 10^{2}\;\text{calories} = 0.13125 \times 10^{2}\;\text{calories} = 13.125\;\text{calories}. $$

The specific heat of the bullet’s material is given as $$c = 0.030\;\text{cal}\,(\text{gram}\,^{\circ}\text{C})^{-1}$$. The temperature rise $$\Delta T$$ is obtained from the heat equation

$$ Q = m c \Delta T \quad\Longrightarrow\quad \Delta T = \dfrac{Q}{m c}. $$

Substituting the numerical values:

$$ \Delta T = \dfrac{13.125\;\text{cal}}{(5\;\text{g})(0.030\;\text{cal g}^{-1}{}^{\circ}\text{C}^{-1})} = \dfrac{13.125}{0.15} = 87.5^{\circ}\text{C}. $$

Hence, the correct answer is Option A.

We have $$m = 1\ \text{kg}$$ of water whose initial temperature is $$20^\circ\text{C}$$. To evaporate it completely, two distinct amounts of heat are required:

1. Heat to raise the temperature from $$20^\circ\text{C}$$ to the boiling point $$100^\circ\text{C}$$.
2. Heat to convert the water at $$100^\circ\text{C}$$ into steam at the same temperature (latent heat of vaporisation).

First, we evaluate the sensible heat needed for the temperature rise. We state the formula for heat absorbed when there is a temperature change:

$$Q_1 = mc\Delta T$$

Here $$c = 4200\ \text{J kg}^{-1}{}^\circ\text{C}^{-1}$$ and $$\Delta T = 100^\circ\text{C} - 20^\circ\text{C} = 80^\circ\text{C}$$. Substituting the values,

$$Q_1 = (1\ \text{kg})(4200\ \text{J kg}^{-1}{}^\circ\text{C}^{-1})(80^\circ\text{C})$$

$$Q_1 = 4200 \times 80\ \text{J}$$

$$Q_1 = 336000\ \text{J}$$

Next, we compute the latent heat required for the phase change. The latent heat of vaporisation is given as $$L = 2260\ \text{kJ kg}^{-1} = 2260 \times 10^{3}\ \text{J kg}^{-1}$$. The formula for the heat of phase change is

$$Q_2 = mL$$

For $$m = 1\ \text{kg}$$,

$$Q_2 = 1 \times 2260 \times 10^{3}\ \text{J}$$

$$Q_2 = 2260000\ \text{J}$$

The total heat required is the sum of these two quantities:

$$Q_{\text{total}} = Q_1 + Q_2$$

$$Q_{\text{total}} = 336000\ \text{J} + 2260000\ \text{J}$$

$$Q_{\text{total}} = 2596000\ \text{J}$$

Now we determine the rate at which the electric kettle can supply energy. The kettle is connected to a mains supply of rms voltage $$V_{\text{rms}} = 200\ \text{V}$$, and the heating element has an average resistance $$R = 20\ \Omega$$. We recall Joule’s law for electrical power:

$$P = \dfrac{V_{\text{rms}}^{2}}{R}$$

Substituting the given values,

$$P = \dfrac{(200\ \text{V})^{2}}{20\ \Omega}$$

$$P = \dfrac{40000\ \text{V}^2}{20\ \Omega}$$

$$P = 2000\ \text{W}$$

So the kettle delivers $$2000\ \text{J s}^{-1}$$ of energy. The time required to supply the total heat is obtained from

$$t = \dfrac{Q_{\text{total}}}{P}$$

$$t = \dfrac{2596000\ \text{J}}{2000\ \text{J s}^{-1}}$$

$$t = 1298\ \text{s}$$

We convert this time into minutes for easier comprehension:

$$t = \dfrac{1298\ \text{s}}{60\ \text{s min}^{-1}}$$

$$t \approx 21.6\ \text{min}$$

This value is closest to the option stating $$22\ \text{min}$$.

Hence, the correct answer is Option C.

We recall the steady-state heat conduction law, also called Fourier’s law of thermal conduction. It states that the energy flux $$\phi$$ (heat current density) through a homogeneous slab is given by

$$\phi \;=\; k \,\dfrac{\Delta T}{L}$$

where $$k$$ is the thermal conductivity of the material, $$\Delta T$$ is the temperature difference between the two faces of the slab, and $$L$$ is the thickness of the slab.

For the present problem we have the following data:

Thermal conductivity: $$k = 0.1\; \text{W K}^{-1}\text{ m}^{-1}$$

Temperatures of the reservoirs: $$T_\text{hot} = 10^{3}\; \text{K}$$ and $$T_\text{cold} = 10^{2}\; \text{K}$$

Therefore the temperature difference is

$$\Delta T = T_\text{hot} - T_\text{cold} = 10^{3} - 10^{2} = 1000 - 100 = 900\ \text{K}$$

The thickness of the copper slab is given as $$L = 1\ \text{m}$$.

Now we substitute these values into Fourier’s law:

$$\phi = k \,\dfrac{\Delta T}{L} = 0.1\;\text{W K}^{-1}\text{ m}^{-1} \times \dfrac{900\ \text{K}}{1\ \text{m}}$$

Carrying out the multiplication and division step by step, we first note that dividing by 1 m leaves the numerator unchanged in magnitude and only attaches the correct units:

$$\phi = 0.1 \times 900\ \text{W m}^{-2}$$

Multiplying $$0.1$$ and $$900$$ gives

$$\phi = 90\ \text{W m}^{-2}$$

Thus the steady-state energy flux through the copper slab is $$90\ \text{W m}^{-2}$$.

Looking at the provided options, this value corresponds to Option C.

Hence, the correct answer is Option C.

We have an unknown metal whose specific heat we call $$c_{m}\,( \text{J kg}^{-1}\text{K}^{-1})$$.

First, every mass must be written in kilograms because the S.I. unit of specific heat is J kg$$^{-1}$$K$$^{-1}$$.

$$m_{m}=192\;\text{g}=0.192\;\text{kg}$$ (mass of metal)

$$m_{w}=240\;\text{g}=0.240\;\text{kg}$$ (mass of water)

$$m_{c}=128\;\text{g}=0.128\;\text{kg}$$ (mass of brass calorimeter)

The temperatures are

$$T_{m}^{\,(i)} = 100^{\circ}\text{C},\qquad T_{w}^{\,(i)} = 8.4^{\circ}\text{C},\qquad T_{f}=21.5^{\circ}\text{C}.$$

The specific heats given or known are

$$c_{w}=4186\;\text{J kg}^{-1}\text{K}^{-1}\;(\text{water}),\qquad c_{c}=394\;\text{J kg}^{-1}\text{K}^{-1}\;(\text{brass}).$$

Energy conservation in calorimetry states:

$$\text{Heat lost by hot body} = \text{Heat gained by all cooler bodies}.$$

So,

$$m_{m}\,c_{m}\,(T_{m}^{\,(i)}-T_{f}) \;=\; m_{w}\,c_{w}\,(T_{f}-T_{w}^{\,(i)}) \;+\; m_{c}\,c_{c}\,(T_{f}-T_{w}^{\,(i)}).$$

Now we substitute every known quantity step by step.

The temperature fall of the metal is

$$T_{m}^{\,(i)}-T_{f}=100-21.5=78.5\;\text{K}.$$

The common temperature rise of water and the calorimeter is

$$T_{f}-T_{w}^{\,(i)}=21.5-8.4=13.1\;\text{K}.$$

Heat lost by the metal:

$$Q_{m}=0.192\,c_{m}\times 78.5 =0.192 \times 78.5 \; c_{m} =15.072\,c_{m}\;(\text{J}).$$

Heat gained by water:

$$Q_{w}=0.240 \times 4186 \times 13.1.$$

First multiply $$4186\times 13.1=54836.6,$$ then multiply by $$0.240$$ giving

$$Q_{w}=0.240 \times 54836.6 = 13160.784\;\text{J}.$$

Heat gained by the brass calorimeter:

$$Q_{c}=0.128 \times 394 \times 13.1.$$

Compute $$394\times 13.1 = 5161.4,$$ then multiply by $$0.128$$ giving

$$Q_{c}=0.128 \times 5161.4 = 660.6592\;\text{J}.$$

Total heat gained:

$$Q_{w}+Q_{c}=13160.784 + 660.6592 = 13821.4432\;\text{J}.$$

Set heat lost equal to heat gained:

$$15.072\,c_{m}=13821.4432.$$

Solve for $$c_{m}$$ by dividing both sides by 15.072:

$$c_{m}= \dfrac{13821.4432}{15.072}\;\text{J kg}^{-1}\text{K}^{-1}.$$

Carrying out the division,

$$c_{m}\approx 916\;\text{J kg}^{-1}\text{K}^{-1}.$$

Hence, the correct answer is Option A.

We begin with energy conservation because the vessel is thermally insulated. Whatever heat is absorbed to evaporate some water must be exactly supplied by the heat released when another part of the water freezes. No heat can enter or leave the vessel.

Let

$$m \text{ g}$$ be the mass of water that evaporates,

$$x \text{ g}$$ be the mass of water that freezes into ice.

The latent heat values given are

$$L_v = 2.10 \times 10^{6}\ \text{J kg}^{-1}=2.10 \times 10^{3}\ \text{J g}^{-1},$$

$$L_f = 3.36 \times 10^{5}\ \text{J kg}^{-1}=3.36 \times 10^{2}\ \text{J g}^{-1}.$$

First we write the formulae for the heats involved:

Heat absorbed in vaporising $$m$$ grams of water at 0 °C:

$$Q_{\text{vap}} = m\,L_v.$$

Heat released when $$x$$ grams of water freeze at 0 °C:

$$Q_{\text{freeze}} = x\,L_f.$$

Because the vessel is adiabatic, these two quantities must be equal:

$$Q_{\text{vap}} = Q_{\text{freeze}}.$$

So we have

$$m\,L_v = x\,L_f.$$

Substituting the numerical values of the latent heats:

$$m\,(2.10 \times 10^{3}) = x\,(3.36 \times 10^{2}).$$

Now we isolate $$x$$ in terms of $$m$$:

$$x = \frac{2.10 \times 10^{3}}{3.36 \times 10^{2}}\,m.$$

Evaluating the fraction,

$$\frac{2.10 \times 10^{3}}{3.36 \times 10^{2}}=\frac{2100}{336}\approx 6.25.$$

Hence

$$x \approx 6.25\,m.$$

Next we use the fact that the total mass of water present initially was 150 g. After the process finishes we have evaporated mass $$m$$, frozen mass $$x$$, and whatever remains as liquid. Therefore the sum of the masses that disappeared from the liquid phase must not exceed 150 g:

$$x + m \le 150.$$

Substituting $$x = 6.25\,m$$ into this inequality gives

$$6.25\,m + m \le 150,$$

$$7.25\,m \le 150.$$

Solving for $$m$$,

$$m \le \frac{150}{7.25}.$$

Carrying out the division,

$$\frac{150}{7.25} \approx 20.7\ \text{g}.$$

The amount that actually evaporates will be this maximum value because once all available heat from freezing has been used, no further evaporation can occur while still maintaining 0 °C. Thus the evaporated mass is about 20 g, which matches option B most closely.

Hence, the correct answer is Option B.

Let us denote the specific heats of liquids A and B by $$c_A$$ and $$c_B$$ respectively (in $$\text{J g}^{-1}\,{}^{\circ}\text{C}^{-1}$$). We shall assume there is no heat loss to the surroundings, so the heat lost by the hotter liquid equals the heat gained by the colder liquid.

For any mixing process we shall use the principle of conservation of energy in the form

$$m_{\text{hot}}\,c_{\text{hot}}\,(T_{\text{hot}}-T_f)\;=\;m_{\text{cold}}\,c_{\text{cold}}\,(T_f-T_{\text{cold}}),$$

where the subscripts ‘hot’ and ‘cold’ refer to the initially warmer and cooler liquids, and $$T_f$$ is the final (equilibrium) temperature.

We first employ the given data of the first mixing:

Mass of A $$=100\text{ g},\;T_A=100^{\circ}\text{C},$$ Mass of B $$=50\text{ g},\;T_B=75^{\circ}\text{C},$$ Final temperature $$T_f=90^{\circ}\text{C}.$$

Applying the formula, the heat lost by A equals the heat gained by B:

$$100\,c_A\,(100-90)\;=\;50\,c_B\,(90-75).$$

Simplifying each side, we get

$$100\,c_A\,(10)\;=\;50\,c_B\,(15).$$

That is

$$1000\,c_A\;=\;750\,c_B.$$ Dividing by $$250$$ gives

$$4\,c_A\;=\;3\,c_B$$ or $$c_B=\frac{4}{3}\,c_A.$$

This relation between the specific heats will now be used for the second mixing.

In the second experiment we again take 100 g of A at $$100^{\circ}\text{C}$$, but the 50 g of B is now at $$50^{\circ}\text{C}$$. Let the new final temperature be $$T^{\prime}$$.

Using the conservation-of-energy equation once more:

$$100\,c_A\,(100-T^{\prime})\;=\;50\,c_B\,(T^{\prime}-50).$$

Substituting $$c_B=\dfrac{4}{3}\,c_A$$ obtained earlier:

$$100\,c_A\,(100-T^{\prime})\;=\;50\left(\frac{4}{3}c_A\right)(T^{\prime}-50).$$

The factor $$c_A$$ cancels from both sides, leaving

$$100\,(100-T^{\prime})\;=\;50\left(\frac{4}{3}\right)(T^{\prime}-50).$$

Compute the constant on the right:

$$50\left(\frac{4}{3}\right)=\frac{200}{3}.$$

Thus

$$100\,(100-T^{\prime})=\frac{200}{3}(T^{\prime}-50).$$

Eliminating the fraction by multiplying every term by 3:

$$300\,(100-T^{\prime})=200\,(T^{\prime}-50).$$

Expanding each side,

$$30000-300\,T^{\prime}=200\,T^{\prime}-10000.$$

Collecting the temperature terms on one side and the constants on the other:

$$-300\,T^{\prime}-200\,T^{\prime}=-10000-30000,$$ $$-500\,T^{\prime}=-40000.$$

Dividing by $$-500$$:

$$T^{\prime}=\frac{-40000}{-500}=80.$$

So the equilibrium temperature of the mixture in the second case is

$$80^{\circ}\text{C}.$$

Hence, the correct answer is Option C.

We apply the principle of calorimetry, which tells us that in an isolated system the heat lost by the hotter body equals the heat gained by the colder bodies:

$$\text{Heat lost by metal ball} = \text{Heat gained by water} + \text{Heat gained by vessel}.$$

First we write the general expressions. For any substance, the heat exchanged is given by $$Q = mc\Delta T$$ where $$m$$ is mass, $$c$$ is specific heat capacity and $$\Delta T$$ is the change in temperature. For a calorimeter (the vessel) whose heat capacity is already given as a whole number $$C$$, the heat taken is $$Q = C\Delta T.$$

Let the common final temperature after mixing be $$T_f$$ (in $$^\circ\text{C}$$ or K; the difference is the same). Initial data:

Metal ball: $$m_m = 0.1\ \text{kg}, \; c_m = 400\ \text{J kg}^{-1}\text{K}^{-1}, \; T_{mi}=500^\circ\text{C}.$$ Water: $$m_w = 0.5\ \text{kg}, \; c_w = 4200\ \text{J kg}^{-1}\text{K}^{-1}, \; T_{wi}=30^\circ\text{C}.$$ Vessel: $$C_v = 800\ \text{J K}^{-1}, \; T_{vi}=30^\circ\text{C}.$$

Because the water and the vessel start at the same temperature, their temperature rise is $$T_f - 30$$. The metal ball cools from $$500^\circ\text{C}$$ to $$T_f$$, so its temperature fall is $$500 - T_f$$.

Now we set up the energy balance:

$$ m_m c_m (500 - T_f) = m_w c_w (T_f - 30) + C_v(T_f - 30). $$

Substituting every numerical value:

$$ 0.1 \times 400 \,(500 - T_f) = 0.5 \times 4200 \,(T_f - 30) + 800\,(T_f - 30). $$

Compute each coefficient step by step:

Left hand coefficient: $$0.1 \times 400 = 40.$$ Right hand first term coefficient: $$0.5 \times 4200 = 2100.$$ Adding the vessel’s 800 gives a total right-hand coefficient $$2100 + 800 = 2900.$$

Thus the equation simplifies to

$$ 40(500 - T_f) = 2900(T_f - 30). $$

Expanding both sides:

$$ 40 \times 500 - 40T_f = 2900T_f - 2900 \times 30. $$

Calculate the plain numbers:

$$ 20000 - 40T_f = 2900T_f - 87000. $$

Collect all terms containing $$T_f$$ on the right and the pure numbers on the left:

$$ 20000 + 87000 = 2900T_f + 40T_f. $$

$$ 107000 = 2940T_f. $$

Solve for the final temperature:

$$ T_f = \frac{107000}{2940} \approx 36.4^\circ\text{C}. $$

The initial temperature of the water was $$30^\circ\text{C}$$, so the rise in temperature is

$$ \Delta T = 36.4 - 30 = 6.4^\circ\text{C}. $$

The percentage increment relative to the original $$30^\circ\text{C}$$ is

$$ \text{Percentage rise} = \frac{6.4}{30} \times 100 \approx 21.3\%. $$

The nearest option provided is $$20\%$$.

Hence, the correct answer is Option D.

We have a solid cylinder of radius $$R$$ made of a material of thermal conductivity $$K_1$$. All around it, in perfect contact, there is a coaxial hollow cylindrical shell whose inner radius is also $$R$$ and whose outer radius is $$2R$$. The shell is made of another material whose thermal conductivity is $$K_2$$. Heat is allowed to flow only along the axis (length) of the composite cylinder, and we assume that there is no loss of heat sideways. Our aim is to find one single number $$K_{\text{eff}}$$ that can replace the two conductors so that the heat flow calculated with $$K_{\text{eff}}$$ is exactly the same as the actual heat flow through the two materials taken together.

First, recall Fourier’s law of steady-state heat conduction along a length:

$$Q = \dfrac{K\,A\,\Delta T}{L},$$

where

$$Q$$ = heat current (rate of heat flow),
$$K$$ = thermal conductivity of the material,
$$A$$ = cross-sectional area perpendicular to the heat flow,
$$\Delta T$$ = temperature difference applied between the two ends, and
$$L$$ = length of the conductor in the direction of heat flow.

For axial (longitudinal) flow, each cylindrical region has its flat circular cross-section perpendicular to the flow. Because the two regions extend from the same hot face to the same cold face, the temperature difference $$\Delta T$$ and the length $$L$$ are identical for both materials. Therefore both regions behave as parallel thermal conductors. In a parallel arrangement, individual heat currents simply add:

$$Q_{\text{total}} = Q_1 + Q_2.$$

Using Fourier’s law for each part we write

$$Q_1 = \dfrac{K_1\,A_1\,\Delta T}{L}, \qquad Q_2 = \dfrac{K_2\,A_2\,\Delta T}{L}.$$

Substituting these into the addition rule gives

$$Q_{\text{total}} = \dfrac{K_1\,A_1\,\Delta T}{L} + \dfrac{K_2\,A_2\,\Delta T}{L}.$$

Now we define the effective conductivity $$K_{\text{eff}}$$ through

$$Q_{\text{total}} = \dfrac{K_{\text{eff}}\,A_{\text{total}}\,\Delta T}{L},$$

where $$A_{\text{total}}$$ is the total cross-sectional area of the whole composite cylinder.

Equating the two expressions for $$Q_{\text{total}}$$ and cancelling the common factors $$\Delta T/L$$ on both sides, we obtain the relation

$$K_{\text{eff}}\,A_{\text{total}} = K_1\,A_1 + K_2\,A_2.$$

All that remains is to insert the actual areas. Because the cross-sections are circular, each area equals $$\pi \times (\text{radius})^2$$.

Area of the inner solid cylinder:

$$A_1 = \pi R^2.$$

Area of the outer hollow shell (annulus):

We subtract the inner area from the outer area: $$A_2 = \pi (2R)^2 \;-\; \pi R^2 = \pi (4R^2 - R^2) = 3\pi R^2.$$

Total cross-sectional area of the composite cylinder:

$$A_{\text{total}} = \pi (2R)^2 = 4\pi R^2.$$

Substituting $$A_1, A_2$$ and $$A_{\text{total}}$$ into the earlier relation gives

$$K_{\text{eff}}\,(4\pi R^2) = K_1\,(\pi R^2) + K_2\,(3\pi R^2).$$

We now cancel the common factor $$\pi R^2$$ from every term:

$$4\,K_{\text{eff}} = K_1 + 3K_2.$$

Solving for $$K_{\text{eff}}$$ we divide both sides by $$4$$:

$$K_{\text{eff}} = \dfrac{K_1 + 3K_2}{4}.$$

Therefore the effective thermal conductivity of the entire assembly is

$$K_{\text{eff}} = \frac{K_1 + 3K_2}{4}.$$

Hence, the correct answer is Option D.

Let us denote the original length of each rod by $$L$$. Both rods initially are at the same temperature $$30^\circ\text{C}$$, so their initial lengths are equal.

When a solid rod is heated through a temperature difference $$\Delta T$$, its new length is given by the linear expansion formula

$$L_\text{new}=L\,(1+\alpha\,\Delta T),$$

where $$\alpha$$ is the coefficient of linear expansion for that material. We now apply this formula separately to rods A and B.

For rod A we have:

$$\Delta T_A = 180^\circ\text{C}-30^\circ\text{C}=150^\circ\text{C},$$

so its new length becomes

$$L_A = L\,(1+\alpha_A\,\Delta T_A).$$

For rod B, its final temperature is $$T^\circ\text{C}$$, hence

$$\Delta T_B = T^\circ\text{C}-30^\circ\text{C},$$

and its new length is

$$L_B = L\,(1+\alpha_B\,\Delta T_B).$$

According to the problem, these new lengths are the same, so

$$L_A = L_B.$$

Substituting the two expressions, we get

$$L\,(1+\alpha_A\,\Delta T_A) = L\,(1+\alpha_B\,\Delta T_B).$$

Because the original length $$L$$ is common and non-zero, we can cancel it, giving

$$1+\alpha_A\,\Delta T_A = 1+\alpha_B\,\Delta T_B.$$

Now subtract 1 from both sides:

$$\alpha_A\,\Delta T_A = \alpha_B\,\Delta T_B.$$

The ratio of the coefficients of linear expansion is given to be

$$\frac{\alpha_A}{\alpha_B} = \frac{4}{3}.$$

Substituting this ratio into the equality of expansions, we find

$$\frac{4}{3}\,\Delta T_A = \Delta T_B.$$

We already know $$\Delta T_A = 150^\circ\text{C}$$, so

$$\Delta T_B = \frac{4}{3}\times 150^\circ\text{C} = 200^\circ\text{C}.$$

Finally, the actual final temperature of rod B is obtained by adding its temperature rise to the initial temperature:

$$T = 30^\circ\text{C} + 200^\circ\text{C} = 230^\circ\text{C}.$$

Hence, the correct answer is Option A.

We are told that the thermometer follows a linear (uniform) scale, so its reading $$x$$ and the actual Celsius temperature $$T$$ are related by the straight-line equation

$$x = kT + c$$

where $$k$$ is the scale factor (slope) and $$c$$ is the intercept (the reading when the temperature is 0 °C). We will determine $$k$$ and $$c$$ using the two fixed points, and then find the unknown temperature.

First fixed point (boiling water): the temperature is $$100\,^{\circ}\text{C}$$ and the thermometer reads $$x_0$$. Substituting in the linear equation gives

$$x_0 = k(100) + c \quad\Longrightarrow\quad 100k + c = x_0 \quad -(1)$$

Second fixed point (ice): the temperature is $$0\,^{\circ}\text{C}$$ and the thermometer reads $$\dfrac{x_0}{3}$$. Substituting gives

$$\frac{x_0}{3} = k(0) + c \quad\Longrightarrow\quad c = \frac{x_0}{3} \quad -(2)$$

Now we substitute the value of $$c$$ from (2) into equation (1) to find $$k$$:

$$100k + \frac{x_0}{3} = x_0$$

Subtract $$\dfrac{x_0}{3}$$ from both sides:

$$100k = x_0 - \frac{x_0}{3}$$

Write the right side with a common denominator:

$$x_0 - \frac{x_0}{3} = \frac{3x_0}{3} - \frac{x_0}{3} = \frac{2x_0}{3}$$

So

$$100k = \frac{2x_0}{3}$$

Divide by 100:

$$k = \frac{\frac{2x_0}{3}}{100} = \frac{2x_0}{300} = \frac{x_0}{150}$$

Now the thermometer is placed in contact with the unknown object and reads $$\dfrac{x_0}{2}$$. Let the corresponding true temperature be $$T$$. Substituting $$x = \dfrac{x_0}{2}$$, $$k = \dfrac{x_0}{150}$$ and $$c = \dfrac{x_0}{3}$$ into the linear relation $$x = kT + c$$ gives

$$\frac{x_0}{2} = \left(\frac{x_0}{150}\right)T + \frac{x_0}{3}$$

Divide every term by $$x_0$$ to simplify:

$$\frac{1}{2} = \frac{T}{150} + \frac{1}{3}$$

Subtract $$\dfrac{1}{3}$$ from $$\dfrac{1}{2}$$:

$$\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$

Hence

$$\frac{1}{6} = \frac{T}{150}$$

Multiply both sides by 150:

$$T = 150 \times \frac{1}{6} = 25$$

So the object’s temperature is $$25\,^{\circ}\text{C}$$.

Hence, the correct answer is Option A.

We have 50 g of liquid water whose initial temperature is $$40^{\circ}\text{C}$$. Some mass of ice, say $$m\; \text{g}$$, is added at $$-20^{\circ}\text{C}$$. After thermal equilibrium is attained, the common temperature becomes $$0^{\circ}\text{C}$$ and 20 g of ice still remains unmelted. Therefore only $$m-20\;\text{g}$$ of the originally added ice has actually melted.

The process involves heat exchange between two parts:

1. Cooling of the warm water from $$40^{\circ}\text{C}$$ down to $$0^{\circ}\text{C}$$.
   Using the formula for sensible heat $$Q = mc\Delta T$$, the heat released is

2. Warming and partial melting of the ice added at $$-20^{\circ}\text{C}$$.

   (a) First, the ice must be warmed from $$-20^{\circ}\text{C}$$ to $$0^{\circ}\text{C}$$.
     Again with $$Q = mc\Delta T$$,

   (b) Next, only $$m-20\;\text{g}$$ of this ice actually melts at $$0^{\circ}\text{C}$$. The latent heat needed is given by

The total heat absorbed by the ice is therefore

Because no heat is lost to the surroundings, the heat released by the water must equal the heat gained by the ice:

Substituting the expressions we have just found,

Now we solve algebraically for $$m$$. First add 6680 J to both sides:

Dividing both sides by 376,

This mass is closest to 40 g among the given choices.

Hence, the correct answer is Option D.

The situation is a classic calorimetry problem, so we apply the principle that in an isolated system the total heat lost by the hotter part equals the total heat gained by the colder part.

Here the hotter part is the water that starts at 50 °C and cools to 0 °C. The colder part is the ice that starts at -10 °C, first warms up to 0 °C and then melts completely. At the end everything is liquid water at 0 °C, so energy balance gives

$$\text{Heat lost by water} \;=\; \text{Heat gained by ice}.$$

We evaluate each side one step at a time.

Heat lost by the water. The formula for sensible heat is $$Q = m c \Delta T.$$ Mass of water is $$M_2\ \text{g},$$ its specific heat is $$1\ \text{cal g}^{-1}\,^\circ\text{C}^{-1},$$ and the fall in temperature is $$\Delta T = 50^\circ\text{C} - 0^\circ\text{C} = 50^\circ\text{C}.$$ So

$$Q_{\text{water}} \;=\; M_2 \times 1 \times 50 \;=\; 50\,M_2\ \text{cal}.$$

Heat gained by the ice. The ice first has to warm from -10 °C to 0 °C and then undergo the phase change.

1. Warming the ice: again using $$Q = m c \Delta T,$$ with mass $$M_1,$$ specific heat of ice $$0.5\ \text{cal g}^{-1}\,^\circ\text{C}^{-1},$$ and temperature rise $$\Delta T = 0^\circ\text{C} - (-10^\circ\text{C}) = 10^\circ\text{C},$$ we get

$$Q_{\text{warming}} \;=\; M_1 \times 0.5 \times 10 \;=\; 5\,M_1\ \text{cal}.$$

2. Melting the ice: the heat needed is $$Q = mL,$$ where $$L$$ is the latent heat of fusion of ice. Thus

$$Q_{\text{melting}} \;=\; M_1 \, L.$$

The total heat absorbed by the ice is the sum of these two contributions:

$$Q_{\text{ice}} \;=\; 5\,M_1 \;+\; M_1\,L.$$

Setting heat lost equal to heat gained gives

$$50\,M_2 \;=\; 5\,M_1 \;+\; M_1\,L.$$

Now we solve for the latent heat $$L.$$ First subtract $$5\,M_1$$ from both sides:

$$50\,M_2 \;-\; 5\,M_1 \;=\; M_1\,L.$$

Finally divide both sides by $$M_1$$:

$$L \;=\; \frac{50\,M_2}{M_1} \;-\; 5.$$

This matches exactly the expression given in Option D.

Hence, the correct answer is Option D.

We apply Newton’s law of cooling which states first: $$\frac{dT}{dt}=-k\,(T-T_s),$$ where $$T$$ is the temperature of the body at time $$t$$, $$T_s$$ is the constant temperature of the surroundings and $$k$$ is a positive constant characteristic of the system.

Integrating this differential equation, we obtain the standard formula

$$T-T_s=(T_0-T_s)\,e^{-kt},$$

where $$T_0$$ is the initial temperature at $$t=0$$.

For the data given, the temperature of the surroundings is $$T_s=25^\circ\text{C}$$, and the body cools from $$T_0=60^\circ\text{C}$$ to $$T_1=50^\circ\text{C}$$ in the first $$10$$ minutes. We substitute these values into the formula to determine the constant $$k$$.

At $$t_1=10\ \text{min}$$, we have

$$T_1-T_s=(T_0-T_s)\,e^{-k t_1}.$$

Explicitly,

$$50-25=(60-25)\,e^{-10k}.$$

Simplifying the numbers gives

$$25=35\,e^{-10k}.$$

Dividing both sides by $$35$$,

$$\frac{25}{35}=e^{-10k}\quad\Longrightarrow\quad e^{-10k}=\frac{5}{7}.$$

Taking natural logarithm on both sides,

$$-10k=\ln\!\left(\frac{5}{7}\right)\quad\Longrightarrow\quad k=-\frac{1}{10}\,\ln\!\left(\frac{5}{7}\right)=\frac{1}{10}\,\ln\!\left(\frac{7}{5}\right).$$

Now we wish to find the temperature $$T_2$$ of the body after another $$10$$ minutes, i.e., at $$t_2=20\ \text{min}$$ from the start. Using the same cooling law,

$$T_2-T_s=(T_0-T_s)\,e^{-k t_2}.$$

Because $$t_2=20$$ and $$e^{-k t_2}=(e^{-10k})^2,$$ we write

$$T_2-T_s=(T_0-T_s)\,(e^{-10k})^2.$$

We already have $$e^{-10k}=\dfrac{5}{7}$$, so

$$(e^{-10k})^2=\left(\frac{5}{7}\right)^2=\frac{25}{49}.$$

Also, $$T_0-T_s=60-25=35.$$ Substituting these numbers,

$$T_2-25=35\;\frac{25}{49}.$$

Multiplying,

$$T_2-25=\frac{35\times25}{49}=\frac{35}{49}\times25=\frac{5}{7}\times25=\frac{125}{7}\approx17.857.$$

Adding the surroundings temperature back,

$$T_2=25+17.857\approx42.857^\circ\text{C}.$$

Rounding to the nearest whole number gives $$T_2\approx43^\circ\text{C}.$$

Hence, the correct answer is Option A.

We first recall the basic principle of calorimetry: in an isolated system, the total heat lost by the hotter objects equals the total heat gained by the colder ones, that is

$$\text{Heat lost} \;=\; \text{Heat gained}.$$

The data given are

$$\begin{aligned} \text{Mass of hot copper ball} & = 100 \,\text{g},\\[4pt] \text{Specific heat of copper} & = 0.1 \,\text{cal g}^{-1}{}^{\circ}\!{\rm C}^{-1},\\[4pt] \text{Initial temperature of water and calorimeter} & = 30^{\circ}\!{\rm C},\\[4pt] \text{Mass of copper calorimeter} & = 100 \,\text{g},\\[4pt] \text{Mass of water} & = 170 \,\text{g},\\[4pt] \text{Equilibrium temperature} & = 75^{\circ}\!{\rm C}. \end{aligned}$$

We take the specific heat of water to be the standard value

$$s_{\text{water}} = 1 \,\text{cal g}^{-1}{}^{\circ}\!{\rm C}^{-1}.$$

Heat lost by the hot copper ball as it cools from the unknown temperature $$T$$ to $$75^{\circ}\!{\rm C}$$:

$$ Q_{\text{lost}} = m_{\text{ball}}\;s_{\text{Cu}}\;(T - 75) = 100 \times 0.1 \times (T - 75) = 10\,(T - 75)\ \text{cal}. $$

Heat gained by the copper calorimeter as it warms from $$30^{\circ}\!{\rm C}$$ to $$75^{\circ}\!{\rm C}$$:

$$ Q_{\text{cal}} = m_{\text{cal}}\;s_{\text{Cu}}\;(75 - 30) = 100 \times 0.1 \times 45 = 450\ \text{cal}. $$

Heat gained by the water over the same temperature rise:

$$ Q_{\text{water}} = m_{\text{water}}\;s_{\text{water}}\;(75 - 30) = 170 \times 1 \times 45 = 7650\ \text{cal}. $$

Total heat gained by the colder parts of the system is therefore

$$ Q_{\text{gained}} = Q_{\text{cal}} + Q_{\text{water}} = 450 + 7650 = 8100\ \text{cal}. $$

Applying the calorimetry principle, we equate the two heats:

$$ Q_{\text{lost}} = Q_{\text{gained}} \;\;\Longrightarrow\;\; 10\,(T - 75) = 8100. $$

Solving this simple linear equation step by step, we divide by 10:

$$ T - 75 = \frac{8100}{10} = 810. $$

Adding $$75^{\circ}\!{\rm C}$$ to both sides gives the initial temperature of the copper ball:

$$ T = 810 + 75 = 885^{\circ}\!{\rm C}. $$

Thus the copper ball was originally at a temperature of $$885^{\circ}\!{\rm C}$$.

Hence, the correct answer is Option C.

We have to use the principle of calorimetry, namely that in an isolated system

$$\text{Heat lost by the hot body}= \text{Heat gained by the cold bodies}.$$

The heat absorbed or released by any body is given by the well-known formula

$$Q = m\,c\,\Delta T,$$

where $$m$$ is the mass, $$c$$ is the specific heat capacity and $$\Delta T$$ is the change in temperature.

Data for the aluminium sphere

Mass $$m_1 = 0.20\;\text{kg}$$, initial temperature $$T_1 = 150^{\circ}\text{C}$$, final temperature $$T_f = 40^{\circ}\text{C}$$.

So the temperature fall is

$$\Delta T_1 = T_1 - T_f = 150^{\circ}\text{C} - 40^{\circ}\text{C} = 110^{\circ}\text{C}.$$

Data for the water in the calorimeter

Volume given is $$150\;\text{cc}.$$ Because the density of water is $$1\;\text{g cm}^{-3},$$ the mass is

$$m_2 = 150\;\text{g} = 0.150\;\text{kg}.$$

Initial temperature $$T_2 = 27^{\circ}\text{C},$$ final temperature $$T_f = 40^{\circ}\text{C}$$; hence

$$\Delta T_2 = T_f - T_2 = 40^{\circ}\text{C} - 27^{\circ}\text{C} = 13^{\circ}\text{C}.$$

Data for the calorimeter itself

The water equivalent is $$0.025\;\text{kg}$$. This means its heat capacity is the same as that of $$0.025\;\text{kg}$$ of water. Therefore we can treat it as additional water of mass

$$m_3 = 0.025\;\text{kg},$$

with the same temperature rise

$$\Delta T_3 = \Delta T_2 = 13^{\circ}\text{C}.$$

Specific heat of water

For water we use $$c_w = 4200\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}.$$ (The conversion given, $$4.2\;\text{J} = 1\;\text{cal},$$ is equivalent to this value.)

Heat gained by water

$$Q_2 = m_2\,c_w\,\Delta T_2 = 0.150\;\text{kg}\times 4200\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}\times 13^{\circ}\text{C}.$$

First multiply the mass and specific heat:

$$0.150 \times 4200 = 630\;\text{J}\,{}^{\circ}\text{C}^{-1}.$$

Then multiply by the temperature rise:

$$630 \times 13 = 8190\;\text{J}.$$

Heat gained by the calorimeter

$$Q_3 = m_3\,c_w\,\Delta T_3 = 0.025\;\text{kg}\times 4200\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}\times 13^{\circ}\text{C}.$$

Calculating similarly:

$$0.025 \times 4200 = 105\;\text{J}\,{}^{\circ}\text{C}^{-1},$$

and

$$105 \times 13 = 1365\;\text{J}.$$

Total heat gained by water plus calorimeter

$$Q_{\text{gained}} = Q_2 + Q_3 = 8190\;\text{J} + 1365\;\text{J} = 9555\;\text{J}.$$

Heat lost by the aluminium sphere

Let $$c$$ be the specific heat capacity of aluminium (in $$\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}$$). Then

$$Q_{\text{lost}} = m_1\,c\,\Delta T_1 = 0.20\;\text{kg}\times c \times 110^{\circ}\text{C}.$$

Multiplying the numerical factors first:

$$0.20 \times 110 = 22,$$

so

$$Q_{\text{lost}} = 22\,c.$$

Applying the heat balance

Because no heat escapes the system,

$$Q_{\text{lost}} = Q_{\text{gained}}.$$

Therefore

$$22\,c = 9555.$$

Solving for $$c$$

$$c = \frac{9555}{22}\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}.$$

Carrying out the division:

$$c = 434.318\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}\;(\text{approximately}).$$

Rounding to three significant figures gives

$$c \approx 434\;\text{J kg}^{-1}\,{}^{\circ}\text{C}^{-1}.$$

Hence, the correct answer is Option A.

We have been asked to calculate the change in internal energy of water when its temperature rises from 40 °C to 60 °C. The mass of the water given is 200 g. We are told to ignore the very small volume expansion of water in this temperature interval, so we can safely treat the specific heat at constant pressure $$c_p$$ as effectively equal to the specific heat at constant volume $$c_v$$. That means the heat supplied translates directly into a change in internal energy.

First, we rewrite every quantity in convenient SI units.

The specific heat of water is supplied as

$$c = 4184\ \text{J kg}^{-1}\ \text{K}^{-1}.$$

The mass is 200 g. Since $$1\ \text{kg}=1000\ \text{g}$$, we convert:

$$m = 200\ \text{g} = \frac{200}{1000}\ \text{kg} = 0.2\ \text{kg}.$$

The temperature change is simply the difference between the final and the initial temperature. We note that a temperature interval in degrees Celsius is numerically the same as in kelvin, so

$$\Delta T = T_{\text{final}} - T_{\text{initial}} = 60^\circ\text{C} - 40^\circ\text{C} = 20\ \text{K}.$$

We now recall the relation connecting heat, specific heat and temperature change for a fixed mass of substance:

$$Q = m\,c\,\Delta T.$$

Because we are neglecting expansion, the heat $$Q$$ supplied goes entirely into increasing the internal energy $$\Delta U$$ of the water, so we can write

$$\Delta U = Q.$$

Substituting the known values, we get

$$\Delta U = m\,c\,\Delta T = (0.2\ \text{kg})(4184\ \text{J kg}^{-1}\ \text{K}^{-1})(20\ \text{K}).$$

Let us multiply step by step. First multiply the specific heat by the temperature change:

$$4184\ \text{J kg}^{-1}\ \text{K}^{-1} \times 20\ \text{K} = 83680\ \text{J kg}^{-1}.$$

Now multiply by the mass:

$$\Delta U = 0.2\ \text{kg} \times 83680\ \text{J kg}^{-1} = 16736\ \text{J}.$$

Finally, convert joules to kilojoules, since the answer choices are expressed that way. We know $$1\ \text{kJ}=1000\ \text{J}$$, so

$$\Delta U = \frac{16736\ \text{J}}{1000} = 16.736\ \text{kJ}.$$

Rounding to three significant figures, the change in internal energy is approximately $$16.7\ \text{kJ}$$.

Among the given options, this value matches Option D.

Hence, the correct answer is Option D.

Let the mass of water present in the container be $$m$$ grams. The heater supplies heat at a perfectly uniform rate, so the amount of heat given to the water is proportional to the time for which the heater operates.

First, the water is heated from $$0^\circ{\rm C}$$ to $$100^\circ{\rm C}$$. We use the basic formula for heat absorbed during a temperature change:

$$Q = mc\Delta T,$$

where $$Q$$ is the heat absorbed, $$m$$ is the mass, $$c$$ is the specific heat capacity and $$\Delta T$$ is the rise in temperature. Here, $$c = 1\;{\rm cal}\,( {\rm g}^\circ{\rm C} )^{-1}$$ and $$\Delta T = 100^\circ{\rm C} - 0^\circ{\rm C} = 100^\circ{\rm C}$$. Hence the heat needed for this stage is

$$Q_1 = m \times 1 \times 100 = 100m\;{\rm cal}.$$

This stage takes 10 minutes. Because the heater’s power is constant, the heat supplied per minute is the same throughout the whole experiment. Denote this unknown constant rate by $$P$$ (cal per minute). Then, for the first stage,

$$Q_1 = P \times (\text{time for stage 1}) = P \times 10.$$

Equating the two expressions for $$Q_1$$ gives us

$$P \times 10 = 100m \quad\rightarrow\quad P = \frac{100m}{10} = 10m\;{\rm cal\;min^{-1}}.$$

Next, the entire mass of water at $$100^\circ{\rm C}$$ is converted into steam at the same temperature. The heat required for a complete change of state is the latent heat:

$$Q_2 = mL,$$

where $$L$$ is the heat of vaporization per gram, which we wish to find. The time taken for this stage is 55 minutes, so by the same constant-power reasoning,

$$Q_2 = P \times (\text{time for stage 2}) = P \times 55.$$

Substituting the value of $$P$$ obtained earlier,

$$mL = (10m)\times 55.$$

Now we divide both sides by $$m$$ to eliminate the mass (since it is the same in both stages and non-zero):

$$L = 10 \times 55 = 550\;{\rm cal\,g^{-1}}.$$

Hence, the measured heat of vaporization from this experiment is $$550\;{\rm cal\,g^{-1}}$$.

Hence, the correct answer is Option B.

We are asked to find an expression for the rate of heat transfer per unit area, written as $$\dfrac{Q}{A}$$. Because this is a rate of energy flow through a unit surface, its physical dimension is that of power per unit area.

First let us write down the basic dimensions in the $$M\;L\;T$$ system:

Power $$P$$ has dimension $$[P]=M L^2 T^{-3}$$ (since $$P=\dfrac{\text{energy}}{\text{time}}$$ and energy is work $$F\! \cdot\! d = (M L T^{-2})L$$). Dividing by area $$A\;(=[L^2])$$ gives

$$\left[\dfrac{Q}{A}\right]=M L^0 T^{-3}=M T^{-3}.$$

Now we list the three quantities on which the student believes $$\dfrac{Q}{A}$$ depends and write their dimensions.

1. Dynamic viscosity $$\eta$$  Formula for its dimension: $$\eta=\dfrac{\text{shear stress}}{\text{velocity gradient}}.$$  Shear stress has dimension $$M L^{-1} T^{-2}$$, and velocity gradient has dimension $$T^{-1}$$,  so $$[\eta]=M L^{-1} T^{-1}.$$

2. The factor $$\dfrac{S\Delta\theta}{h}$$  Specific heat $$S$$ is defined by $$Q=m S \Delta\theta \;.$$  Re-arranging, $$S=\dfrac{Q}{m\,\Delta\theta}$$, hence  $$[S]=\dfrac{M L^2 T^{-2}}{M\,\Theta}=L^{2} T^{-2} \Theta^{-1},$$  where $$\Theta$$ stands for temperature.  Multiplying by the temperature difference $$\Delta\theta$$ removes the temperature dimension, so  $$\left[S\Delta\theta\right]=L^{2} T^{-2}.$$  Dividing by height $$h\;(=[L])$$ gives  $$\left[\dfrac{S\Delta\theta}{h}\right]=L T^{-2}.$$

3. The factor $$\dfrac{1}{\rho g}$$  Density $$\rho$$ has dimension $$M L^{-3}$$.  Acceleration due to gravity $$g$$ has dimension $$L T^{-2}$$.  Hence $$\left[\dfrac{1}{\rho g}\right]=\dfrac{1}{(M L^{-3})(L T^{-2})}=M^{-1} L^{2} T^{2}.$$

We now search for a product of the above three quantities, each possibly raised to some power, that reproduces $$M T^{-3}$$. Inspection of the options shows that only direct products have been proposed, so we can test them one by one.

Option A : $$\left(\dfrac{S\Delta\theta}{h}\right)\eta$$

Combining the dimensions we have

$$\left[L T^{-2}\right]\left[M L^{-1} T^{-1}\right] = M\,L^{(1-1)}\,T^{(-2-1)} = M\,T^{-3}.$$

This exactly matches the required $$M T^{-3}$$, so Option A is dimensionally consistent.

Option B : $$\eta\left(\dfrac{S\Delta\theta}{h}\right)\left(\dfrac{1}{\rho g}\right)$$

$$\left[M L^{-1} T^{-1}\right] \left[L T^{-2}\right] \left[M^{-1} L^{2} T^{2}\right] = M^{(1-1)}\,L^{(-1+1+2)}\,T^{(-1-2+2)} = L^{2}\,T^{-1}.$$

This gives $$L^{2} T^{-1}$$, which is not $$M T^{-3}$$, so Option B is incorrect.

Option C : $$\left(\dfrac{S\Delta\theta}{\eta h}\right)\left(\dfrac{1}{\rho g}\right)$$

First compute $$\dfrac{S\Delta\theta}{\eta h}$$:

$$\dfrac{[S\Delta\theta]}{[\eta][h]} = \dfrac{L^{2} T^{-2}}{(M L^{-1} T^{-1})(L)} = \dfrac{L^{2} T^{-2}}{M L^{0} T^{-1}} = M^{-1} L^{2} T^{-1}.$$

Multiplying this by $$\left[M^{-1} L^{2} T^{2}\right]$$ from $$\dfrac{1}{\rho g}$$ gives

$$M^{-2} L^{4} T^{1},$$

which again fails to match $$M T^{-3}$$, so Option C is incorrect.

Option D : $$\dfrac{S\Delta\theta}{\eta h}$$

We already found above that its dimension is $$M^{-1} L^{2} T^{-1}$$, clearly not $$M T^{-3}$$, hence Option D is also incorrect.

Among all the choices, only Option A reproduces the correct physical dimension for heat flux. Therefore the student’s dimensional reasoning leads to

$$\dfrac{Q}{A}\;=\;\eta\left(\dfrac{S\Delta\theta}{h}\right).$$

Hence, the correct answer is Option A.

The sphere cools by radiation since it is in a vacuum. The net power radiated by a black body is given by the Stefan-Boltzmann law. The net power radiated per unit area is $$\sigma (T^4 - T_0^4)$$, where $$T$$ is the temperature of the sphere and $$T_0$$ is the surrounding temperature. The surface area of the sphere is $$4\pi R^2$$, so the total net power radiated is:

This power loss equals the rate of decrease of the internal energy of the sphere. Therefore:

The internal energy $$U$$ depends on the temperature. Given the specific heat per unit mass is $$\alpha T^3$$, the heat capacity per unit mass is $$c = \alpha T^3$$. For a mass $$M$$, the change in internal energy for a temperature change $$dT$$ is:

Thus, the rate of change of internal energy with time is:

Substituting into the power equation:

Rearranging for $$dt$$:

To find the time taken for the sphere to cool from $$3T_0$$ to $$2T_0$$, integrate from $$T = 3T_0$$ to $$T = 2T_0$$. Since the temperature decreases, the negative sign is handled by reversing the limits:

Factor the denominator using $$T^4 - T_0^4 = (T^2)^2 - (T_0^2)^2 = (T^2 - T_0^2)(T^2 + T_0^2)$$. However, a substitution simplifies the integral. Let $$u = T^4 - T_0^4$$, then $$du = 4T^3 dT$$, so $$T^3 dT = \frac{du}{4}$$.

Change the limits: when $$T = 2T_0$$, $$u = (2T_0)^4 - T_0^4 = 16T_0^4 - T_0^4 = 15T_0^4$$. When $$T = 3T_0$$, $$u = (3T_0)^4 - T_0^4 = 81T_0^4 - T_0^4 = 80T_0^4$$.

Substitute:

The integral of $$\frac{du}{u}$$ is $$\ln|u|$$, and since $$u > 0$$, it is $$\ln u$$:

Simplify using logarithm properties:

Reduce $$\frac{80}{15}$$ by dividing numerator and denominator by 5:

Thus:

Comparing with the options, this matches Option B.

Hence, the correct answer is Option B.

Newton's law of cooling states that the rate of cooling of a body is proportional to the difference between its temperature and the ambient temperature. The ambient temperature is given as 30°C. The differential equation is:

$$\frac{dT}{dt} = -k(T - T_s)$$

where $$ T $$ is the temperature of the body at time $$ t $$, $$ T_s = 30^\circ \text{C} $$ is the ambient temperature, and $$ k $$ is a positive constant. Separating variables and integrating:

$$\int \frac{dT}{T - T_s} = -k \int dt$$

$$\ln|T - T_s| = -kt + C$$

where $$ C $$ is the integration constant. Since the body's temperature is above the ambient temperature, we can write:

$$T - T_s = e^{-kt + C} = e^C e^{-kt}$$

Let $$ A = e^C $$, so:

$$T - T_s = A e^{-kt}$$

$$T = T_s + A e^{-kt} = 30 + A e^{-kt}$$

At time $$ t = 0 $$, the temperature is 80°C:

$$80 = 30 + A e^{0} \Rightarrow 80 = 30 + A \Rightarrow A = 50$$

Thus, the temperature equation is:

$$T = 30 + 50 e^{-kt}$$

It takes 5 minutes for the temperature to drop from 80°C to 40°C. At $$ t = 5 $$ minutes, $$ T = 40^\circ \text{C} $$:

$$40 = 30 + 50 e^{-k \cdot 5}$$

$$10 = 50 e^{-5k}$$

$$\frac{10}{50} = e^{-5k}$$

$$0.2 = e^{-5k}$$

Taking natural logarithm on both sides:

$$\ln(0.2) = \ln(e^{-5k})$$

$$\ln(0.2) = -5k$$

Since $$ \ln(0.2) = \ln\left(\frac{1}{5}\right) = -\ln(5) $$ and given $$ \ln(5) = 1.609 $$:

$$-1.609 = -5k$$

$$k = \frac{1.609}{5} = 0.3218$$

Now, we need to find the time taken to cool from 62°C to 32°C. Let $$ t_1 $$ be the time when the temperature is 62°C and $$ t_2 $$ be the time when it is 32°C. The time required is $$ t_2 - t_1 $$. Using the temperature equation:

At $$ T = 62^\circ \text{C} $$:

$$62 = 30 + 50 e^{-k t_1}$$

$$32 = 50 e^{-k t_1}$$

$$e^{-k t_1} = \frac{32}{50} = 0.64$$

At $$ T = 32^\circ \text{C} $$:

$$32 = 30 + 50 e^{-k t_2}$$

$$2 = 50 e^{-k t_2}$$

$$e^{-k t_2} = \frac{2}{50} = 0.04$$

Dividing the two expressions:

$$\frac{e^{-k t_1}}{e^{-k t_2}} = \frac{0.64}{0.04} = 16$$

$$\frac{e^{-k t_1}}{e^{-k t_2}} = e^{-k(t_1 - t_2)} = e^{k(t_2 - t_1)}$$

So:

$$e^{k(t_2 - t_1)} = 16$$

Taking natural logarithm:

$$k(t_2 - t_1) = \ln(16)$$

$$\ln(16) = \ln(2^4) = 4 \ln 2 = 4 \times 0.693 = 2.772$$

Substituting $$ k = \frac{\ln(5)}{5} = \frac{1.609}{5} $$:

$$t_2 - t_1 = \frac{\ln(16)}{k} = \frac{2.772}{0.3218} \approx 8.614$$

Alternatively, using the exact expression:

$$t_2 - t_1 = \frac{\ln(16)}{k} = \frac{4 \ln 2}{\frac{\ln 5}{5}} = \frac{4 \ln 2 \times 5}{\ln 5} = \frac{20 \ln 2}{\ln 5}$$

Substituting the given values:

$$t_2 - t_1 = \frac{20 \times 0.693}{1.609} = \frac{13.86}{1.609} \approx 8.614$$

Rounding to one decimal place, the time is 8.6 minutes.

Hence, the correct answer is Option B.

According to Newton's law of cooling, the rate of cooling is proportional to the difference between the temperature of the body and the surroundings. The formula for the average rate of cooling over a time interval is given by:

$$\frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right)$$

where $$ T_1 $$ is the initial temperature, $$ T_2 $$ is the final temperature, $$ t $$ is the time interval, $$ T_s $$ is the temperature of the surroundings, and $$ k $$ is a constant.

For the first 10 minutes:

• Initial temperature $$ T_1 = 60^\circ \text{C} $$
• Final temperature $$ T_2 = 50^\circ \text{C} $$
• Time $$ t = 10 $$ minutes
• Rate of cooling $$ \frac{60 - 50}{10} = 1^\circ \text{C per minute} $$
• Average temperature $$ \frac{60 + 50}{2} = 55^\circ \text{C} $$

So, the equation becomes:

$$1 = k (55 - T_s) \quad \text{(Equation 1)}$$

For the next 10 minutes:

• Initial temperature $$ T_1 = 50^\circ \text{C} $$
• Final temperature $$ T_2 = 42^\circ \text{C} $$
• Time $$ t = 10 $$ minutes
• Rate of cooling $$ \frac{50 - 42}{10} = 0.8^\circ \text{C per minute} $$
• Average temperature $$ \frac{50 + 42}{2} = 46^\circ \text{C} $$

So, the equation becomes:

$$0.8 = k (46 - T_s) \quad \text{(Equation 2)}$$

We now have two equations:

$$1 = k (55 - T_s)$$

$$0.8 = k (46 - T_s)$$

To solve for $$ T_s $$, divide Equation 1 by Equation 2:

$$\frac{1}{0.8} = \frac{k (55 - T_s)}{k (46 - T_s)}$$

Simplify the left side:

$$\frac{1}{0.8} = 1.25$$

So:

$$1.25 = \frac{55 - T_s}{46 - T_s}$$

Cross-multiply:

$$1.25 \times (46 - T_s) = 55 - T_s$$

Calculate $$ 1.25 \times 46 $$:

$$1.25 \times 46 = 1.25 \times (40 + 6) = (1.25 \times 40) + (1.25 \times 6) = 50 + 7.5 = 57.5$$

So:

$$57.5 - 1.25 T_s = 55 - T_s$$

Bring all terms involving $$ T_s $$ to one side and constants to the other. Add $$ 1.25 T_s $$ to both sides:

$$57.5 = 55 - T_s + 1.25 T_s$$

$$57.5 = 55 + 0.25 T_s$$

Subtract 55 from both sides:

$$57.5 - 55 = 0.25 T_s$$

$$2.5 = 0.25 T_s$$

Solve for $$ T_s $$:

$$T_s = \frac{2.5}{0.25} = \frac{2.5 \times 100}{0.25 \times 100} = \frac{250}{25} = 10$$

So, the temperature of the surroundings is $$ 10^\circ \text{C} $$.

Verification: Substitute $$ T_s = 10 $$ into Equation 1:

$$1 = k (55 - 10) = k \times 45 \implies k = \frac{1}{45}$$

Now substitute into Equation 2:

$$0.8 = \frac{1}{45} \times (46 - 10) = \frac{1}{45} \times 36 = \frac{36}{45} = \frac{4}{5} = 0.8$$

This matches, confirming the solution.

Hence, the correct answer is Option B.

In steady-state heat conduction, the heat current through a uniform rod is given by the formula

$$Q \;=\; \dfrac{k\,A\,(\theta_1-\theta_2)}{L},$$

where $$k$$ is the thermal conductivity, $$A$$ the cross-sectional area, $$L$$ the length of the rod and $$(\theta_1-\theta_2)$$ the temperature difference between its ends.

All three rods have the same cross-section $$A = 4\ \text{cm}^2$$. The given data are

$$\begin{aligned} k_{\text{Cu}} &= 0.92, & L_{\text{Cu}} &= 46\ \text{cm}, & \theta_{\text{hot}} &= 100^\circ\text{C},\\[2mm] k_{\text{Br}} &= 0.26, & L_{\text{Br}} &= 13\ \text{cm}, & \theta_{\text{cold}} &= 0^\circ\text{C},\\[2mm] k_{\text{St}} &= 0.12, & L_{\text{St}} &= 12\ \text{cm}. & \end{aligned}$$

Let the common junction temperature be $$T^\circ\text{C}$$. Because the system is insulated elsewhere and has reached steady state,

$$\text{(heat entering through copper)}\;=\;\text{(heat leaving through brass)}\;+\;\text{(heat leaving through steel)}.$$

First we calculate the thermal conductance of each rod, $$G = \dfrac{kA}{L}$$.

For copper:

$$G_{\text{Cu}} \;=\; \dfrac{0.92 \times 4}{46} \;=\; \dfrac{3.68}{46} \;=\; 0.08\ \text{(cal/s)}/^\circ\text{C}.$$

For brass:

$$G_{\text{Br}} \;=\; \dfrac{0.26 \times 4}{13} \;=\; \dfrac{1.04}{13} \;=\; 0.08\ \text{(cal/s)}/^\circ\text{C}.$$

For steel:

$$G_{\text{St}} \;=\; \dfrac{0.12 \times 4}{12} \;=\; \dfrac{0.48}{12} \;=\; 0.04\ \text{(cal/s)}/^\circ\text{C}.$$

Now we write the heat-balance equation:

$$G_{\text{Cu}}\,(100 - T) \;=\; G_{\text{Br}}\,T \;+\; G_{\text{St}}\,T.$$

Substituting the conductances,

$$0.08\,(100 - T) \;=\; 0.08\,T \;+\; 0.04\,T.$$

Combining the terms on the right:

$$0.08\,(100 - T) \;=\; 0.12\,T.$$

Expanding the left side:

$$8 \;-\; 0.08\,T \;=\; 0.12\,T.$$

Bringing like terms together:

$$8 \;=\; 0.12\,T + 0.08\,T \;=\; 0.20\,T.$$

Hence

$$T \;=\; \dfrac{8}{0.20} \;=\; 40^\circ\text{C}.$$

We now find the required heat current through the copper rod:

$$Q_{\text{Cu}} \;=\; G_{\text{Cu}}\,(100 - T) \;=\; 0.08\,(100 - 40) \;=\; 0.08 \times 60 \;=\; 4.8\ \text{cal}\,\text{s}^{-1}.$$

Hence, the correct answer is Option C.

We are given that the volume of water is 2 liters. Since the density of water is 1 kg/L, the mass of water is 2 kg.

The heating coil provides energy at a rate of 1 kW. Since 1 kW equals 1000 W and 1 W is 1 J/s, the coil supplies 1000 J/s.

The container loses energy at a rate of 160 J/s. Therefore, the net energy absorbed by the water per second is the energy supplied by the coil minus the energy lost by the container:

$$ \text{Net energy per second} = 1000 \text{J/s} - 160 \text{J/s} = 840 \text{J/s} $$

The temperature needs to rise from 27°C to 77°C. The change in temperature is:

$$ \Delta T = 77^\circ \text{C} - 27^\circ \text{C} = 50^\circ \text{C} $$

The specific heat of water is given as 4.2 kJ/kg·°C. Converting to joules: 4.2 kJ/kg·°C = 4.2 × 1000 = 4200 J/kg·°C.

The total heat energy required to raise the temperature of the water is given by:

$$ Q = m \cdot c \cdot \Delta T $$

Substituting the values:

$$ Q = 2 \text{kg} \times 4200 \text{J/kg·°C} \times 50^\circ \text{C} $$

First, multiply the mass and specific heat:

$$ 2 \times 4200 = 8400 $$

Then multiply by the temperature change:

$$ 8400 \times 50 = 420000 \text{J} $$

So, the total heat required is 420,000 J.

The net energy supplied per second is 840 J/s. The time $$ t $$ in seconds required to supply 420,000 J is:

$$ t = \frac{\text{Total heat}}{\text{Net energy per second}} = \frac{420000}{840} $$

Simplifying the fraction:

$$ \frac{420000}{840} = \frac{420000 \div 120}{840 \div 120} = \frac{3500}{7} = 500 \text{seconds} $$

Alternatively, note that 840 × 500 = 420,000, so $$ t = 500 $$ seconds.

Converting 500 seconds to minutes and seconds: since 1 minute = 60 seconds,

$$ \text{Minutes} = \left\lfloor \frac{500}{60} \right\rfloor = \left\lfloor 8.333\ldots \right\rfloor = 8 \text{minutes} $$

The remaining seconds are:

$$ 500 - (8 \times 60) = 500 - 480 = 20 \text{seconds} $$

Thus, the time is 8 minutes and 20 seconds.

Comparing with the options, Option A is 8 min 20 s.

Hence, the correct answer is Option A.

The ratio of the coefficient of volume expansion of the glass container to that of the viscous liquid is given as 1 : 4. Let the coefficient of volume expansion of the glass container be $$\gamma_g$$ and that of the liquid be $$\gamma_l$$. Therefore, $$\gamma_g / \gamma_l = 1/4$$, which implies $$\gamma_l = 4\gamma_g$$.

Let the initial inner volume of the container at temperature $$T_0$$ be $$V_0$$. We need to find the fraction of $$V_0$$ that the liquid should occupy at $$T_0$$ so that the volume of the vacant space remains constant at all temperatures. Let this fraction be $$k$$, so the initial volume of the liquid is $$k V_0$$ and the initial vacant space is $$(1 - k) V_0$$.

When the temperature changes by $$\Delta T$$, the new volume of the container is $$V_0 (1 + \gamma_g \Delta T)$$. The new volume of the liquid is $$k V_0 (1 + \gamma_l \Delta T) = k V_0 (1 + 4\gamma_g \Delta T)$$, since $$\gamma_l = 4\gamma_g$$. The vacant space at the new temperature is the difference between the container volume and the liquid volume:

$$\text{Vacant space} = V_0 (1 + \gamma_g \Delta T) - k V_0 (1 + 4\gamma_g \Delta T)$$

This vacant space must remain constant and equal to the initial vacant space $$(1 - k) V_0$$ at all temperatures. Therefore, we set up the equation:

$$V_0 (1 + \gamma_g \Delta T) - k V_0 (1 + 4\gamma_g \Delta T) = (1 - k) V_0$$

Dividing both sides by $$V_0$$ (assuming $$V_0 \neq 0$$):

$$1 + \gamma_g \Delta T - k (1 + 4\gamma_g \Delta T) = 1 - k$$

Expanding the left side:

$$1 + \gamma_g \Delta T - k - 4k\gamma_g \Delta T = 1 - k$$

Subtracting 1 from both sides:

$$\gamma_g \Delta T - k - 4k\gamma_g \Delta T = -k$$

Adding $$k$$ to both sides:

$$\gamma_g \Delta T - 4k\gamma_g \Delta T = 0$$

Factoring out $$\gamma_g \Delta T$$:

$$\gamma_g \Delta T (1 - 4k) = 0$$

Since $$\gamma_g \neq 0$$ and $$\Delta T$$ can be arbitrary (not always zero), the term $$(1 - 4k)$$ must be zero:

$$1 - 4k = 0$$

Solving for $$k$$:

$$4k = 1$$

$$k = \frac{1}{4}$$

Thus, the liquid should occupy $$\frac{1}{4}$$ of the inner volume of the container at the initial temperature. This fraction corresponds to the ratio 1 : 4.

Verifying the solution: with $$k = \frac{1}{4}$$, the initial liquid volume is $$\frac{V_0}{4}$$ and the initial vacant space is $$\frac{3V_0}{4}$$. At a new temperature, the container volume is $$V_0 (1 + \gamma_g \Delta T)$$, and the liquid volume is $$\frac{V_0}{4} (1 + 4\gamma_g \Delta T)$$. The vacant space is:

$$V_0 (1 + \gamma_g \Delta T) - \frac{V_0}{4} (1 + 4\gamma_g \Delta T) = V_0 + V_0\gamma_g \Delta T - \frac{V_0}{4} - V_0\gamma_g \Delta T = V_0 - \frac{V_0}{4} = \frac{3V_0}{4}$$

This is constant and equal to the initial vacant space, confirming the solution.

Hence, the correct answer is Option B.

We are given two cooling experiments: one with water and one with another liquid. Both are in identical vessels with identical surroundings and cool from 30°C to 25°C in the same time of 2 minutes. The water equivalent of the vessel is 30 g, meaning the vessel behaves as if it were 30 g of water in terms of heat capacity.

First, recall Newton's law of cooling: the rate of cooling is proportional to the temperature difference between the body and its surroundings. For both systems, the initial and final temperatures are the same, and the cooling time is identical. This implies that the cooling curves are identical, leading to the same decay constant in the exponential cooling equation.

The decay constant $$b$$ is given by $$b = \frac{k}{C}$$, where $$k$$ is a constant depending on the surface area and conditions (same for both vessels), and $$C$$ is the total heat capacity of the system (including the vessel). Since the cooling curves are identical, $$b$$ must be the same for both systems. Therefore, $$\frac{k}{C_{\text{water}}} = \frac{k}{C_{\text{liquid}}}$$, which implies $$C_{\text{water}} = C_{\text{liquid}}$$.

Now, calculate the total heat capacity for the water system. The specific heat of water is $$1 \, \text{cal/g°C}$$. The mass of water is 50 g, and the water equivalent of the vessel is 30 g. The heat capacity of the water is $$50 \times 1 = 50 \, \text{cal/°C}$$, and the heat capacity of the vessel is $$30 \times 1 = 30 \, \text{cal/°C}$$. Thus, the total heat capacity for the water system is:

$$C_{\text{water}} = 50 + 30 = 80 \, \text{cal/°C}$$

For the liquid system, let the specific heat of the liquid be $$c_l \, \text{cal/g°C}$$. The mass of the liquid is 100 g, so the heat capacity of the liquid is $$100 \times c_l \, \text{cal/°C}$$. The vessel has the same water equivalent of 30 g, contributing $$30 \times 1 = 30 \, \text{cal/°C}$$. Thus, the total heat capacity for the liquid system is:

$$C_{\text{liquid}} = 100c_l + 30 \, \text{cal/°C}$$

Since $$C_{\text{water}} = C_{\text{liquid}}$$, we set them equal:

$$80 = 100c_l + 30$$

Solve for $$c_l$$:

Subtract 30 from both sides:

$$80 - 30 = 100c_l$$

$$50 = 100c_l$$

Divide both sides by 100:

$$c_l = \frac{50}{100} = 0.5 \, \text{cal/g°C}$$

The options are in kcal/kg. Note that $$1 \, \text{cal/g°C} = 1 \, \text{kcal/kg°C}$$ because:

• 1 kcal = 1000 cal
• 1 kg = 1000 g

Thus, $$0.5 \, \text{cal/g°C} = 0.5 \, \text{kcal/kg°C}$$.

Comparing with the options:

A. 2.0 kcal/kg

B. 7 kcal/kg

C. 3 kcal/kg

D. 0.5 kcal/kg

Hence, the correct answer is Option D.

$$\frac{dT}{dt} = -k(T - \theta_0)$$

$$\int \frac{dT}{T - \theta_0} = -k \int dt$$

$$\ln(T - \theta_0) = -kt + C$$

At $$t = 0$$, the initial temperature $$T = \theta$$. Thus, the constant $$C = \ln(\theta - \theta_0)$$.

The temperature $$T$$ at any time $$t$$ is given by:

 $$T - \theta_0 = (\theta - \theta_0)e^{-kt}$$

$$T = \theta_0 + (\theta - \theta_0)e^{-kt}$$

Graph (A) correctly shows this.

We are given two temperature scales: Kelvin (K) and a linear scale Y. The freezing and boiling points of water are known on both scales:

• On Kelvin scale: freezing point = 273 K, boiling point = 373 K
• On Y scale: freezing point = -160°Y, boiling point = -50°Y

We need to find the temperature on the Y scale that corresponds to 340 K. Since both scales are linear, the relationship between them can be expressed as:

$$T_Y = m \cdot T_K + c$$

where $$T_Y$$ is the temperature in °Y, $$T_K$$ is the temperature in K, $$m$$ is the slope, and $$c$$ is the intercept.

To find $$m$$ and $$c$$, we use the two known points:

• Point 1: (273 K, -160°Y)
• Point 2: (373 K, -50°Y)

The slope $$m$$ is calculated as the change in Y divided by the change in K:

$$m = \frac{\Delta T_Y}{\Delta T_K} = \frac{(-50) - (-160)}{373 - 273} = \frac{-50 + 160}{100} = \frac{110}{100} = 1.1$$

So, $$m = 1.1$$.

Now, substitute one point into the equation to find $$c$$. Using point 1 (273 K, -160°Y):

$$-160 = 1.1 \cdot 273 + c$$

First, compute $$1.1 \cdot 273$$:

$$1.1 \cdot 273 = 1.1 \cdot (200 + 73) = 1.1 \cdot 200 + 1.1 \cdot 73 = 220 + 80.3 = 300.3$$

So,

$$-160 = 300.3 + c$$

Solving for $$c$$:

$$c = -160 - 300.3 = -460.3$$

Thus, the equation is:

$$T_Y = 1.1 \cdot T_K - 460.3$$

Now, find $$T_Y$$ when $$T_K = 340$$ K:

$$T_Y = 1.1 \cdot 340 - 460.3$$

Compute $$1.1 \cdot 340$$:

$$1.1 \cdot 340 = 1.1 \cdot (300 + 40) = 1.1 \cdot 300 + 1.1 \cdot 40 = 330 + 44 = 374$$

Then,

$$T_Y = 374 - 460.3 = -86.3$$

Therefore, 340 K corresponds to -86.3°Y on the Y scale.

Comparing with the options:

• A: -73.7°Y
• B: -233.7°Y
• C: -86.3°Y
• D: -106.3°Y

Hence, the correct answer is Option C.