Chapter 10: Thermal Properties of Matter

Consider a rectangular sheet of the solid of length $$a$$ and breadth $$b$$. Its initial area is

$$A = ab.$$

When the temperature rises by $$\Delta T$$, each linear dimension changes according to the linear-expansion law. The new length and breadth are

$$a' = a(1 + \alpha_l \Delta T), \qquad b' = b(1 + \alpha_l \Delta T).$$

The new area is

$$A' = a'b' = ab(1 + \alpha_l \Delta T)^2 = ab\left(1 + 2\alpha_l \Delta T + \alpha_l^2 (\Delta T)^2\right).$$

For ordinary solids $$\alpha_l$$ is of the order of $$10^{-5} \, \mathrm{K^{-1}}$$, so the term $$\alpha_l^2 (\Delta T)^2$$ is negligible compared with $$2\alpha_l \Delta T$$. Therefore

$$A' \approx ab\left(1 + 2\alpha_l \Delta T\right),$$

$$\Delta A = A' - A = ab \cdot 2\alpha_l \Delta T = 2\alpha_l A \, \Delta T.$$

Hence the coefficient of area expansion is

$$\frac{\Delta A / A}{\Delta T} = 2\alpha_l.$$

Thus the area expansivity is twice the linear expansivity, as required.

When the iron ring is heated its diameter increases. The ring will just fit on the rim when its diameter, on heating, becomes equal to the diameter of the rim.

Let $$d_1 = 5.231 \, \mathrm{m}$$ be the diameter of the ring at $$T_1 = 27 \, \mathrm{^\circ C}$$, and let $$d_2 = 5.243 \, \mathrm{m}$$ be the required diameter (equal to the rim).

Applying the linear-expansion law to the diameter,

$$d_2 = d_1\left(1 + \alpha_l (T_2 - T_1)\right),$$

so the change in diameter is

$$\Delta d = d_2 - d_1 = d_1\,\alpha_l\,(T_2 - T_1).$$

For iron, $$\alpha_l = 1.20 \times 10^{-5} \, \mathrm{K^{-1}}$$. Substituting,

$$5.243 - 5.231 = 5.231 \times (1.20 \times 10^{-5}) \times (T_2 - 27).$$

$$0.012 = (6.277 \times 10^{-5})\,(T_2 - 27).$$

$$T_2 - 27 = \frac{0.012}{6.277 \times 10^{-5}} \approx 191.2 \, \mathrm{^\circ C}.$$

Therefore

$$T_2 = 27 + 191.2 \approx 218 \, \mathrm{^\circ C}.$$

The ring should be heated to about $$218 \, \mathrm{^\circ C}$$ so that it can be slipped on to the rim.

Heat lost by the hot aluminium sphere is gained by the water and the copper calorimeter (principle of calorimetry; the system is isolated).

Data: mass of aluminium $$m_{\mathrm{Al}} = 0.047 \, \mathrm{kg}$$, falling from $$100 \, \mathrm{^\circ C}$$ to $$23 \, \mathrm{^\circ C}$$; mass of water $$m_w = 0.25 \, \mathrm{kg}$$ and mass of copper calorimeter $$m_{\mathrm{Cu}} = 0.14 \, \mathrm{kg}$$, both rising from $$20 \, \mathrm{^\circ C}$$ to $$23 \, \mathrm{^\circ C}$$.

Take the specific heat capacities $$s_w = 4.18 \times 10^3 \, \mathrm{J\,kg^{-1}\,K^{-1}}$$ and $$s_{\mathrm{Cu}} = 0.386 \times 10^3 \, \mathrm{J\,kg^{-1}\,K^{-1}}$$.

Heat lost by the sphere:

$$Q_{\mathrm{lost}} = m_{\mathrm{Al}}\,s_{\mathrm{Al}}\,(100 - 23) = 0.047 \times s_{\mathrm{Al}} \times 77.$$

Heat gained by water and calorimeter:

$$Q_{\mathrm{gained}} = m_w s_w (23 - 20) + m_{\mathrm{Cu}} s_{\mathrm{Cu}} (23 - 20).$$

$$Q_{\mathrm{gained}} = (0.25)(4180)(3) + (0.14)(386)(3) = 3135 + 162.1 = 3297.1 \, \mathrm{J}.$$

Equating $$Q_{\mathrm{lost}} = Q_{\mathrm{gained}}$$:

$$0.047 \times s_{\mathrm{Al}} \times 77 = 3297.1.$$

$$s_{\mathrm{Al}} = \frac{3297.1}{0.047 \times 77} = \frac{3297.1}{3.619} \approx 911 \, \mathrm{J\,kg^{-1}\,K^{-1}}.$$

Hence the specific heat capacity of aluminium is about $$0.911 \, \mathrm{kJ\,kg^{-1}\,K^{-1}}$$.

By the principle of calorimetry, the heat released by the warm water as it cools is absorbed by the ice — first to melt it, then to warm the resulting melt-water up to the final temperature.

Data: $$m_{\mathrm{ice}} = 0.15 \, \mathrm{kg}$$ at $$0 \, \mathrm{^\circ C}$$; $$m_w = 0.30 \, \mathrm{kg}$$ at $$50 \, \mathrm{^\circ C}$$; final temperature $$\theta = 6.7 \, \mathrm{^\circ C}$$; $$s_w = 4186 \, \mathrm{J\,kg^{-1}\,K^{-1}}$$.

Heat lost by the warm water (cooling from $$50 \, \mathrm{^\circ C}$$ to $$6.7 \, \mathrm{^\circ C}$$):

$$Q_{\mathrm{lost}} = m_w s_w (50 - 6.7) = 0.30 \times 4186 \times 43.3 \approx 54376 \, \mathrm{J}.$$

Heat gained by the ice = heat to melt it + heat to warm the melt-water from $$0 \, \mathrm{^\circ C}$$ to $$6.7 \, \mathrm{^\circ C}$$:

$$Q_{\mathrm{gained}} = m_{\mathrm{ice}}\,L_f + m_{\mathrm{ice}}\,s_w\,(6.7 - 0).$$

The warming term is

$$m_{\mathrm{ice}}\,s_w\,(6.7) = 0.15 \times 4186 \times 6.7 \approx 4207 \, \mathrm{J}.$$

Equating $$Q_{\mathrm{lost}} = Q_{\mathrm{gained}}$$:

$$54376 = 0.15\,L_f + 4207.$$

$$0.15\,L_f = 54376 - 4207 = 50169 \, \mathrm{J}.$$

$$L_f = \frac{50169}{0.15} \approx 3.34 \times 10^5 \, \mathrm{J\,kg^{-1}}.$$

The latent heat of fusion of ice is about $$3.34 \times 10^5 \, \mathrm{J\,kg^{-1}}$$.

The conversion of ice at $$-12 \, \mathrm{^\circ C}$$ into steam at $$100 \, \mathrm{^\circ C}$$ takes place in four stages. The total heat is the sum of the heats for each stage. Here $$m = 3 \, \mathrm{kg}$$.

Stage 1 — warm the ice from $$-12 \, \mathrm{^\circ C}$$ to $$0 \, \mathrm{^\circ C}$$:

$$Q_1 = m\,s_{\mathrm{ice}}\,\Delta T = 3 \times 2100 \times 12 = 7.56 \times 10^4 \, \mathrm{J}.$$

Stage 2 — melt the ice at $$0 \, \mathrm{^\circ C}$$:

$$Q_2 = m\,L_f = 3 \times 3.35 \times 10^5 = 1.005 \times 10^6 \, \mathrm{J}.$$

Stage 3 — warm the water from $$0 \, \mathrm{^\circ C}$$ to $$100 \, \mathrm{^\circ C}$$:

$$Q_3 = m\,s_{\mathrm{water}}\,\Delta T = 3 \times 4186 \times 100 = 1.2558 \times 10^6 \, \mathrm{J}.$$

Stage 4 — convert the water at $$100 \, \mathrm{^\circ C}$$ into steam:

$$Q_4 = m\,L_v = 3 \times 2.256 \times 10^6 = 6.768 \times 10^6 \, \mathrm{J}.$$

Total heat required:

$$Q = Q_1 + Q_2 + Q_3 + Q_4 = (0.0756 + 1.005 + 1.2558 + 6.768) \times 10^6 \, \mathrm{J}.$$

$$Q \approx 9.1 \times 10^6 \, \mathrm{J}.$$

About $$9.1 \times 10^6 \, \mathrm{J}$$ of heat is required.

In the steady state the heat current $$H$$ (heat flowing per second) is the same all along the composite rod — no heat accumulates at the junction.

Let $$T$$ be the steady-state temperature of the steel–copper junction. Let the cross-sectional area of the copper rod be $$A$$; then the steel rod has area $$2A$$.

Data: $$L_{\mathrm{steel}} = 0.15 \, \mathrm{m}$$, $$L_{\mathrm{Cu}} = 0.10 \, \mathrm{m}$$, $$K_{\mathrm{steel}} = 50.2 \, \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}$$, $$K_{\mathrm{Cu}} = 385 \, \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}$$.

Heat current through the steel rod (furnace end at $$300 \, \mathrm{^\circ C}$$):

$$H_{\mathrm{steel}} = \frac{K_{\mathrm{steel}}\,(2A)\,(300 - T)}{L_{\mathrm{steel}}}.$$

Heat current through the copper rod (far end at $$0 \, \mathrm{^\circ C}$$):

$$H_{\mathrm{Cu}} = \frac{K_{\mathrm{Cu}}\,A\,(T - 0)}{L_{\mathrm{Cu}}}.$$

In the steady state $$H_{\mathrm{steel}} = H_{\mathrm{Cu}}$$:

$$\frac{50.2 \times 2A \times (300 - T)}{0.15} = \frac{385 \times A \times T}{0.10}.$$

The area $$A$$ cancels. Simplifying each side:

$$669.3\,(300 - T) = 3850\,T.$$

$$200800 - 669.3\,T = 3850\,T.$$

$$200800 = 4519.3\,T.$$

$$T = \frac{200800}{4519.3} \approx 44.4 \, \mathrm{^\circ C}.$$

The steel–copper junction is at about $$44.4 \, \mathrm{^\circ C}$$ in the steady state.

The two bars are joined end to end and carry the same heat current $$H$$ in the steady state. They have equal lengths $$L_1 = L_2 = L = 0.1 \, \mathrm{m}$$ and equal areas $$A_1 = A_2 = A = 0.02 \, \mathrm{m^2}$$. The iron end is at $$T_1 = 373 \, \mathrm{K}$$ and the brass end at $$T_2 = 273 \, \mathrm{K}$$.

(i) Temperature of the junction. Let $$T_0$$ be the junction temperature. Equal heat currents through the two bars:

$$\frac{K_1 A (T_1 - T_0)}{L} = \frac{K_2 A (T_0 - T_2)}{L}.$$

Since $$A$$ and $$L$$ are equal on both sides, they cancel:

$$K_1 (T_1 - T_0) = K_2 (T_0 - T_2).$$

Solving for $$T_0$$:

$$T_0 = \frac{K_1 T_1 + K_2 T_2}{K_1 + K_2}.$$

$$T_0 = \frac{(79)(373) + (109)(273)}{79 + 109} = \frac{29467 + 29757}{188} = \frac{59224}{188} \approx 315.0 \, \mathrm{K}.$$

(ii) Equivalent thermal conductivity. Replace the compound bar by a single bar of length $$2L$$, area $$A$$ and conductivity $$K$$ carrying the same current. Writing the temperature drops across the two bars and adding them,

$$T_1 - T_0 = \frac{HL}{K_1 A}, \qquad T_0 - T_2 = \frac{HL}{K_2 A},$$

$$T_1 - T_2 = \frac{HL}{A}\left(\frac{1}{K_1} + \frac{1}{K_2}\right).$$

For the equivalent bar, $$T_1 - T_2 = \dfrac{H(2L)}{K A}$$. Equating the two expressions and cancelling $$HL/A$$:

$$\frac{2}{K} = \frac{1}{K_1} + \frac{1}{K_2} \quad\Rightarrow\quad K = \frac{2 K_1 K_2}{K_1 + K_2}.$$

$$K = \frac{2 \times 79 \times 109}{79 + 109} = \frac{17222}{188} \approx 91.6 \, \mathrm{W\,m^{-1}\,K^{-1}}.$$

(iii) Heat current. Using the equivalent bar of length $$2L = 0.2 \, \mathrm{m}$$:

$$H = \frac{K A (T_1 - T_2)}{2L} = \frac{91.6 \times 0.02 \times (373 - 273)}{0.2}.$$

$$H = \frac{91.6 \times 0.02 \times 100}{0.2} \approx 916.1 \, \mathrm{W}.$$

(The same value follows from $$H = K_1 A (T_1 - T_0)/L$$ using the iron bar.)

By Newton's law of cooling, while a body cools through a small temperature range the rate of fall of temperature is proportional to the excess of the body's (average) temperature over the surroundings:

$$\frac{\Delta T}{\Delta t} = K\left(\langle T\rangle - T_s\right),$$

where $$\langle T\rangle$$ is the average temperature during the interval and $$T_s = 20 \, \mathrm{^\circ C}$$ is the room temperature.

First cooling (94 °C to 86 °C in 2 min):

Average temperature $$\langle T\rangle = \dfrac{94 + 86}{2} = 90 \, \mathrm{^\circ C}$$, so the excess is $$90 - 20 = 70 \, \mathrm{^\circ C}$$.

$$\frac{94 - 86}{2} = K(70) \quad\Rightarrow\quad 4 = 70\,K \quad\Rightarrow\quad K = \frac{4}{70} \, \mathrm{min^{-1}}.$$

Second cooling (71 °C to 69 °C in time $$t$$):

Average temperature $$\langle T\rangle = \dfrac{71 + 69}{2} = 70 \, \mathrm{^\circ C}$$, so the excess is $$70 - 20 = 50 \, \mathrm{^\circ C}$$.

$$\frac{71 - 69}{t} = K(50) = \frac{4}{70} \times 50.$$

$$\frac{2}{t} = \frac{200}{70} = 2.857 \, \mathrm{min^{-1}}.$$

$$t = \frac{2}{2.857} \approx 0.7 \, \mathrm{min} \approx 42 \, \mathrm{s}.$$

It takes about $$0.7$$ minute (roughly $$42$$ seconds) to cool from $$71 \, \mathrm{^\circ C}$$ to $$69 \, \mathrm{^\circ C}$$.

The Kelvin (absolute), Celsius and Fahrenheit scales are related by

$$t_C = T - 273.15, \qquad t_F = \frac{9}{5}\,t_C + 32,$$

where $$T$$ is in kelvin, $$t_C$$ in $$\mathrm{^\circ C}$$ and $$t_F$$ in $$\mathrm{^\circ F}$$.

Neon (triple point $$T = 24.57 \, \mathrm{K}$$):

$$t_C = 24.57 - 273.15 = -248.58 \, \mathrm{^\circ C}.$$

$$t_F = \frac{9}{5}(-248.58) + 32 = -447.44 + 32 = -415.44 \, \mathrm{^\circ F}.$$

Carbon dioxide (triple point $$T = 216.55 \, \mathrm{K}$$):

$$t_C = 216.55 - 273.15 = -56.60 \, \mathrm{^\circ C}.$$

$$t_F = \frac{9}{5}(-56.60) + 32 = -101.88 + 32 = -69.88 \, \mathrm{^\circ F}.$$

On an absolute (thermodynamic) scale, the temperature assigned to a body is proportional to the chosen physical quantity, and the scale is fixed by assigning a definite number to the triple point of water.

If $$T_A$$ is the reading of a body on scale $$A$$ and $$T_B$$ its reading on scale $$B$$, then on each scale the ratio of a temperature to the triple-point value represents the same physical fraction. Hence

$$\frac{T_A}{(\text{triple point on } A)} = \frac{T_B}{(\text{triple point on } B)}.$$

The triple point of water is $$200 \, \mathrm{A}$$ on scale $$A$$ and $$350 \, \mathrm{B}$$ on scale $$B$$, so

$$\frac{T_A}{200} = \frac{T_B}{350}.$$

This gives the required relation:

$$T_A = \frac{200}{350}\,T_B = \frac{4}{7}\,T_B \qquad \text{or} \qquad T_B = \frac{7}{4}\,T_A.$$

So any given temperature reads $$\tfrac{4}{7}$$ as many units on scale $$A$$ as it does on scale $$B$$.

The resistance varies linearly with temperature: $$R = R_0\left[1 + \alpha (T - T_0)\right]$$. Take the triple point of water as the reference, $$T_0 = 273.16 \, \mathrm{K}$$ with $$R_0 = 101.6 \, \Omega$$.

Find $$\alpha$$ using the lead point. At the melting point of lead, $$R = 165.5 \, \Omega$$ and $$T = 600.5 \, \mathrm{K}$$:

$$165.5 = 101.6\left[1 + \alpha (600.5 - 273.16)\right].$$

$$\frac{165.5}{101.6} = 1 + \alpha (327.34) \quad\Rightarrow\quad 1.6289 = 1 + 327.34\,\alpha.$$

$$\alpha = \frac{0.6289}{327.34} = 1.921 \times 10^{-3} \, \mathrm{K^{-1}}.$$

Find $$T$$ when $$R = 123.4 \, \Omega$$.

$$123.4 = 101.6\left[1 + \alpha (T - 273.16)\right].$$

$$\frac{123.4}{101.6} = 1 + \alpha (T - 273.16) \quad\Rightarrow\quad 1.2146 = 1 + \alpha (T - 273.16).$$

$$\alpha (T - 273.16) = 0.2146.$$

$$T - 273.16 = \frac{0.2146}{1.921 \times 10^{-3}} \approx 111.7 \, \mathrm{K}.$$

$$T \approx 273.16 + 111.7 \approx 384.8 \, \mathrm{K}.$$

The temperature at which the resistance is $$123.4 \, \Omega$$ is about $$384.8 \, \mathrm{K}$$.

Actual length on the hot day (45 °C). The measuring tape is itself made of steel and was calibrated (i.e. its centimetre-marks are correct) at $$27.0 \, \mathrm{^\circ C}$$. On the hot day the tape has expanded, so every centimetre-mark on it now spans a slightly larger physical distance:

$$\text{(true length of one tape division)} = 1 \, \mathrm{cm} \times (1 + \alpha\,\Delta T).$$

Here $$\Delta T = 45.0 - 27.0 = 18.0 \, \mathrm{^\circ C}$$ and $$\alpha = 1.20 \times 10^{-5} \, \mathrm{K^{-1}}$$, so

$$1 + \alpha\,\Delta T = 1 + (1.20 \times 10^{-5})(18.0) = 1 + 2.16 \times 10^{-4}.$$

The tape reads $$63.0 \, \mathrm{cm}$$, so the actual length of the rod at $$45.0 \, \mathrm{^\circ C}$$ is

$$L_{45} = 63.0 \times (1 + 2.16 \times 10^{-4}) = 63.0 + 63.0 \times 2.16 \times 10^{-4}.$$

$$L_{45} = 63.0 + 0.0136 \approx 63.0136 \, \mathrm{cm}.$$

Length of the rod at 27 °C. The rod is also made of steel. On cooling from $$45.0 \, \mathrm{^\circ C}$$ back to $$27.0 \, \mathrm{^\circ C}$$ it contracts:

$$L_{27} = \frac{L_{45}}{1 + \alpha\,\Delta T} = \frac{63.0136}{1 + 2.16 \times 10^{-4}} \approx 63.0136 - 0.0136 \approx 63.0 \, \mathrm{cm}.$$

This result is expected: at $$27.0 \, \mathrm{^\circ C}$$ the tape is correctly calibrated, and since the rod and tape are made of the same material, the tape simply reads the rod's true length, $$63.0 \, \mathrm{cm}$$.

The wheel will slip on to the shaft only when the shaft is cooled enough for its outer diameter to shrink to the diameter of the central hole in the wheel.

Let $$d_1 = 8.70 \, \mathrm{cm}$$ be the shaft diameter at $$T_1 = 27 \, \mathrm{^\circ C}$$, and let it shrink to $$d_2 = 8.69 \, \mathrm{cm}$$ at the required temperature $$T_2$$.

The change in diameter due to linear contraction is

$$d_2 - d_1 = d_1\,\alpha\,(T_2 - T_1).$$

Substituting $$\alpha = 1.20 \times 10^{-5} \, \mathrm{K^{-1}}$$:

$$8.69 - 8.70 = 8.70 \times (1.20 \times 10^{-5}) \times (T_2 - 27).$$

$$-0.01 = (1.044 \times 10^{-4})\,(T_2 - 27).$$

$$T_2 - 27 = \frac{-0.01}{1.044 \times 10^{-4}} \approx -95.8 \, \mathrm{^\circ C}.$$

$$T_2 = 27 - 95.8 \approx -68.8 \, \mathrm{^\circ C}.$$

So the shaft must be cooled to about $$-69 \, \mathrm{^\circ C}$$ (roughly $$204 \, \mathrm{K}$$) for the wheel to slip on. Dry ice (solid $$\mathrm{CO_2}$$, at about $$-78 \, \mathrm{^\circ C}$$) is cold enough to achieve this.

When a metal sheet with a hole in it is heated, the hole expands exactly as if it were a disc of the same metal filling the hole. So the diameter of the hole increases according to the linear-expansion law, using the linear expansivity of the metal.

Data: initial diameter $$d = 4.24 \, \mathrm{cm}$$, temperature rise $$\Delta T = 227 - 27.0 = 200 \, \mathrm{^\circ C}$$, and $$\alpha = 1.70 \times 10^{-5} \, \mathrm{K^{-1}}$$.

The change in diameter is

$$\Delta d = d\,\alpha\,\Delta T = 4.24 \times (1.70 \times 10^{-5}) \times 200.$$

$$\Delta d = 4.24 \times 3.40 \times 10^{-3} \approx 1.44 \times 10^{-2} \, \mathrm{cm}.$$

The diameter of the hole increases by about $$1.44 \times 10^{-2} \, \mathrm{cm}$$ (i.e. about $$0.0144 \, \mathrm{cm}$$).

If the wire could contract freely on cooling, its length would decrease by $$\Delta L = L\,\alpha\,|\Delta T|$$. But the rigid supports prevent this contraction, so the wire is effectively held stretched back to its original length. This produces a tensile strain, and hence a tension, in the wire.

Temperature change: $$\Delta T = (-39) - 27 = -66 \, \mathrm{^\circ C}$$, so the wire is prevented from contracting by an amount corresponding to $$|\Delta T| = 66 \, \mathrm{^\circ C}$$.

The prevented (tensile) strain is

$$\text{strain} = \frac{\Delta L}{L} = \alpha\,|\Delta T| = (2.0 \times 10^{-5})(66) = 1.32 \times 10^{-3}.$$

From the definition of Young's modulus, $$Y = \dfrac{\text{stress}}{\text{strain}}$$, the tensile stress is

$$\text{stress} = Y \times \text{strain} = (0.91 \times 10^{11})(1.32 \times 10^{-3}) \approx 1.20 \times 10^{8} \, \mathrm{Pa}.$$

The cross-sectional area of the wire (diameter $$2.0 \, \mathrm{mm}$$, so radius $$r = 1.0 \times 10^{-3} \, \mathrm{m}$$) is

$$A = \pi r^2 = \pi (1.0 \times 10^{-3})^2 \approx 3.14 \times 10^{-6} \, \mathrm{m^2}.$$

The tension developed is

$$F = \text{stress} \times A = (1.20 \times 10^{8})(3.14 \times 10^{-6}) \approx 3.8 \times 10^{2} \, \mathrm{N}.$$

Equivalently, $$F = Y A \alpha\,|\Delta T| \approx 377 \, \mathrm{N}$$. So a tension of about $$3.8 \times 10^{2} \, \mathrm{N}$$ develops in the wire.

Since the ends of the combined rod are free to expand, the brass part and the steel part each expand independently, and the total change in length is simply the sum of the two.

Both rods have original length $$L = 50 \, \mathrm{cm} = 0.50 \, \mathrm{m}$$, and the temperature rise is $$\Delta T = 250 - 40.0 = 210 \, \mathrm{^\circ C}$$.

Expansion of the brass rod ($$\alpha_b = 2.0 \times 10^{-5} \, \mathrm{K^{-1}}$$):

$$\Delta L_b = L\,\alpha_b\,\Delta T = 0.50 \times (2.0 \times 10^{-5}) \times 210 = 2.1 \times 10^{-3} \, \mathrm{m}.$$

Expansion of the steel rod ($$\alpha_s = 1.2 \times 10^{-5} \, \mathrm{K^{-1}}$$):

$$\Delta L_s = L\,\alpha_s\,\Delta T = 0.50 \times (1.2 \times 10^{-5}) \times 210 = 1.26 \times 10^{-3} \, \mathrm{m}.$$

Total change in length:

$$\Delta L = \Delta L_b + \Delta L_s = 2.1 \times 10^{-3} + 1.26 \times 10^{-3} = 3.36 \times 10^{-3} \, \mathrm{m}.$$

$$\Delta L \approx 0.34 \, \mathrm{cm}.$$

Thermal stress: Because the ends of the rod are free to expand, neither part is prevented from expanding — there is no constraint on the change of length. Hence no thermal stress is developed at the junction. (Note that the diameter of the rods does not enter the calculation at all.)

The density of a fixed mass $$m$$ of glycerine is $$\rho = m/V$$. When the temperature rises by $$\Delta T$$, the volume increases by $$\Delta V = \gamma V \Delta T$$, where $$\gamma$$ is the coefficient of volume expansion. The mass is unchanged, so the density decreases.

The new density is

$$\rho' = \frac{m}{V + \Delta V} = \frac{m}{V(1 + \gamma \Delta T)} = \frac{\rho}{1 + \gamma \Delta T}.$$

The fractional change in density is

$$\frac{\Delta \rho}{\rho} = \frac{\rho' - \rho}{\rho} = \frac{1}{1 + \gamma \Delta T} - 1 = \frac{-\gamma \Delta T}{1 + \gamma \Delta T}.$$

Since $$\gamma \Delta T \ll 1$$, to a very good approximation

$$\frac{\Delta \rho}{\rho} \approx -\gamma \Delta T.$$

Substituting $$\gamma = 49 \times 10^{-5} \, \mathrm{K^{-1}}$$ and $$\Delta T = 30 \, \mathrm{^\circ C}$$:

$$\frac{\Delta \rho}{\rho} = -(49 \times 10^{-5}) \times 30 = -1.47 \times 10^{-2}.$$

The density decreases by a fraction of about $$1.5 \times 10^{-2}$$, i.e. by roughly $$1.5\%$$.

The drilling machine has a power of $$P = 10 \, \mathrm{kW} = 10^4 \, \mathrm{W}$$ and runs for $$t = 2.5 \, \mathrm{min} = 150 \, \mathrm{s}$$.

Total energy supplied by the machine:

$$E = P\,t = 10^4 \times 150 = 1.5 \times 10^6 \, \mathrm{J}.$$

Only $$50\%$$ of this energy actually heats the aluminium block (the other half heats the machine or is lost to the surroundings), so the heat absorbed by the block is

$$Q = 0.50 \times E = 0.50 \times 1.5 \times 10^6 = 7.5 \times 10^5 \, \mathrm{J}.$$

The rise in temperature follows from $$Q = m\,s\,\Delta T$$, with $$m = 8.0 \, \mathrm{kg}$$ and $$s = 0.91 \, \mathrm{J\,g^{-1}\,K^{-1}} = 910 \, \mathrm{J\,kg^{-1}\,K^{-1}}$$:

$$\Delta T = \frac{Q}{m\,s} = \frac{7.5 \times 10^5}{8.0 \times 910} = \frac{7.5 \times 10^5}{7280}.$$

$$\Delta T \approx 103 \, \mathrm{^\circ C}.$$

The temperature of the aluminium block rises by about $$103 \, \mathrm{^\circ C}$$.

The hot copper block is placed on a large block of ice at $$0 \, \mathrm{^\circ C}$$. The maximum amount of ice melts when the copper gives up all the heat it possibly can — that is, when it cools all the way down to $$0 \, \mathrm{^\circ C}$$.

Heat released by the copper block in cooling from $$500 \, \mathrm{^\circ C}$$ to $$0 \, \mathrm{^\circ C}$$:

$$Q = m\,s_{\mathrm{Cu}}\,\Delta T,$$

with $$m = 2.5 \, \mathrm{kg} = 2500 \, \mathrm{g}$$, $$s_{\mathrm{Cu}} = 0.39 \, \mathrm{J\,g^{-1}\,K^{-1}}$$ and $$\Delta T = 500 \, \mathrm{^\circ C}$$:

$$Q = 2500 \times 0.39 \times 500 = 4.875 \times 10^5 \, \mathrm{J}.$$

This heat is used to melt the ice. If $$m_{\mathrm{ice}}$$ is the mass melted and $$L_f = 335 \, \mathrm{J\,g^{-1}}$$ is the heat of fusion,

$$Q = m_{\mathrm{ice}}\,L_f \quad\Rightarrow\quad m_{\mathrm{ice}} = \frac{Q}{L_f} = \frac{4.875 \times 10^5}{335}.$$

$$m_{\mathrm{ice}} \approx 1455 \, \mathrm{g} \approx 1.45 \, \mathrm{kg}.$$

So at most about $$1.45 \, \mathrm{kg}$$ of ice can melt.

By the principle of calorimetry, the heat lost by the hot metal block equals the heat gained by the water together with the copper calorimeter.

Data: mass of metal $$m = 0.20 \, \mathrm{kg}$$, cooling from $$150 \, \mathrm{^\circ C}$$ to $$40 \, \mathrm{^\circ C}$$; volume of water $$150 \, \mathrm{cm^3}$$, so mass of water $$m_w = 0.150 \, \mathrm{kg}$$; water equivalent of the calorimeter $$w = 0.025 \, \mathrm{kg}$$; the water and calorimeter both warm from $$27 \, \mathrm{^\circ C}$$ to $$40 \, \mathrm{^\circ C}$$. Take $$s_w = 4186 \, \mathrm{J\,kg^{-1}\,K^{-1}}$$.

The 'water equivalent' means the calorimeter absorbs heat just like $$0.025 \, \mathrm{kg}$$ of water, so the water and calorimeter together behave as an effective mass

$$m_w + w = 0.150 + 0.025 = 0.175 \, \mathrm{kg}$$

of water.

Heat gained by water and calorimeter:

$$Q_{\mathrm{gained}} = (m_w + w)\,s_w\,(40 - 27) = 0.175 \times 4186 \times 13 \approx 9523 \, \mathrm{J}.$$

Heat lost by the metal block:

$$Q_{\mathrm{lost}} = m\,s\,(150 - 40) = 0.20 \times s \times 110 = 22\,s.$$

Setting $$Q_{\mathrm{lost}} = Q_{\mathrm{gained}}$$:

$$22\,s = 9523 \quad\Rightarrow\quad s = \frac{9523}{22} \approx 433 \, \mathrm{J\,kg^{-1}\,K^{-1}}.$$

So the specific heat of the metal is about $$0.43 \, \mathrm{J\,g^{-1}\,K^{-1}}$$ ($$\approx 433 \, \mathrm{J\,kg^{-1}\,K^{-1}}$$).

Effect of heat losses. If heat losses to the surroundings are not negligible, then part of the heat given out by the metal escapes to the surroundings instead of going into the water. The water and calorimeter therefore receive less heat and reach a lower final temperature than they otherwise would. Using this lower final temperature in the calculation makes $$Q_{\mathrm{gained}}$$ — and hence the computed $$s$$ — come out smaller. Therefore the calculated value is smaller (an underestimate) than the actual specific heat of the metal.

According to the law of equipartition of energy, each active degree of freedom of a molecule contributes $$\tfrac{1}{2}R$$ to the molar specific heat at constant volume.

Monatomic gases: a single atom has only $$3$$ translational degrees of freedom, so

$$C_v = 3 \times \tfrac{1}{2}R = \tfrac{3}{2}R \approx 2.92 \, \mathrm{cal\,mol^{-1}\,K^{-1}}.$$

Diatomic gases: hydrogen, nitrogen, oxygen, nitric oxide and carbon monoxide are all diatomic. In addition to $$3$$ translational degrees of freedom, a diatomic molecule has $$2$$ rotational degrees of freedom (rotation about the two axes perpendicular to the line joining the atoms). This gives $$5$$ active degrees of freedom, so

$$C_v = 5 \times \tfrac{1}{2}R = \tfrac{5}{2}R \approx 4.95 \, \mathrm{cal\,mol^{-1}\,K^{-1}}.$$

This is why the measured values for these five gases (about $$4.9$$–$$5.0 \, \mathrm{cal\,mol^{-1}\,K^{-1}}$$) are close to $$5$$ and markedly larger than the monatomic value $$2.92$$.

Chlorine: its molar specific heat ($$6.17 \, \mathrm{cal\,mol^{-1}\,K^{-1}}$$) is noticeably larger than $$\tfrac{5}{2}R$$. This indicates that, besides translation and rotation, the chlorine molecule also has appreciably excited vibrational degrees of freedom at room temperature. (The chlorine atoms are heavy, so the molecule vibrates at a relatively low frequency; this vibrational mode is therefore partly active even at ordinary temperatures and adds an extra contribution to $$C_v$$.)

The fall in the child's body temperature is $$101 \, \mathrm{^\circ F} - 98 \, \mathrm{^\circ F} = 3 \, \mathrm{^\circ F}$$. We must convert this temperature interval to the Celsius scale. Since a change of $$180 \, \mathrm{^\circ F}$$ corresponds to a change of $$100 \, \mathrm{^\circ C}$$,

$$\Delta T = 3 \times \frac{100}{180} = 3 \times \frac{5}{9} = \frac{5}{3} \approx 1.67 \, \mathrm{^\circ C}.$$

The heat lost by the child's body (taking the specific heat equal to that of water, $$s = 1000 \, \mathrm{cal\,kg^{-1}\,{}^\circ C^{-1}}$$, and mass $$m = 30 \, \mathrm{kg}$$) is

$$Q = m\,s\,\Delta T = 30 \times 1000 \times \frac{5}{3} = 5.0 \times 10^{4} \, \mathrm{cal}.$$

All of this heat is removed by the evaporation of sweat. If $$m_e$$ is the mass of sweat evaporated and $$L = 580 \, \mathrm{cal\,g^{-1}}$$ is the latent heat of evaporation,

$$Q = m_e\,L \quad\Rightarrow\quad m_e = \frac{Q}{L} = \frac{5.0 \times 10^{4}}{580} \approx 86.2 \, \mathrm{g}.$$

This evaporation takes place over $$t = 20 \, \mathrm{min}$$, so the average rate of extra evaporation is

$$\text{rate} = \frac{m_e}{t} = \frac{86.2 \, \mathrm{g}}{20 \, \mathrm{min}} \approx 4.3 \, \mathrm{g\,min^{-1}}.$$

Heat leaks into the icebox by conduction through all six faces of the cube. Each face has area $$(0.30 \, \mathrm{m})^2 = 0.09 \, \mathrm{m^2}$$, so the total surface area is

$$A = 6 \times 0.09 = 0.54 \, \mathrm{m^2}.$$

The wall thickness is $$L = 5.0 \, \mathrm{cm} = 0.05 \, \mathrm{m}$$, and the temperature difference between the outside ($$45 \, \mathrm{^\circ C}$$) and the ice inside ($$0 \, \mathrm{^\circ C}$$) is $$\Delta T = 45 \, \mathrm{^\circ C}$$.

Over a time $$t = 6 \, \mathrm{h} = 6 \times 3600 = 21600 \, \mathrm{s}$$, the heat conducted into the box is

$$Q = \frac{K\,A\,\Delta T\,t}{L}.$$

With $$K = 0.01 \, \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}$$:

$$Q = \frac{(0.01)(0.54)(45)(21600)}{0.05}.$$

$$Q = \frac{5248.8}{0.05} \approx 1.05 \times 10^{5} \, \mathrm{J}.$$

This heat melts some of the ice. With heat of fusion $$L_f = 335 \times 10^3 \, \mathrm{J\,kg^{-1}}$$, the mass melted is

$$m_{\mathrm{melted}} = \frac{Q}{L_f} = \frac{1.05 \times 10^{5}}{335 \times 10^3} \approx 0.313 \, \mathrm{kg}.$$

The ice remaining after $$6 \, \mathrm{h}$$ is therefore

$$m_{\mathrm{left}} = 4.0 - 0.313 \approx 3.7 \, \mathrm{kg}.$$

In the steady state, the heat conducted through the base of the boiler per second equals the heat needed per second to convert the boiling water into steam.

Heat current required. Water boils away at the rate $$\dot{m} = 6.0 \, \mathrm{kg/min} = \dfrac{6.0}{60} = 0.10 \, \mathrm{kg/s}$$. With heat of vaporisation $$L_v = 2256 \times 10^3 \, \mathrm{J\,kg^{-1}}$$, the heat needed per second is

$$H = \dot{m}\,L_v = 0.10 \times 2256 \times 10^3 = 2.256 \times 10^{5} \, \mathrm{J\,s^{-1}}.$$

Conduction through the base. Let $$T$$ be the temperature of the flame side of the base; its inner side is in contact with boiling water at $$100 \, \mathrm{^\circ C}$$. With base area $$A = 0.15 \, \mathrm{m^2}$$, thickness $$L = 1.0 \, \mathrm{cm} = 0.01 \, \mathrm{m}$$ and $$K = 109 \, \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}$$,

$$H = \frac{K\,A\,(T - 100)}{L}.$$

Equating the two expressions for $$H$$:

$$2.256 \times 10^{5} = \frac{109 \times 0.15 \times (T - 100)}{0.01}.$$

$$2.256 \times 10^{5} = 1635\,(T - 100).$$

$$T - 100 = \frac{2.256 \times 10^{5}}{1635} \approx 138 \, \mathrm{^\circ C}.$$

$$T \approx 100 + 138 = 238 \, \mathrm{^\circ C}.$$

The part of the flame in contact with the boiler is at about $$238 \, \mathrm{^\circ C}$$.

Newton's law of cooling in differential form is

$$\dfrac{dT}{dt} = -k\,(T - T_{s}),$$

where $$T_{s} = 20^{\circ}\mathrm{C}$$ is the temperature of the surroundings. Integrating from initial temperature $$T_{i}$$ to final $$T_{f}$$ over time $$t$$:

$$\ln\dfrac{T_{i} - T_{s}}{T_{f} - T_{s}} = k\,t.$$

First cooling: $$80^{\circ}\mathrm{C} \to 50^{\circ}\mathrm{C}$$ in $$5$$ min.

$$\ln\dfrac{80 - 20}{50 - 20} = 5k \;\Rightarrow\; \ln\dfrac{60}{30} = 5k \;\Rightarrow\; \ln 2 = 5k \;\Rightarrow\; k = \dfrac{\ln 2}{5}\,\mathrm{min^{-1}}.$$

Second cooling: $$60^{\circ}\mathrm{C} \to 30^{\circ}\mathrm{C}$$ in time $$t$$.

$$\ln\dfrac{60 - 20}{30 - 20} = k\,t \;\Rightarrow\; \ln\dfrac{40}{10} = k\,t \;\Rightarrow\; \ln 4 = k\,t \;\Rightarrow\; 2\ln 2 = k\,t.$$

Substitute $$k = (\ln 2)/5$$:

$$t = \dfrac{2 \ln 2}{(\ln 2)/5} = 2 \times 5 = 10\,\mathrm{min}.$$

(The simple averaging shortcut $$\Delta T / \Delta t = k\,(\langle T\rangle - T_{s})$$ gives an approximate $$9\,\mathrm{min}$$, but it is only accurate for small temperature differences; for this 30 K drop the exponential form above is needed.)

The body takes $$10$$ minutes to cool from $$60^{\circ}\mathrm{C}$$ to $$30^{\circ}\mathrm{C}$$.