Two identical beakers A and B contain equal volumes of two different liquids at 60°C each and left to cool down. Liquid in A has density of $$8 \times 10^{2}$$ kg m$$^{-3}$$ and specific heat of 2000 J kg$$^{-1}$$K$$^{-1}$$ while the liquid in B has density $$10^{3}$$ kg m$$^{-3}$$ and specific heat of 4000 J kg$$^{-1}$$K$$^{-1}$$. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

Problem Analysis

The question compares the rate of cooling of two spheres ($$A$$ and $$B$$) using Stefan-Boltzmann's Law / Newton's Law of Cooling.

• Rate of cooling ($$-\frac{dT}{dt}$$): Represents how fast a body cools down.
• Mass ($$m$$): Expressed as $$\text{Density } (\rho) \times \text{Volume } (V)$$. For identical solid spheres, the surface area ($$A$$) and volume ($$V$$) are the same.
• Specific heat capacity ($$s$$ or $$S$$): Quantity of heat required to change temperature.
• For body A: $$\rho_A = 800$$, $$s_A = 2000$$
• For body B: $$\rho_B = 10^3 = 1000$$, $$s_B = 4000$$
• For Sphere A:
• For Sphere B:

Given parameters from the problem:

Step-by-Step Solution

Step 1: Formulate the Proportionality

The rate of loss of temperature (cooling rate) is given by:

$$-\frac{dT}{dt} = \frac{e\sigma A}{ms}(T^4 - T_0^4) \approx \frac{e\sigma A}{ms} \cdot 4T_0^3(T - T_0)$$

Since the spheres are identical in size, their surface area ($$A$$) and volume ($$V$$) are constant. Substituting mass $$m = \rho \cdot V$$:

$$-\frac{dT}{dt} = \frac{e\sigma A}{(\rho V)s} \cdot 4T_0^3(T - T_0)$$

Thus, the rate of cooling is inversely proportional to the product of density ($$\rho$$) and specific heat ($$s$$):

$$-\frac{dT}{dt} \propto \frac{1}{\rho \cdot s}$$

Step 2: Calculate the Product ($$\rho \cdot s$$) for Both Spheres

$$(\rho s)_A = 800 \times 2000 = 16 \times 10^5$$

$$(\rho s)_B = 10^3 \times 4000 = 40 \times 10^5$$

Step 3: Compare the Rates of Cooling

Comparing the values calculated in Step 2:

$$(\rho s)_B > (\rho s)_A$$

Since the rate of cooling is inversely proportional to this product:

$$\left(-\frac{dT}{dt}\right)_B < \left(-\frac{dT}{dt}\right)_A$$

Conclusion

Sphere A cools down faster than Sphere B because it has a smaller product of density and specific heat capacity.

option (B)