Question

Two identical systems, with heat capacity at constant volume that varies as $C_v = bT^3$ (where $b$ is a constant) are thermally isolated. Initially, one system is at a temperature 100 K and the other is at 200 K. The systems are then brought into thermal contact and the combined system is allowed to reach thermal equilibrium. The final temperature (in K) of the combined system will be

• A. 171
• B. 141
• C. 150
• D. 180
• E. 125

Hint:

When heat capacity depends on temperature, the final equilibrium temperature is not a simple arithmetic mean. It shifts toward the higher initial temperature due to the increased heat capacity at higher temperatures.

Answer 1

The Correct Option is A

Concept:
Since the systems are isolated, the total heat lost by the hotter system must equal the heat gained by the colder system. Because $C_v$ is temperature-dependent, we must integrate. \[ \int_{100}^{T_f} C_v dT = \int_{T_f}^{200} C_v dT \]

Step 1: Set up the integral equation.
\[ \int_{100}^{T_f} bT^3 dT = \int_{T_f}^{200} bT^3 dT \] \[ \left[ \frac{bT^4}{4} \right]_{100}^{T_f} = \left[ \frac{bT^4}{4} \right]_{T_f}^{200} \] \[ T_f^4 - 100^4 = 200^4 - T_f^4 \]

Step 2: Solve for $T_f$.
\[ 2T_f^4 = 200^4 + 100^4 \] \[ 2T_f^4 = 16 \times 10^8 + 1 \times 10^8 = 17 \times 10^8 \] \[ T_f^4 = 8.5 \times 10^8 \implies T_f = \sqrt[4]{8.5} \times 100 \] Since $\sqrt[4]{8.5} \approx 1.707$, $T_f \approx 171$ K.