Discussion
One block at temperature \(T_{1}\) and another identical block at temperature \(T_{2}\) are placed in thermal contact and come to an equilibrium, the system being isolated from surroundings. If the heat capacities at constant volume C depend on temperature as C = aT, the entropy change of the system would be :
Answer: A
The entropy change is \(a\Big[2\Big(\frac{T_{1}^{2}+T_{2}^{2}}{2}\Big)^{1/2}-T_{1}-T_{2}\Big]\).
Energy is conserved, so the heat lost by one block equals the heat gained by the other. With \(C = aT\):
\[ \begin{aligned} \int_{T_{1}}^{T_{f}} aT\,dT + \int_{T_{2}}^{T_{f}} aT\,dT &= 0 \\ T_{f} &= \left(\frac{T_{1}^{2} + T_{2}^{2}}{2}\right)^{1/2} \end{aligned} \]
Then \(dS = C\,dT/T = a\,dT\), so each block changes entropy by \(a(T_{f} - T_{i})\):
\[ \Delta S = a(2T_{f} - T_{1} - T_{2}) \]
The ln forms belong to a constant heat capacity, where \(T_{f}\) is the arithmetic mean; they do not apply when C grows with T.
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