Discussion
A solid with heat capacity \(C_{A}\) at temperature \(T_{A}\) is placed in contact with the heat bath having P heat capacity \(C_{B}\) at temperature \(T_{B}\) such that \(C_{B} \gg C_{A}\). The change in entropy of the combined system after a solid reached a thermal equilibrium with a heat bath is \(\Delta S\) = .....
Answer: B
The entropy change is \(C_{A}\left[\ln\frac{T_{B}}{T_{A}} + \frac{T_{A}}{T_{B}} - 1\right]\).
Because \(C_{B} \gg C_{A}\), the bath stays at \(T_{B}\) and the solid ends at \(T_{B}\). The solid absorbs heat \(Q = C_{A}(T_{B} - T_{A})\), which the bath gives up at constant temperature:
\[ \begin{aligned} \Delta S_{A} &= \int_{T_{A}}^{T_{B}}\frac{C_{A}dT}{T} \\ &= C_{A}\ln\frac{T_{B}}{T_{A}} \\ \Delta S_{B} &= -\frac{C_{A}(T_{B} - T_{A})}{T_{B}} \\ &= C_{A}\left(\frac{T_{A}}{T_{B}} - 1\right) \end{aligned} \]
Adding them gives the keyed result, which is never negative (since \(\ln x \ge 1 - 1/x\)), as the second law requires.
The options with \(C_{B}\ln\) terms treat the bath as if its temperature changed. The option with \(T_{B}/T_{A}\) inverts the bath's ratio and does not follow from \(Q/T_{B}\).
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