From Exercise 4.5-10 of Thermodynamics and an Introduction to Thermostatistics by Callen:

Two identical bodies each have heat capacities (at constant volume) of $$C(T)=a / T$$

The initial temperatures are $ T_{10} $ and $ T_{20} $, with $ T_{20}>T_{10} $. The two bodies are to be brought to thermal equilibrium with each other (maintaining both volumes constant) while delivering as much work as possible to a reversible work source. What is the final equilibrium temperature and what is the maximum work delivered to the reversible work source?

I have been suggested using $$\bigg(\frac{\partial S}{\partial T}\bigg)_V = \bigg(\frac{\partial U}{\partial T}\bigg)_V \bigg(\frac{\partial S}{\partial U} \bigg)_V = \frac{C(T)}{T},$$ calculating $$\Delta S = \int_{T_i}^{T_f} \frac{\partial S}{\partial T} dT = \int_{T_i}^{T_f} \frac{C(T)}{T} dT,$$ and setting $\Delta S = 0$ as by the theorem of maximal work.

But I can't get comfortable with the fact that the system is supposed to perform work while simultaneously maintaining constant volume. In other words, if $P dV = 0$, how can $\Delta U$ be nonzero if the system is thermally isolated?

  • 1
    $\begingroup$ The two bodies are supposed to be like heat reservoirs that are supplying heat to a reversible engine (whose volume can change). It is the engine that does the work. $\endgroup$ Nov 16, 2019 at 15:19
  • $\begingroup$ Why are we then applying the maximum work theorem to the heat reservoir rather then the entire system? $\endgroup$
    – coolcat29
    Nov 16, 2019 at 18:49
  • $\begingroup$ Who said anything about a maximum work theorem? The total change in entropy of the two bodies is going to be zero. This will determine the heat delivered from the hot body to the engine and the heat removed from the engine by the cold body. The difference will be the maximum work. $\endgroup$ Nov 16, 2019 at 21:33
  • $\begingroup$ The change in entropy per cycle of the engine working fluid is, of course, zero. $\endgroup$ Nov 16, 2019 at 21:49

1 Answer 1


If we apply the first and second laws of thermodynamics to the combined system consisting of the two bodies and the engine, we obtain: $$\Delta U=\Delta U_H+\Delta U_C+\Delta U_E=-W_E\tag{1}$$and$$\Delta S=\Delta S_H+\Delta S_C+\Delta S_E=0\tag{2}$$ where the subscripts H, C, and E refer to the hot body, the cold body, and the engine working fluid, respectively. Since the engine is assumed to be working in a cycle, we must have that $\Delta U_E=0$ and $\Delta S_E=0$, so Eqns. 1 and 2 reduce to: $$\Delta U_H+\Delta U_C=-W_E\tag{3}$$and$$\Delta S_H+\Delta S_C=0\tag{4}$$If $T_f$ is the final temperature of the hot and cold bodies when no more work can be done, for the specified heat capacity equation, we have that:$$\Delta U_H=\int_{T_{20}}^{T_f}{CdT}=a\ln{(T_f/T_{20})}\tag{5a}$$$$\Delta U_C=\int_{T_{10}}^{T_f}{CdT}=a\ln{(T_f/T_{10})}\tag{5b}$$$$\Delta S_H=\int_{T_{20}}^{T_f}{\frac{C}{T}dT}=a\left(\frac{1}{T_{20}}-\frac{1}{T_f}\right)\tag{6a}$$and$$\Delta S_C=\int_{T_{10}}^{T_f}{\frac{C}{T}dT}=a\left(\frac{1}{T_{10}}-\frac{1}{T_f}\right)\tag{6b}$$Eqns. 6 together with Eqn. 4 determines the final temperature of the bodies. This can then be substituted into Eqns. 5 and 3 to determine the maximum work done by the engine on the surroundings.


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