A heat engine operates between a body with finite heat capacity 𝑐 at initial temperature $𝑇_1$ and a reservoir at fixed temperature $T_2$. Show that the maximum amount of work that can be done is given by $𝑊$ where: $W=c\left|T_2-T_1\right|-cT_2\ln\left(T_2/T_1\right)$.

The way this was shown used results of a Carnot engine. From what I understand the Carnot cycle is reversible, because it operates between 2 reservoirs. Here, the body with finite heat capacity will eventually end up at a temperature of the reservoir, $T_2$.

Once this happens, how can you go back the other way? Even if you input work into the engine, you can not force heat to flow between two bodies already at the same temperature, can you? Why can we assume the results derived from considering a Carnot cycle and apply them to this question?

  • $\begingroup$ To be clear, which temperature is the higher? $\endgroup$
    – Bob D
    Dec 18, 2019 at 12:28
  • $\begingroup$ Also, the Carnot cycle operates between two FIXED temperature reservoirs. $\endgroup$
    – Bob D
    Dec 18, 2019 at 12:30
  • $\begingroup$ Assume $T_1$ is larger $\endgroup$ Dec 18, 2019 at 12:35
  • $\begingroup$ @BobD So am I correct in saying we can not use the fact the efficiency of a cycle = $T_2/T_1$? If so I can check with my professor and ask why they used that result. $\endgroup$ Dec 18, 2019 at 12:36
  • 1
    $\begingroup$ Who says you can't force the heat to flow back between the two bodies already at the same temperature using an engine operating in a cycle? $\endgroup$ Dec 18, 2019 at 12:55

1 Answer 1


The change in entropy of the body is $c\ln{(T_2/T_1)}$ and the change in entropy of the reservoir is $Q_R/T_2$, where $Q_R$ is the heat transferred from the engine to the reservoir; the change in entropy of the engine is zero, since it is assumed to be operating in a cycle. So the change in entropy of the combination of body, engine working fluid, and reservoir is $$\Delta S=c\ln{(T_2/T_1)}+Q_R/T_2$$If the process is carried out reversibly to give maximum work, $\Delta S=0$. That results in the heat transferred to the reservoir as: $$Q_R=cT_2\ln{(T_1/T_2)}$$The heat transferred from the body to the engine working fluid is $Q_H=c(T_1-T_2)$. So the reversible work done by the system is $$W=Q_H-Q_R=c(T_1-T_2)-cT_2\ln{(T_1/T_2)}$$This assumes that $T_1>T_2$. If $T_2>T_1$, the correct answer is the answer given in the book.


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