We have a body with total heat capacity $C$ and initial temperature $T_1$. It works as a heat source.

It is in an infinite surrounding of temperature $T_0$, which works as a cooler.

Finally, we have a heat engine with half of Carnot efficiency: $\eta = \frac{1 - \frac{T_0}{T_1}}{2}$.

We have to calculate the change of entropy when the body reaches the temperature of surroundings. So, we use the body to "power" the engine.

To begin, I don't know what kind of a cycle it is? The efficiency is half of Carnot engine, but that doesn't prove it is Carnot cycle. Even then, how to approach such a problem? I can imagine that $Q_{in} = \frac{W_{out}}{\eta}$, but what then?

EDIT + $\mathbb{EDIT2}$:

$\begin{gather} Q_H = C(T_1 - T_0) \\ \eta = 1-\frac{Q_C}{Q_H} =\frac{1-\frac{T0}{T1}}{2} \\ Q_C = (1-\eta)Q_H = \dots \end{gather}$

Then, $\ dS_{body} = \frac{dQ}{T}$, so$\ \ \Delta S_{body} = \int_{T_1}^{T_0}\frac{CdT}{T}=C\log \frac{T_0}{T_1}$.

Then, $\Delta S_{surroundings} = \frac{Q_C}{T}$

Finally, I add these two results together and I think this is the answer. Thanks to Chet Miller.

  • $\begingroup$ It is not a Carnot cycle because a Carnot cycle operates between two fixed thermal reservoirs. In your example, the heat provided to the engine does not come from a high temperature thermal reservoir (constant temperature source), but from a body whose temperature starts at $T_1$ and ends at $T_0$. $\endgroup$
    – Bob D
    Nov 17, 2019 at 22:48
  • $\begingroup$ Okay, so I know how to calculate the entropy change if the body and the surroundings are in direct contact, but how to plug in this "engine" between them? $\endgroup$
    – lkky7
    Nov 17, 2019 at 23:23

1 Answer 1


You know the amount of heat removed from the body $Q_H$, so, together with the efficiency of the engine, you can get the amount of heat rejected from the engine to the surroundings $Q_C$: $$Q_H=C(T_1-T_0)$$ and $$\eta = \frac{(Q_H-Q_C)}{Q_H}=\frac{1-\frac{T_0}{T_1}}{2}$$Knowing the heat rejected to the surroundings, you can calculate the entropy change of the surroundings. And, knowing the initial and final temperatures of the body, you know the entropy change of the body. So the entropy generated is equal to the sum of the entropy change of the body and of the surroundings.

  • $\begingroup$ Thanks, I've posted a part of the solution in my question, using your notation. Could you look at it? $\endgroup$
    – lkky7
    Nov 18, 2019 at 0:06
  • $\begingroup$ I don't think you did the algebra correctly to get $Q_C$. I get $$Q_C=\frac{C}{2}T_1\left[1-\left(\frac{T_0}{T_1}\right)^2\right]$$The surroundings are treated as an ideal infinite reservoir, $Q_C/T_0$ is correct. $\endgroup$ Nov 18, 2019 at 0:25

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