Confused with Entropy and Clausius inequality

I am confused with a question in my thermodynamics class and would like to seek some clarification.

The problem I was given:

A reversible heat engine operates on a Carnot Cycle between a heat source (at initial temperature, $$T_A$$) and a heat sink (at initial temperature, $$T_B$$).

Unlike the infinite thermal reservoirs, the heat source and the heat sink in this case contain ideal gases with the same finite masses. After a certain period of time, the temperatures of both the thermal reservoirs will be equalized to $$T_2$$. For the surroundings, assume that there is no heat transfer and no temperature change. The pressures of both the thermal reservoirs remain constant.

Assuming that all processes are ideal, $$Prove:\quad T_2=\sqrt{T_AT_B}$$

My Approach:

Using Clausius equality of a reversible cycle, $$\frac{dQ}{T} = constant$$ $$\frac{Q_1}{T_1} = -\frac{Q_2}{T_2}$$ $$\frac{mC_p(T_2-T_A)}{T_2} = -\frac{mC_p(T_2-T_B)}{T_2}$$ $$\frac{T_2-T_A}{T_2} = \frac{-T_2+T_B}{T_2}$$ $$T_2 = \frac{T_A+T_B}{2}$$

The solution I was given uses the summation of entropy in an isolated system.

$$S_A = -S_B$$ $$ln\frac{T_2}{T_A} = -ln\frac{T_2}{T_B}$$ $$\frac{T_2}{T_A} = \frac{T_B}{T_2}$$ $$T_2=\sqrt{T_AT_B}$$

My question:

Isn't $$S_A$$ the same as $$\frac{Q_1}{T_1}$$ if the process is reversible? And if so, why does my value differ? Or is it because $$S_A$$ is from the perspective of reservoir A while $$\frac{Q_1}{T_1}$$ is from the perspective of the heat engine. Thus, applying Clausius inequality does not consider the respective reservoirs heat transfers.

P.S. I am quite lost in this subject, would appreciate a bit of help in highlighting my wrong assumptions/ understanding.

• why do you think that $dS=\delta Q/T$ is a "constant" in a reversible cycle? It is true that $dS_A+dS_B=0$ but both $T$ and $\delta Q$ are variable and so can be $dS$. Commented Mar 15, 2020 at 16:34
• What is $T_1$ in your approach? Commented Mar 15, 2020 at 17:04
• I think you are imagining that this process is taking place in a single Carnot cycle, with all the heat being transferred (to each reservoir) in this one cycle. This is not how the reversible process would be carried out for the change involved here. There would be an infinite number of Carnot cycles, with tiny amounts of heat being transferred in each of these. So in each cycle, the hot reservoir would cool by $dT_H$ and the cold reservoir would heat by $dT_C$. Commented Mar 15, 2020 at 20:11

$$\frac{mC_p(T_2-T_A)}{T_2} = -\frac{mC_p(T_2-T_B)}{T_2}$$

This is incorrect. The mistake you are making in your approach is assuming all the heat transfer is occurring at the final temperature $$T_2$$ of the heat source and sink.

Heat transfer $$Q_1$$ is not occurring at constant temperature. It is occurring over a range of temperatures from $$T_A$$ to $$T_2$$. Likewise heat transfer $$Q_2$$ occurs over a range of temperatures from $$T_B$$ to $$T_2$$.

For a reversible process we have

$$\Delta S_{A}=-\Delta S_{B}$$

and for an ideal gas, constant pressure process we have

$$\Delta S_{A}=c_{p}$$ ln $$\frac{T_2}{T_A}$$

and

$$\Delta S_{B}=c_{p}$$ ln $$\frac{T_2}{T_B}$$

$$ln\frac{T_2}{T_A}=-ln\frac{T_2}{T_B}$$

From which the book solution follows.

Hope this helps.

• Thanks for the clarification!
– Wb10
Commented Mar 16, 2020 at 1:56
• @Wb16 You are very welcome! Commented Mar 16, 2020 at 2:10