Confused with Entropy and Clausius inequality

Question

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.

Answer 1

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

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.