Confusion with Carnot Cycle Problem

Question

Consider two fluids $F_1$ and $F_2$ with equal volume and heat capacity as well as $T_1$ and $T_2$ respectively, whereby $T_1 > T_2$. One uses a Carnot cycle to transfer heat from $F_1$ to $F_2$ in small cycles such that the temperatures after a certain amount of cycles are equal $T_1 = T_2 = T_0$. Now, I want to find this temperature $T_0$ in terms of $T_1$, $T_2$ and $C_V$.

My confusion is that the problem requires pumping from hot to cold temperature. Isn't this a spontaneous process? How would this be different from simply putting the two fluids in direct contact and figuring out their equilibrium temperature?

Could an approach maybe be to consider the problem of pumping hot to cold like in a refrigeration process, and then take the negative of that process?

Any hints would be appreciated!

Edit- Equation for Entropy:

$$\begin{align}\Delta S =& \int_{T_1}^{T_0} C_v\frac{dT}{T} + \int_{T_2}^{T_0} C_v\frac{dT}{T} \\ =& \ C_v\ln(\frac{T_0}{T_1}) + C_v\ln(\frac{T_0}{T_2}) \\ =& \ C_v\ln(\frac{T_0^2}{T_1T_2})\end{align} $$

Answer 1

Here's a hint. If you just allow the fluids to equilibrate, the amount of heat that F1 loses is equal to the amount of heat that F2 gains. But, if you run a Carnot engine between them, the amount of heat that F1 loses exceeds the amount of heat that F2 gains. The difference is the amount of work that the Carnot engine does.

Answer 2

EDIT: This answer is incorrect...

The efficiency of carnot engine $\eta = 1- \frac{T2}{T1} = 1 - \frac{|Q_{rejected of system}|}{|Q_{absorbed of system }|} $

$= 1- \frac{-CdT_1}{CdT_2} \implies -T1dT1= T2dT2 \implies \frac{T_1^2-T_0^2}{2}$

$=\frac{T_0^2-T_2^2}{2} \implies T_0=\sqrt{\frac{T_1^2+T_2^2}{2}}$.