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:

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

  • $\begingroup$ It’s not clear what you’re asking. When one talks about “pumping” heat one is normally talking about a heat pump and heat pumps move heat from a low temperature environment to a high temperature environment or, in your case from $F_2$ to $F_1$, not the other way around. Since heat does not spontaneously go from low to high temperature, the heat pump requires work to cause the transfer $\endgroup$
    – Bob D
    Commented Aug 5, 2020 at 15:42

2 Answers 2


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.

  • $\begingroup$ Thanks for your answer, when using your hint I thought I had it figured out, but on second glance I think I dont't. What I did now was write the difference of the changes in heats $Q_1 - Q_2 = C_v(T_1-T_0) - C_v(T_2 - T_0) = W$ and tried to solve it from there. However, this seems to yield strange results. Is the approach wrong, or am I messing up somewhere? $\endgroup$
    Commented Aug 7, 2020 at 11:53
  • $\begingroup$ Check your signs. Also, you need to do an entropy balance. $\endgroup$ Commented Aug 7, 2020 at 15:46
  • $\begingroup$ I'm sorry but I am still somewhat stuck. I have found an expression for the change in entropy, but I am not really sure how to proceed from there. Could you maybe give me an idea of what the equation ,that I need to solve, is supposed to look like? $\endgroup$
    Commented Aug 7, 2020 at 17:48
  • 1
    $\begingroup$ For a reversible process it should be zero, so $\Delta S = 0$, which would mean that $T_0 = \sqrt{T_1 T_2}$? $\endgroup$
    Commented Aug 7, 2020 at 21:36
  • 1
    $\begingroup$ Problem solved. $\endgroup$ Commented Aug 7, 2020 at 22:45

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 }|} $

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

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

  • $\begingroup$ With that value for the final temperature, the sum of the entropy changes of the hotter fluid and the colder fluid do not add up to zero for reversible operation (as they must). $\endgroup$ Commented Aug 9, 2020 at 3:07
  • $\begingroup$ @Chet Miller yes you are right sir thanks $\endgroup$
    – user600016
    Commented Aug 11, 2020 at 13:35

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