0
$\begingroup$

How to find the maximum work that can be extracted by two objects which have variable temperatures $T_1$ and $T_2$ with $T_1 > T_2$?

I've thought that the maximum work obtainable is the one produced by a reversible machine, but since the temperatures are not constant I can't use the efficiency of a Carnot heat engine. In any case I can write $\Delta S = 0$ where $S$ is the entropy. So it becomes:

$\Delta S = 0 \Leftrightarrow \Delta S_1 + \Delta S_2 = 0 \Leftrightarrow \int\frac{\delta Q_1}{T_1} + \int\frac{\delta Q_2}{T_2} = 0$

Which should allow me to find the heat put into the system ($Q_2$) and the one taken out ($Q_1$) and so the maximum work. But both $T_1$ and $T_2$ are variables, plus, I have as unknown the final temperature $T_f$ as well, so I don't how to integrate.

Am I missing something? Any help is appreciated.

Improve this question
$\endgroup$
10
  • $\begingroup$ Hint: What is the change in entropy for a body of mass M and heat capacity C whose temperature changes from temperature $T_1$ to temperature $T_f$? $\endgroup$
    – Chet Miller
    Commented Jun 3, 2020 at 13:40
  • $\begingroup$ $M c ln\frac{T_f}{T_i}$ right? Actually I've thought it could be: $M_1 c_1 ln\frac{T_f}{T_i} + M_2 c_2 ln\frac{T_f}{T_i} = 0$ but in this way I should find $T_f$ but then I don't know how to continue $\endgroup$
    – Schiele
    Commented Jun 3, 2020 at 13:49
  • $\begingroup$ If you know Tf, then you know the amounts of heat exchanged by each body with the working fluid. $\endgroup$
    – Chet Miller
    Commented Jun 3, 2020 at 13:54
  • $\begingroup$ I don't know $T_f$ since it's even different from $T_e$ the equilibrium temperature which I could obtain considering the two objects in contact, but in this case, with $T_e$, I don't have any work extracted. So in my problem $T_f \ne T_e$ $\endgroup$
    – Schiele
    Commented Jun 3, 2020 at 14:03
  • 1
    $\begingroup$ Oh okay I think I got it, thank you very much for your help $\endgroup$
    – Schiele
    Commented Jun 3, 2020 at 16:19

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.