0
$\begingroup$

For an irreversible heat flow, from an object $A$ at temperature $T_A$ to another object $B$ at temperature $T_B < T_A$, I'd like to know how to evaluate the change in entropy using the following expression:

$$ \Delta S =\sum_{i} \frac {\Delta Q_i}{T_i} $$

I know it is an irreversible adiabatic process, thus I have to select a reversible transformation which proceeds from the initial to the final state in order to find the entropy change with respect to $A$, and then sum it with the change in entropy of $B$ ($\Delta S = \Delta {S_a} + \Delta {S_b}$), but I don't know how to calculate this expression.

If someone could explain generically how it can be done, I'd appreciate it.

$\endgroup$
  • $\begingroup$ This question is dealt with in great detail by RICHARD C. TOLMAN, PAUL C. FINE: "On the Irreversible Production of Entropy," REV MOD PHYS VOLUME 20, NUMBER 1 JANUARY, 1948 $\endgroup$ – hyportnex Nov 26 '14 at 19:27
1
$\begingroup$

If it is adiabatic then $ \Delta Q_i $ will be always zero. The fact that it is irreversible doesnt matter. Any path that thakes you from A to B will result in the same change of entropy, as both initial and final states are in equilibrium. If you choose what is called a quasistatic path, which is idealized as a tranformation that occurrs slow enough so that each intermediate state is in equilibrium, then you can integrate your equation and the result will be as expected: $\Delta S =0$ (you expected that because it is an adiabatic process)

$\endgroup$
1
$\begingroup$

The problem you are trying to solve is different from the standard textbook situation where $A$ and $B$ are two heat reservoirs. When considering equilibration of temperatures, you have to account for the finite heat capacity of the two objects. This allows you to consider a series of quasi-static reversible heat exchanges with a series of heat reservoirs, and calculate the total entropy change from there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.