0
$\begingroup$

Consider the ideal Carnot cycle consisting of two ideal reversible isothermal stages at $T_0$ and $T_1$ and two ideal adiabatic reversible (isentropic) stages connecting them; assume that $T_0 > T_1$. Denote the corresponding isothermally absorbed and rejected heat transfers, resp., by $Q_0$ and $Q_1$ so the work delivered by the cycle is $W=Q_0-Q_1$.

During the higher temperature isothermal stage the engine receives $Q_0$ thermal energy at temperature $T_0$ and at the same time also absorbs $S_0=\frac{Q_0}{T_0}$ entropy. Similarly, at the lower temperature $T_1$ it rejects $Q_1$ thermal energy and $S_1=\frac{Q_1}{T_1} entropy$. Since there is no entropy change during the two adiabatic reversible (isentropic) stages, not just globally but also locally, we must have $S_1=S_0$.

Now it is usually claimed with Clausius that the incoming heat is simultaneously converted to work. Since the delivered work is $W=Q_0-Q_1$ and at the lower temperature $T_1$ isothermal stage energy is expelled, the conversion is to occur at the high temperature $T_0$ isothermal stage. By this interpretation, the "heat" as flows in is being converted to work and one can only have this by assuming that the total absorbed entropy, $S_0$, inside the engine, is being reduced by $S_x=\frac{W}{T_0}=\frac{Q_0-Q_1}{T_0}$ as the heat to work conversion is taking place.

In other words, by the time the adiabatic stage starts only $S_y=S_0-S_x=\frac{Q_0}{T_0}-\frac{Q_0-Q_1}{T_0}=\frac{Q_1}{T_0}$ excess entropy left relative to what it was at the beginning of the cycle. This does not change in the isentropic stage but it is further reduced by the expelled entropy in the isothermal stage at $T_1$ after which inside the engine there remains $\Delta S= \frac{Q_1}{T_0}-\frac{Q_1}{T_1}$. But this $\Delta S < 0$ because $T_1<T_0$, and since this stage is followed by an isentropic stage to restore the cycle there must be some positive entropy source to compensate for this "missing" entropy; this is the uncompensated heat of Clausius.

In this cycle then there must be an excess production of $-\Delta S$ entropy during the isothermal stage at $T_1$ during which the entropy $S_1=S_0$ along with heat $Q_1$ are expelled. So here the isothermal stage while it is supposedly reversible also generates $-\Delta S >0$ entropy as compensation in addition to transferring reversibly $S_0$ entropy from the engine to the lower temperature reservoir.

Am I the only one here who finds this explanation strange? So, how does compensation really work?

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ "Now it is usually claimed with Clausius that the incoming heat is simultaneously converted to work". Just to be clear, you are saying the incoming heat is converted to the isothermal expansion work, not the net work $Q_{o}-Q_{1}$, correct? $\endgroup$
    – Bob D
    Dec 10, 2022 at 21:06
  • $\begingroup$ @Bob_D I am not claiming anything I am just trying to make sense out of certain interpretation according to which "heat is converted to work". Since we have $W=Q_0-Q_1$ work there must be some source to that work in a cycle, and the way I have been interpreted the predominant teaching is that an equivalent amount of absorbed "heat" to $W$ is converted to work. Since I do not understand that statement, there is my question. $\endgroup$
    – hyportnex
    Dec 11, 2022 at 8:19
  • $\begingroup$ @Bob_D In fact my hope is that you will agree with me that in a reversible process "heat" is not converted to work but that the drop of absorbed entropy (that is always conserved in a reversible cycle) over a temperature differential is the one that does the work, and the two explanations didactically speaking are not the same except in the very special case of an ideal gas. There is no difference in the final result or in anything observable of the different interpretations, it is strictly a matter of interpretation of the results. Which is easier to understand or simpler? $\endgroup$
    – hyportnex
    Dec 11, 2022 at 8:24
  • $\begingroup$ OK, sorry for having misunderstood you, then please, explain how does Clauisus balance the entropy (entropy in = entropy out) in the Carnot cycle with the uncompensated heat; what is the physical mechanism? $\endgroup$
    – hyportnex
    Dec 11, 2022 at 16:28
  • $\begingroup$ @I have no idea what you mean by “uncompensated heat”. $\endgroup$
    – Bob D
    Dec 11, 2022 at 16:31

1 Answer 1

Reset to default
1
$\begingroup$

The work $W_0=Q_0$ is the work done by the working fluid on the surroundings during the isothermal expansion at $T_0$. The work $W_1=Q_1$ is the work done by the surroundings on the working fluid during the isothermal compression at $T_1$. So the net work W done by the working fluid is $W=W_0-W_1=Q_0-Q_1$.

Improve this answer
$\endgroup$
9
  • $\begingroup$ Exactly what my point is (well, Carnot's) because to you and to me and to Carnot "heat in" and "heat out" is just another word for $T_0 S_0$ and $T_1 S_1$, and since in a reversible process $S_0=S_1$ all work is done by "dropping" i.e, moving that $S_0$ from $T_0$ to $T_1$ during the adiabatic stage, there is no heat to work "conversion" involved,. And everything else in the isothermal stages is just to make sure that the entropy absorption/entropy rejection is done reversibly. The other adiabatic stage is to move back the original internal entropy to the starting temperature of the cycle. $\endgroup$
    – hyportnex
    Dec 11, 2022 at 8:33
  • $\begingroup$ Then I don't really understand what you are asking. $\endgroup$
    – Chet Miller
    Dec 11, 2022 at 13:14
  • $\begingroup$ @hyportnex Neither do I $\endgroup$
    – Bob D
    Dec 11, 2022 at 13:24
  • $\begingroup$ @Bob_D If you two agree with my above comment then perforce you have agreed with the statement that isothermal heat transfer does no work. Indeed, because in Carnot's view thermal work $S_0(T_0-T_1)$ is obtained by moving (dropping) entropy from a high to a low temperature, cycle or no cycle. In my question (actually this is Bronsted's argument), I asked for an explanation as to how Clausius justifies entropy conservation in the reversible cycle by introducing "uncompensated heat", a concept he needs because he does believe in the conversion of heat to work. $\endgroup$
    – hyportnex
    Dec 11, 2022 at 15:08
  • $\begingroup$ @Bob_D and once you make the "mental conversion" by saying that "heat" is just $TS$ in any kind of process or state, and is not only a form of work in transfer, all of the sudden a much clearer picture arises but it comes with the view that reversibly transferred "heat", ie, $TS$ does not convert to work for in it the $S$ stays the same, and the act of dropping of that thermal mass , or thermal charge, $S$ is what works just as dropping a gravitational mass from a higher position to a lower one, or electric charge dropped between to electric potentials is the act that works. $\endgroup$
    – hyportnex
    Dec 11, 2022 at 15:20

Your Answer

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

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