In the paper, Clausius and Darwin Can Both Be Right, the author claims to show an apparent paradox in the status of the heat that is not converted into work in a modified version of the Carnot Cycle. In this modified version, after one reversible cycle (at which point the transfer of heat to the cold bath is reversible since the work done by the cycle is available to reverse the heat flow), the available work is then dissipated into heat that flows into the cold bath, e.g, via driving a paddle in the cold bath causing friction.

The author claims that the entire change in entropy in this overall process is due to the increase in entropy caused by the dissipation of the work into heat, which is irreversible. But if so, this would apparently show that the heat that was not converted to work caused no increase in entropy. But since this latter heat flow can no longer be reversed since the work done has been dissipated, the author claims that this shows an irreversible heat flow with no increase in entropy.

This seems wrong to me, since presumably someone would have shown that there are irreversible processes that do not increase entropy long ago if the reasoning were correct. But I can't quite put my finger on where his argument breaks down in the paper.

Is it simply that the status of the heat that was not converted to work incrementally changes from reversible to irrevesible as the work is dissipated? If so, this feels a little strange, almost like entanglement's spooky action at a distance. The dissipation of the work produced by a local cycle in some remote system (e.g., in the lifting of a weight) can change the status of the heat that flowed into cold reservoir.

My apologies for not including the equations in my question, I'm not proficient in creating such equations.

The relevant discussion takes place in section 3.1. I am not concerned with any other part of the paper or any of the other claims made in other sections, eg claims about evolution vs entropy. Also, note that the flow of the text on p.56 is a little strange. It seems that Fig.2 should have been a page break. The right column above Fig.2 should be read after the left column above Fig.2. Then the right column below the figure should be read after the left.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Nov 8 '19 at 0:58

The author never explains why in the graph on the right $Q_{rev}$ is now $Q_{irr}$. That is not the case in a Carnot engine and it is the origin of his statement that $S_{rev}=S_{irr}$. Just ignore the work path, which is not adding anything to the argument.

An easy way to see this is:

1) If you do not change $Q_{rev}$ into $Q_{irr}$ the conclusion no longer holds.

2) If you make W=0, you still get the same result.

Now let us do the same argument but with a different system. We have two separate containers that do not interact with each other. They are both connected to two heat sources. Let us say that the first subsystem is a Carnot engine, so it absorbs $Q_{hot}$ from the hot reservoir and releases $Q_{cold}$ to the cold reservoir. This entire process being reversible. At the same time, the second subsystem also absorbs heat from the hot reservoir and releases some or all to the cold reservoir in an irreversible way. Now, the system as a whole suffered an irreversible transformation. Would you say that because of this fact, the Carnot cycle had become irreversible? I am sure you will not, and for the same reason you cannot rename $Q_{rev}$ to $Q_{irr}$ in the article's diagram.

| cite | improve this answer | |
  • $\begingroup$ He explains why at the bottom of p.56: "The very fact that the reversible work in above transformation has been irreversibly degraded into heat makes the whole process irreversible. This has been indicated on the right-hand side of the diagram in Figure 2 by attaching the label “irr” to the respective energy changes involved. The physical reason for this shift is the fact that the reversion of Q or Qc or both back to the hot source from the cold sink demands the expenditure of work that we don’t have available here." $\endgroup$ – Nick Gall Nov 8 '19 at 17:47
  • $\begingroup$ As for W=0, that certainly gets rid of any paradox/problem, but it doesn't provide any insight into the issue. Assume we restrict the author's claim slightly: "The entropy of the universe may increase as a result of some spontaneous (or irreversible) process, in which at least some work is done". [bold indicates the added constraint. Now the issue becomes: If some work is done, it appears that an irreversible process (via Path A) does not increase entropy. That seems wrong, but why? Where is the flaw in the reasoning in section 3.1? $\endgroup$ – Nick Gall Nov 8 '19 at 17:53
  • $\begingroup$ comment 1: It is incorrect to say that because now you made the cycle irreversible that part of the heat transfer that was reversible before is now irreversible. it shows a misunderstanding of what reversible means. Comment 2: when W=0 you still obtain the result $dS_{rev}=dS_{irr}$, which shows that dissipating the work irreversibly is not what leads to his result. $\endgroup$ – Wolphram jonny Nov 8 '19 at 19:01
  • $\begingroup$ his argument works even if he has no cycles, just a system with two subsystems, one of reversible transfer of heat (path 1) next to an irreversible transfer of heat, which because it is irreversible you decide to also call the first one irreversible, because the system as a whole cannot be reversibly bring back to the original state. $\endgroup$ – Wolphram jonny Nov 8 '19 at 19:04
  • $\begingroup$ If the comments are not clear I will rewrite it and add it to the answer for the sake of clarity $\endgroup$ – Wolphram jonny Nov 8 '19 at 19:05

I see a different flaw in the conversion that was made. The heat generated is dumped directly and completely into the cold bath. It does not flow from a hotter source temperature. How can such a step be called irreversible? Spontaneous, irreversible heat flow occurs only with a temperature difference, ostensibly across a boundary between a system and a surroundings.

Otherwise, conversion of work goes first to make a change in the internal energy in a system. We never convert work to heat directly. These are flows in/out of system from/to its surroundings. Internal energy has no basis to leave a system as (irreversible) heat flow unless the surroundings to the system are at a colder temperature.

In short, we cannot convert work to irreversible heat flow at a given temperature. We can convert work to internal energy change at a given temperature. Then, we convert internal energy an irreversible heat flow by having it leave the system a hot temperature compared to a colder temperature in the surroundings.

The picture for the proposition must appear as below. The second step through an intermediate temperature $T_i$ is required, and that is a second system in the larger box.

System Picture

The heat $q_i$ is along an irreversible path. The existence of the flow of heat from $T_i$ as an irreversible step has no impact that can permit an unprincipled change in the designation of the reversibility for the engine to the left of it.

The above is where the analysis fails for me.

In addition, the discussion is propagating a second fatal flaw. A reversible process is one where the system and surroundings are in exact thermal and mechanical equilibrium at all points during the process. A reversible cycle is one that can be reversed through the entire cycle and experience no change (either direction has a net zero internal energy, entropy, enthalpy ... change). A reversible cycle is by definition a combination of reversible paths (processes). The minute that a forward cycle is complete, you cannot remove part of the cycle, claim the cycle fails because you have removed part of it, and thereby state that it is now an irreversible cycle. In fact, it is NOT EVEN A CYCLE anymore. You also cannot claim that a reversible cycle becomes irreversible because the heat or work that enters/leaves the cycle goes somewhere/comes from somewhere special.

So, we cannot say that because the work is dissipated, that work is no longer able to cause the cycle to make heat flow back to the hot reservoir, and therefore the cycle that was reversible going clockwise is now suddenly irreversible in all cases even going forward. The minute you take away the work in the first system, you have no cycle that can allow you to make or change labels!

| cite | improve this answer | |
  • $\begingroup$ @JeffreyJWieimer I'm not sure which of the two heat flows in Fig.2 you're talking about: Path A or Path C? I assume C, because A is definitely a flow of heat from a hotter to a colder temperature. As for C, it seems clearly an irreversible process to me, since transforms work into heat. The author says this as well, "the work W generated by the reversible Carnot engine (cf. Fig. 1) is irreversibly degraded, via some frictional mechanism, into an equivalent amount of heat Q at the temperature Tc of the cold reservoir (cf. Fig. 2 below)." Are you claiming that Path C is not irreversible? $\endgroup$ – Nick Gall Nov 12 '19 at 15:00
  • $\begingroup$ @JefferyJWeimer Your diagram helps clarify the issue, thanks. I still say your Ti-qi diagram to the right (roughly a relabeling of Path C does affect the reversibility of Path A). Think of the author's Fig.2 engine (F2E) as having 5 phases: the 4 normal Carnot-cycle phases + a work-to-heat conversion phase (where the work produced in phase 2 is dissipated as heat into the cold bath). After phase 4, F2E is reversible, since the entropy decrease of the hot bath ='s the entropy increase of the cold bath. However, after phase 5, F2E is not reversible because the latter is now > than the former. $\endgroup$ – Nick Gall Nov 12 '19 at 23:52
  • $\begingroup$ @NickGall A wealthy donor hands you \$100. You keep \$20 and dump \$80 in a reversible river. You can expect to get back \$80 when you show \$20 and promise to return $100 to the original donor. But, in a fit of heated rage, you give away the \$20 to the devil. You go back to river, pull out a different \$20, get your \$80, and return \$100 to the original donor. You are out \$20. What the devil does with the \$20 you gave has no affect on the fact that the river remains reversible. It demands nothing more than a promise to put \$20 you have + \$80 it gives you = \$100 to the original donor. $\endgroup$ – Jeffrey J Weimer Nov 13 '19 at 2:55
  • $\begingroup$ I like your analogy. A definition of reversible process is "a process that, after it has taken place, can be reversed and, when reversed, returns the system and its surroundings to their initial states". "You are out 20" means the surroundings (me) are not in their initial states at the end of things. So the F2E cycle and all its processes/flows (incl. Path A - throwing 80 in a river) become irreversible when I give the $20 to the devil (dissipate work in Path C). $\endgroup$ – Nick Gall Nov 13 '19 at 17:47
  • $\begingroup$ @NickGall No! When you give the \$20 to the devil, you give the \$20 to a different system. The system of you + the river + the donor have no clue and are not affected. You cannot think any other way. It is wrong to believe that what you can do in a reversible system can be changed by what you do in a different system at some later point in time. You can propose an analysis that has the entire process that involves the donor, you, the river, and the devil to be irreversible OVERALL. But you cannot blame the donor and the river for doing their part to avoid your foolishness. $\endgroup$ – Jeffrey J Weimer Nov 13 '19 at 20:34

Heat and work are path dependant quantities. Saying "an amount of heat $Q$ has been transfered from the hot to the cold resovoir" does not define the final state of the system, and so I can find many values for its final entropy. So you can follow the path laid out in the paper and generate an entropy of $\frac{Q}{T_c}$ or I could find something monstrously inefficent and generate $\frac{10,000 Q}{T_c}$ or I could set up a Carnot engine and run it until an amount of heat $Q$ has been deposited in the cold resovoir while depositing the produced work back in the hot resovoir (as reversible work) and generate no entropy at all.

Really the paper should be talking about the amount of energy transfered, since that is the relavent state function, and the lower limit on entropy generatin to transfer the energy is 0.

The second law is an inquallity; you can always find a way to generate more entropy than nessarsery (and this paper has found one). You cannot generate less.

| cite | improve this answer | |
  • $\begingroup$ Heat/work are path dependent, but entropy is a thermo. state variable. In the first stage of the cycle in Fig.1, the entropy increase in the working medium is Qh/Th. So at the end of cycle, entropy in the medium must have decreased by an equal amount. This decrease is the sum of Q/Th, from the conversion of heat into work (Path B), and Qc/Tc, from the dissipation of heat into the cold bath (Path A). In the cycle of Fig.2, the author has stipulated a fixed amount of additional entropy is produced in Path C: Q/Tc. Therefore it is not legitimate to imagine other paths generating more entropy. $\endgroup$ – Nick Gall Nov 8 '19 at 18:14

I think I figured out an answer that seems intuitive to me. It is related to @BobD's observation that the apparent flow of heat (Qc) directly from the hot bath to the cold bath via Path A in Fig.1 seems fishy. I think the mistake in logic begins with equation:

dSirr,engine = dSA,irr + dSD, irr = dSA,rev + dSB,rev + dSC, irr = 0 + 0 + W/Tc =Q/Tc (7)

dSA,rev and dSB,rev do not equal zero; they equal Qc/Tc and -Q/Th, respectively. The author's justification for asserting they are both zero is equation (4):

dSA,rev + dSB,rev = 0 (4)

While this is true, it does not imply that dSA,rev and dSB,rev are individually zero. At the end of one reversible cycle of a Carnot engine, the change in entropy of the working medium is 0 (the right side of equation (4)). The decrease in entropy in the system that work is done on is -Q/Th and the increase in entropy of the cold bath is Qc/Tc, such that they balance one another. A more accurate rendition of (4) would be something like this:

Qc/Tc + Q/Th = 0 (4')

So equation (7) should be written as:

dSirr,engine = dSA,irr + dSD, irr = dSA,rev + dSB,rev + dSC, irr = Qc/Tc - Q/Th + W/Tc = Q/Tc (7')

This means that equations (8)-(11) are also all incorrect because they rely on eliminating dSA,rev (or equivalently, dSA,irr) because it is supposedly equal to 0. But it is not; dSA,rev is equal to Qc/Tc.

To summarize: the increase of entropy of the cold bath, Qc/Tc (via Path A) is an increase of entropy whether or not such increase is reversible. If the work produced by the cycle is still available, the entropy is reversible; if the work has been dissipated, it is irreversible.

| cite | improve this answer | |
  • $\begingroup$ > If the work produced by the cycle is still available, the entropy is reversible; if the work has been dissipated, it is irreversible. ... This is the crux of a seriously false argument. Work is not available or not available. We can always find work to input to any type of cycle. Work is not a commodity with an ID or serial number or color that must be found only with that specific ID or serial number or color. Certainly it is not defined as this way in order to certify that a cycle or process is or is not reversible. The definition of reversibility does not care where work is obtained! $\endgroup$ – Jeffrey J Weimer Nov 15 '19 at 23:40

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.