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I was reading about Carnot cycle in "Thermodynamics by Cengel & Boles", now it is stated that Carnot cycle is a totally reversible cycle, i.e it is an internally as well as externally reversible cycle (in all of the processes involved) and it is very well explained how it is like that(pages attached) but what I don't understand is why then we still get an entropy change in the T-S diagram of Carnot cycle, that is if it is completely reversible then shouldn't entropy change of the system must be zero?

The four processes explained as reversible.]1[It says in the first line in section 9-2 that Carnot is composed of four totally reversible processes ]2

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The thing is that reversibility is not the same thing as isentropy. The carnot cycle is reversible not isentropic. Indeed the variation of entropy for a closed system is $$ \Delta S = \frac{Q}{T} +S_c$$ Where Q is the heat transfer and T the temperature of the source with which you are exchanging. $S_c$ is the created entropy. When your transformation is reversible it means that $S_c = 0$ but as you can see $\Delta S$ can still be not equal to zero if you have a transfer. And it happens that you are exchanging heat therefore you are exchanging entropy !

If you consider the system composed of the sources and the engine of the cycle then the variation of entropy is null. Yes the entropy changes in the engine but any increase or decrease in entropy matches with an opposite evolution in the sources.

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  • $\begingroup$ Yes the expression of the entropy change of a closed system is as you mentioned, but in this expression the entropy contribution by Q/T is due to heat transfer between our Carnot system and heat source, and as mentioned in pages attached, the heat transfer which would have produced external irreversiblility in the system is also externally reversible because it is infinitesimally slow by keeping system boundary at source temperature. Hence the total irreversibility of the system is zero & therefore it's entropy change too. Shouldn't it be? Would Q/T still exist even if it is infinitely slow? $\endgroup$ – shashank tyagi Aug 20 '18 at 13:57
  • $\begingroup$ Yes, I know it's purely theoretical & all the graphs & derivations related to Carnot are, but even so, just in theory, does this mean that Q/T (i.e entropy change due to external irreversibility) would still have some value even if the heat exchange is infinitely slow with both system and source at same temperature? If yes, then it seems strange, because wouldn't it be equivalent to an isolated heat source producing entropy by transferring heat around in it's own body(since both system and source are at same temperature),which is impossible? $\endgroup$ – shashank tyagi Aug 20 '18 at 14:16
  • $\begingroup$ @shashank tyagi , You would then have an infinitely small exchange. This value is not existing really defined as you have pointed out, it just a construct. You might want to consider this if you prefer : $\Delta S_{rev} = \int \frac{\delta Q}{T} $. We want this global value but indeed the local value $\delta Q$ doesn't exist. Although you can consider it like a infinitely small exchange. I am really sorry I deleted my previous comment by mistake. $\endgroup$ – Q.Reindeerson Aug 20 '18 at 14:27
  • $\begingroup$ Entropy is like a measure of your missing information about the sytem. It is a measure of the disorder of the system. What happens when you exchange heat without creating entropy is that you exchange the disorder without creating more. For instance if your room is a mess and you put everything in your sister's room, then you have less entropy without creating any. However if you break a glass in your sister room while moving your mess, then there is creation of entropy. But in both cases the value of the entropy of your room is changing. $\endgroup$ – Q.Reindeerson Aug 20 '18 at 14:37
  • $\begingroup$ Thanks Reindeerson, I am satisfied...for now. Really love your explanation of entropy exchange! $\endgroup$ – shashank tyagi Aug 21 '18 at 2:32
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Ah, but if you look at your book states it is a $T$-$s$ diagram - and I wonder if $s$ is used instead of $S$ because $s$ represents the entropy of the gas going around the Carnot cycle and not the entire entropy of the universe. The entropy of the gas considered changes in the Carnot cycle because it heats up and cools down and expands and contracts, but overall the entropy change of the universe will be zero as the processes are reversible.

So because heat is going into and out of the system (the gas considered in the Carnot cycle) and because the system does work and expands and contracts the entropy of the gas in the system changes and the entropy of the surrounding system changes so that overall the entropy change of the universe is equal to zero.

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  • $\begingroup$ Same goes for this, as I mentioned in comment above, heat transfer does produce entropy but only when heat is transferred at a temperature difference, isn't it? If heat transfer is occurring infinitely slowly, while system and source are at same temperature then wouldn't it be like heat as an energy moving inside the same body(source) from one point to another without any temperature difference, hence producing no entropy? $\endgroup$ – shashank tyagi Aug 20 '18 at 14:01
  • $\begingroup$ @shashanktyagi note that during the adiabatic expansion/contraction with no heat flow the change in entropy for the gas is zero, but during heat transfer there is a change in entropy. There is a change in entropy in isothermal expansion and contraction of a gas. --- I don't think this answers your question very well, but it is fundamentally what is happening. The volume of the gas rises at constant temperature and its entropy rises.... unless the expansion is adiabatic (in which case the temperature drops) $\endgroup$ – tom Aug 20 '18 at 15:52

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