# Entropy change and heat transfer in non-ideal heat engines

Please help me, a mathematician, to make sense of entropy. I know it's a topic with about a million questions already, so I understand if this gets ignored. Yet for the life of me, I cannot make sense of the previous answers.

So apparently the definition of entropy change in a system is $$\Delta S = \int_\mathrm{rev} \frac{\mathrm d Q}{T}$$ and this is a state variable. I also know that for irreversible paths, you have to find a reversible one that links the same thermodynamic states to calculate the entropy. Yet in my professor's notes it says

$$\mathrm dS = \oint \frac{\mathrm dQ}{T} < 0 \text{ for irreversible processes }$$

(is this even still an entropy difference? The notation $$dS$$ instead of $$\Delta S$$ confuses me) and elsewhere, talking about the Sterling process, it also says

$$\oint_{irrev} \frac{\mathrm dQ}{T} < 0 \text{ , so the entropy of the machine (not the total entropy) decreases }$$

so it really must be talking about the entropy change in the system (the machine) and not about the total entropy change of system + surroundings. But how can this be when for a system in a circular process the difference in any state variable is 0?

• Pretty sure it should be $>0$, not $<$. May 26, 2021 at 16:25
• Clausius inequality. A bad fridge heats up the room more. coecs.ou.edu/Feng.Chyuan.Lai/thermoweb/Lecture13/lecture13.htm
– Karsten
May 27, 2021 at 11:27
• @buckthorn this is just for the heat engine, not the reservoirs.
– Karsten
May 27, 2021 at 11:29
• @Ritam_Dasgupta It should be < as in the Clausius inequality.
– Karsten
May 27, 2021 at 12:53
• @Ritam_Dasgupta For the entire process, this is true, it has to increase. For selected compartments, it may increase, decrease, or stay the same. In the heat engine case, it increases for the cold reservoir, decreases for the hot reservoir, and stays the same for the engine (after a complete cycle).
– Karsten
May 27, 2021 at 13:44

So apparently the definition of entropy change in a system is $$\Delta S = \int_\mathrm{rev} \frac{\mathrm d Q}{T}$$ and this is a state variable.

This definition is for the surrounding if it is boring, e.g. a temperature bath. The $$_\mathrm{rev}$$ subscript is just for the temperature bath, i.e. there could be irreversible processes elsewhere (e.g. in the system).

Yet in my professor's notes it says $$\mathrm dS = \oint \frac{\mathrm dQ}{T} < 0 \text{ for irreversible processes }$$

The inequality is correct. Equating it to $$\mathrm dS$$ is not. The entropy change for a cyclic process is zero because entropy is a state function.

[OP...] elsewhere, talking about the Sterling process, it also says $$\oint_\mathrm{irrev} \frac{\mathrm dQ}{T} < 0 \text{ , so the entropy of the machine (not the total entropy) decreases }$$

This is the Clausius inequality. If a heat engine is not ideal, it will not extract the maximal energy possible from the heat transfer from hot to cold reservoir. The claim about the entropy of the machine is incorrect. Indeed, the entropy of the machine stays the same (use your argument of taking a reversible path from one cycle to the next, i.e. do nothing). Overall, the entropy increases because the entropy loss in the hot reservoir is no longer matched by the entropy gain of the cold reservoir. Thus, we are not extracting the maximal work in the process.

I found this explanation helpful. They used $$\delta Q$$ instead of $$\mathrm d Q$$ to remind us that there is a path-dependency.