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During an introductory course on thermodynamics it is stated that the thermal efficiency only depends on the temperatures of the hot reservoir ($T_H$) and cold reservoir ($T_C$). So regardless of the processes taking place during a cycle, the thermal efficiency wouldn't change. This bothers me because in order for heat transfers to occur a temperature difference is needed between the reservoirs and cycle. So $Q_H$ could only take place if $T_H$ > $T_{in}$, which is the internal temperature of the cycle. Likewise, $Q_C$ could only occur if $T_{out}$ > $T_C$. So then why are the internal temperatures of a cycle never specified when considering thermal efficiency? It seems that there is a hidden condition that is never mentioned, namely, the internal temperatures of a cycle should always be between $T_H$ and $T_C$ otherwise no heat transfers could occur. However, this suggests that some properties of a cycle are of importance, which are the temperatures inside a cycle. When alluding to 'internal configurations', are all properties and processes meant except the temperatures?

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During an introductory course on thermodynamics it is stated that the thermal efficiency only depends on the temperatures of the hot reservoir ($T_H$) and cold reservoir ($T_C$).

To be more precise, it is the maximum thermal efficiency $\eta$ of a reversible cycle that only depends on temperatures of the reservoirs. Specifically, for the Carnot cycle, it depends on the ratio of $T_C$ to $T_H$ where

$$\eta=1-\frac{T_C}{T_H}$$

So regardless of the processes taking place during a cycle, the thermal efficiency wouldn't change.

Only if all the processes in the cycle are reversible, i.e., carried out so slowly that the system is always in thermal and mechanical equilibrium with the surroundings.

This bothers me because in order for heat transfers to occur a temperature difference is needed between the reservoirs and cycle.

It is true that in order for heat transfer to occur there needs to be a temperature difference. But for a reversible transfer of heat the temperature difference is made infinitesimally small so that, in the limit, the temperatures are the same. So a reversible cycle is an idealization to determine the bounds of efficiency, but it would have to be carried out so slowly as to be impractical.

So $Q_H$ could only take place if $T_H$ > $T_{in}$, which is the internal temperature of the cycle.

Yes, but for a reversible cycle, $T_{H}=T_{int}+dT$, meaning $T_{H}$ is infinitesimally greater than $T_{int}$.

Likewise, $Q_C$ could only occur if $T_{out}$ > $T_C$.

Again, $T_{out}=T_{C}+dT$

So then why are the internal temperatures of a cycle never specified when considering thermal efficiency?

Again, because for a reversible cycle, in the limit, the internal temperature is considered to be the same as the reservoir temperature.

When alluding to 'internal configurations', are all properties and processes meant except the temperatures?

In addition to the temperature, the difference in pressure between the system and surroundings is also kept infinitesimal, so that in the limit the internal pressure is the same as the external pressure.

In short, for a reversible cycle the system and surroundings are always kept in thermal and mechanical equilibrium.

Hope this helps.

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  • $\begingroup$ Did you really mean that the maximum efficiency depends only on the temperature difference between the hot and cold reservoirs? $\endgroup$ Aug 6, 2021 at 13:57
  • $\begingroup$ @ChetMiller For the Carnot efficiency the lower Tc or higher Th the greater the efficiency. Isn't that the same as saying the greater the temperature difference the greater the efficiency? $\endgroup$
    – Bob D
    Aug 6, 2021 at 14:25
  • $\begingroup$ Not in any mathematics I'm familiar with. The only thing that matters is the ratio of the temperatures, not the difference. $\endgroup$ Aug 6, 2021 at 14:43
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    $\begingroup$ @ChetMiller I see now that I jumped to the wrong conclusion. If I increase Tc and Th by the same amount, $\Delta T$ is the same but the ration of Tc To Th increases and thus efficiency decreases. I will correct. Thanks for pointing out. $\endgroup$
    – Bob D
    Aug 6, 2021 at 14:55
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The cycle for which "the thermal efficiency only depends on the temperatures of the hot reservoir (𝑇𝐻) and cold reservoir (𝑇𝐢 )" is a Carnot cycle, ABCD, consisting of two reversible isothermal processes, AB and CD, and two reversible adiabatic processes, BC and DA.

The isothermal processes in this ideal cycle are supposed to take place so slowly that the heat flow rate between working substance and reservoir approaches zero, so the required temperature difference between working substance and reservoir approaches zero. You therefore don't need to distinguish between $T_H$ and $T_{in}$ or between $T_C$ and $T_{out}$ (because you can consider the temperature differences to be as small as you like)!

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  • $\begingroup$ Does that suggest the $T_{in}$ and $T_{out}$ could take any value and is is correct that these two must be in between $T_H$ and $T_C$? $\endgroup$
    – Jelle 3.0
    Aug 6, 2021 at 12:12
  • $\begingroup$ The theoretical maximum efficiency depends only on TH and TC. Actual efficiency can be less than this. And, in a process where finite temperature differences are involved, the location where the entropy (inefficiency) is generated (working fluid or reservoirs) depends on the fluids and flow conditions. $\endgroup$ Aug 6, 2021 at 12:23
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    $\begingroup$ (a) "Does that suggest the 𝑇𝑖𝑛 and π‘‡π‘œπ‘’π‘‘ could take any value [?]" I'm not sure what you mean here. Sorry. (b) "is is correct that these two must be in between 𝑇𝐻 and 𝑇𝐢 ?" Yes. $T_{in}$ must be smaller than $T_H$ and $T_{out}$ must be larger than $T_C$ otherwise heat won't flow as required. But, as I've said, we consider the ideal case in which the temperature differences approach zero. $\endgroup$ Aug 6, 2021 at 12:52

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