# How does a reversible Carnot cycle transform heat to work?

To make clearer what I am asking let me introduce a new terminology to avoid any misunderstandings. Let the heat flow, which is rate, between a reservoir and an engine be denoted by $$\mathfrak J$$ and the absorbed or rejected amount of thermal energy by $$\mathfrak Z$$ for the Greek word "zesti" $$\zeta \epsilon \sigma \tau \eta$$ whose meaning in English is something like "heat". As far as I know nobody else has used this terminology and I know no Greek but here I will use this word, and I apologize to Greek language speakers if this be a wrong usage. Note that $$\mathfrak J$$ is directly observable by calorimetric measurement.

Similarly, let $$\mathfrak w$$ denote the work rate by or on the engine and $$\mathfrak L$$ denote the delivered work for the Italian "lavoro", so it is all scientific sounding now. In this sense, when one writes $$dU = \delta Q + \delta W$$ what is really meant that over an infinitesimal time interval the internal energy increases by $$\mathfrak Z = \delta Q = \mathfrak J dt$$ and by $$\mathfrak L = \delta W = \mathfrak w dt$$.

I do not believe so far anything here is controversial.

This is a question on interpretation of the happenings inside a reversible heat engine performing a Carnot cycle and not about the correctness of the 1st and 2nd laws expressed in the maximum Carnot efficiency, in which I believe as strongly as anybody else can.

Take the usual picture of a heat engine performing a reversible Carnot cycle. Due to heat flow from the high temperature reservoir to the infinitesimally lower temperature engine the internal energy of the engine is increased by a quantity of zesti $$\mathfrak Z_H= \int \mathfrak J_H dt$$ . The engine performs total work $$W$$ and thus in the lower temperature isothermal segment an amount of "zesti" $$\mathfrak Z_a = \int \mathfrak J_a dt$$ will be rejected and per the 1st law where $$\mathfrak Z_a = \mathfrak Z_H -W$$ at the low temperature $$T_a$$.

The entropy absorbed from and delivered to the reservoirs at temperatures $$T_H$$ and $$T_a$$ are equal to $$S=\frac{\mathfrak Z_H}{T_H}=\frac{\mathfrak Z_a}{T_a}\tag{1}\label{1}$$ because all four legs of the cycle is reversible including the adiabatic ones and thus entropy of the engine with its environment is conserved in toto.

Most description of the Carnot cycle uses an ideal gas as the working fluid but that is not necessary because the 1st and 2nd laws and consequent maximality of the Carnot efficiency hold for any reversible or irreversible processes. But we can say with certainty that by the very definition of a reversible adiabatic process in which no change of entropy either inside the engine or outside its environment can happen, all entropy changes are confined to the two isothermal reversible legs of the cycle.

Now as is usually asserted the heat transported and thus the heat flow rate time integrated to an amount of "zesti" and absorbed at the higher temperature $$T_H$$ is transformed by the engine to work. If such transformation takes an arbitrarily small but finite time then it must follow that first we have thermal energy absorbed and then a bit later we get work. Consequently, this must mean that while this happens at $$T_H$$ somehow we will have also "destroyed" some of the engine's internal entropy by an amount of $$\Delta S_H = \frac{W}{T_H}=\frac{\mathfrak Z_H}{T_H}-\frac{\mathfrak Z_a}{T_H}= S\left(1-\frac{T_a}{T_H}\right)\tag{2}\label{2}$$ which by the end of the cycle will have to be produced in the low temperature compression leg to balance the total entropy change to be zero per $$\eqref{1}$$.

Of course not all work of the engine is done at $$T_H$$ because some of the work is also delivered in the reversible adiabatic leg but that leg is isentropic, so the internal entropy change must occur either before or after. But how does the engine know that at $$T_H$$ it is to "destroy" $$\Delta S_H = S \left(1-\frac{T_a}{T_H}\right)$$ an amount that also depends on $$T_a$$?

So how does the engine that performs a reversible Carnot cycle transforms the absorbed "zesti" to work?

• It is not the heat added that causes work do be done in the isothermal expansion. The system does not start at a lower temperature than the hot reservoir but at the exact same temperature. It is the drop in external pressure that controls the process permitting expansion work to be done that in turn results in a decrease in internal energy and temperature. That is followed by a transfer of heat to the system restoring the internal energy. The whole process is controlled by infinitesimal decreases in external pressure. Nov 27, 2022 at 17:09
• Regarding your equation (2), I don’t see how you conclude that the change in entropy of the hot reservoir, $\Delta S_H$ is $W/T_H$, while it is in fact, $\Delta S_{H}=Q_{H}/T_H$. Nov 27, 2022 at 17:09
• I think you are mixing up expansion work and net work of the cycle. W in your equation is net work based on you equating it with heat in minus heat out. Nov 27, 2022 at 17:58
• @BobD Exactly how I see the situation and you have just admitted that Bronsted was right all along when he argued that isothermal entropy ("heat") transfer cannot do work, and the only time heat transfer does work is when entropy is moved through a temperature gradient, here between the high and low temperatures reversible in a Carnot cycle. Nov 27, 2022 at 18:02
• I can't admit Bronsted was right since, as I said when you brought him up in another post, I have no idea what he was saying. I don't know what is meant by the "isothermal entropy ('heat"') transfer cannot do work". Isothermal entropy is not "heat". It is reversible heat divided by constant temperature. I am saying that the expansion work is not "driven" by heat but rather by differential pressure which allows the expansion to occur at the expense of internal energy. Nov 27, 2022 at 19:55

A quantity of heat $$Q_H$$ is absorbed at the high temperature $$T_H$$ and heat $$Q_a$$ remaining is rejected at the low temperature $$T_a$$.

It is not an amount of heat $$Q_a$$ "remaining". You should not think about the system in terms of containing heat. Heat is energy transfer due solely to temperature difference. Things don't "contain" heat. The system contains internal energy. The heat rejected to the low temperature reservoir is due to the isothermal compression work.

Most description of the Carnot cycle uses an ideal gas as the working fluid but that is not necessary because the 1st and 2nd laws and consequent maximality of the Carnot efficiency hold for any reversible or irreversible processes.

The maximum possible efficiency for any heat engine operating between two fixed temperatures is the Carnot efficiency. It applies regardless of the nature of the working fluid, i.e., it does not only apply to an ideal gas. But the Carnot cycle is developed with the working fluid being an ideal gas. The reversible isothermal expansion and compression processes in the cycle assume that the internal energy of the working fluid depends only on temperature. That only applies to an ideal gas.

But we can say with certainty that by the very definition of a reversible adiabatic process in which no change of entropy either inside the engine or outside its environment can happen, all entropy changes are confined to the two isothermal reversible legs of the cycle.

That is correct.

Now as is usually asserted the heat absorbed at the higher temperature $$T_H$$ is transformed by the engine to work. If such transformation takes an arbitrarily small but finite time

To be reversible, the temperature and pressure differences between the system and surroundings have to be infinitesimal, which then means the time for the transformation becomes infinitely slow.

...then it must follow that first we have thermal energy absorbed and then a bit later we get work.

Actually it's the other way around. The expansion work is controlled by gradually reducing the external pressure, not by absorbing heat.

The infinitesimal decrease in external pressure results in an infinitesimal amount of expansion work resulting in an infinitesimal decrease in system temperature followed by an infinitesimal transfer of heat from the hot reservoir restoring the system temperature to that of the reservoir. The sequence is illustrated in the diagrams below for a thought experiment where grains of sand are slowly removed from the top of a vertically oriented piston, one grain at a time.

Consequently, this must mean that while this happens at $$T_H$$ somehow we will have also "destroyed" some of the engine's internal entropy by an amount of...

I'm not sure I follow you here. But if your statement here is based on your previous statement, perhaps you can reconsider it based on my comment to tht statement.

Of course not all work of the engine is done at $$T_H$$ because some of the work is also delivered in the reversible adiabatic leg but that leg is isentropic, so the internal entropy change must occur either before or after.

All the net work done by the system is due to the isothermal processes. The positive work done by the adiabatic expansion exactly equals the negative work done by the adiabatic compression, for a net adiabatic work of zero. Since the change in internal energy for both isothermal processes is zero, the negative change in internal energy for the adiabatic expansion equals the positive change in internal energy for the adiabatic compression, for the total change in internal energy of zero for the cycle, as required. Internal entropy changes only occur in the isothermal processes.

But how does the engine know that at $$T_H$$ it is to "destroy" $$\Delta S_H = S\left(1-\frac{T_a}{T_H}\right)$$ an amount that also depends on $$T_a$$?

Again, I don't understand your logic here.

So how does the engine that performs a reversible Carnot cycle transforms the absorbed heat to work?

Perhaps the diagrams below can help to visualize the reversible isothermal expansion process.

Hope this helps. • please see my updated question with clarification Nov 13, 2022 at 18:25