I'm reading Schroeder's An Introduction to Thermal Physics. Regarding heat engines, it is stated:

Unfortunately, only part of the energy absorbed as heat can be converted to work by a heat engine. The reason is that the heat, as it flows in, brings along entropy, which must somehow be disposed of before the cycle can start over. To get rid of the entropy, every heat engine must dump some waste heat into its environment. The work produced by the engine is the difference between the heat absorbed and the waste heat expelled.

This seems to suggest that to maximize the efficiency of the engine, one should minimize the entropy produced during the process.

Okay, so now let's try to minimize the entropy that is created. The heat that leaves my hot reservoir is $Q_h$. This is the heat that is given to the gas in my engine. So the entropy change for the hot reservoir is $\frac{Q_h}{T_h}$ and the entropy change in the gas is $\frac{Q_h}{T_{gas}}$. Of course, in order for heat to flow, $T_h$ must be bigger than $T_{gas}$. This implies $$\frac{Q_h}{T_h} < \frac{Q_h}{T_{gas}}$$and the total entropy increases during the heat transfer, which is what I would expect. So, to minimize the entropy change, you would want the $T_{gas}$ to be very close to $T_h$. I believe the same argument can be made for the heat transfer from the gas to the cold reservoir.

However, the equation for the efficiency is

$$e = 1 - \frac{Q_c}{Q_h}$$

which can also be written as

$$e \leq 1 - \frac{T_c}{T_h}$$

where $T_c$ is the temperature of the cold reservoir. This equation implies that if the temperatures are similar then the efficiency is essentially zero. If the temperatures are $Q_h = \infty$ and $Q_c = 0$ then the efficiency is maximized. But by the argument I gave above, it seems that this would create a huge amount of entropy.

Am I making an error in my reasoning?

  • $\begingroup$ What I still don't understand is why having an extreme difference in temperature would give an efficiency $\approx 1$. Wouldn't there be entropy created from the heat exchange? If so, how can it be perfectly efficient if there is any excess entropy? $\endgroup$
    – Wise Owl
    Oct 24, 2015 at 19:08
  • $\begingroup$ If you fallow march's answer, in a complete cycle there is no entropy created due to heat exchange. When the temperature of the cold reservoir is very low, the negative entropy change to cancel out the positive entropy change only needs a small amount of heat to be dumped into the cold reservoir. $\endgroup$
    – Siva
    Oct 27, 2015 at 21:29

4 Answers 4


The Short Answer

How is the efficiency of a heat engine related to the entropy produced during the process?

The maximum efficiency for any heat engine operating between two temperature $T_H$ and $T_C$ is the Carnot efficiency, given by $$e_C = 1 -\frac{T_C}{T_H}.$$

Such a heat engine produces no entropy, because we can show that the entropy lost by the hot reservoir is exactly equal to the entropy gain of the cold reservoir, and of course, the system's entropy on the net doesn't change because the system undergoes a cycle.

Any heat engine operating between the same two temperatures whose efficiency is less than $e_C$ necessarily increases the entropy of the universe; in particular, the total entropy of the reservoirs must increase. This increase in entropy of the reservoirs is called entropy generation.

Finally, the efficiency of the perfect engine is less than one, necessarily, because the entropy "flow" into the system from the hot reservoir must be at least exactly balanced by the entropy "flow" out of the system into the cold reservoir (because the net change in system entropy must be zero in the cycle), and this necessitates waste heat from the system into the cold reservoir. The fact that $e_C$ goes to one in the limit of small ratios $T_C/T_H$ is a consequence of the fact that $Q_C$ is small compared to $Q_H$. It is not a consequence of the fact that entropy generation is small in this case, because entropy generation is already zero for the Carnot cycle.


Let's concentrate first on the interaction between the system and the hot reservoir. An amount $\delta Q_H$ of energy flows into the system from the hot reservoir, which means that the system's entropy changes by $$\mathrm dS_\text{sys} = \frac{\delta Q_H}{T_\text{sys}},$$ and correspondingly, the reservoir's entropy changes by $$\mathrm dS_\text{hot} = -\frac{\delta Q_H}{T_{H}}.$$ It is straight-forward to show then, that the total change in entropy of system plus environment satisfies $$\mathrm dS = \mathrm dS_\text{hot}+\mathrm dS_\text{sys} \geq0,$$ with equality holding if and only if the system and environment exchange energy via heating when they have equal temperatures, $T_\text{sys} = T_H$.

As a consequence, in order to minimize entropy production (and, in fact zero it out completely) during this process, we want $T_\text{sys} = T_H$, and the net change in system entropy during this process can then be written as $$\Delta S_\text{sys} = \int \frac{\delta Q_H}{T_\text{sys}} = \frac{Q_H}{T_{H}},$$ since we are assuming that the temperature of the reservoir doesn't change at all during the cycle.

Now, since the system operates on a thermodynamic cycle, and since the system entropy $S_\text{sys}$ is a state variable (state function/$dS$ is an exact differential, etc.), it must be true that $$\mathrm dS_\text{sys,cycle}=0.$$ Therefore, there must be some other process during which the system expels an amount of energy $Q_C$ to some other reservoir via heating in such a way that the change in system entropy during this new process is the negative of the change in system entropy that we calculated before. By the same argument as above, it must be that this change in entropy is $$\Delta S_2 = -\frac{Q_C}{T_C},$$ where $T_C$ is the temperature of the cold reservoir.

Finally, then, since system entropy is a state variable, $$0 = \Delta S + \Delta S_2 = \frac{Q_H}{T_H}-\frac{Q_C}{T_C}.$$ Another way of looking at this equation is that the net change in entropy of the hot reservoir is negative the net change in entropy of the cold reservoir during the cycle, and hence the net change in entropy of the universe is zero during the cycle.

Efficiency and work

Now, none of this seemed related to the fact that efficiency goes to 1 as the ratio of $T_C$ to $T_H$ goes to zero. This comes in in the following way. First, the net work output during one cycle is $$W_\text{out} = Q_H-Q_C,$$ and hence the efficiency of the engine that we've just made is $$e = \frac{W_\text{out}}{Q_H} = 1 - \frac{T_C}{T_H},$$ after some algebra. Based on our calculation above, this must be the maximum efficiency of any engine operating between these two temperatures. However, if we change the temperatures, then we can change the efficiency. The reason the efficiency goes up as the temperature ratio goes down is that $W_\text{out}$, being the difference between the heat flows, must go up if, say, we lower $T_C$ (because then $Q_C$ goes down) or if we raise $T_H$ (because then $Q_H$ goes up).

In some sense, this part really doesn't have much to do with entropy at all, because from the thermodynamic perspective, entropy production (which is the increase in entropy of an isolated system) is a measure of how much work we could have done if we had done the process reversibly, but we have already designed the perfect engine operating between those two particular temperatures above, so entropy doesn't have anything else to say.

  • $\begingroup$ Does the assumption of no entropy change for the Carnot engine depend on the idea that each reservoir is essentially infinite, so they can experience a net transfer of heat in or out on each cycle without any change in temperature? The Carnot cycle does involve a net transfer of heat away from the hot reservoir during one complete cycle, and a net transfer into the cold reservoir, correct? For a finite reservoir any net heat transfer should result in at least a small change in temperature, and thus a change in entropy, so if we assume finite reservoirs does the total entropy increase? $\endgroup$
    – Hypnosifl
    Jan 15, 2016 at 13:47
  • $\begingroup$ @Hypnosifl. Second question: yes. First question: sort of: in order to get reversible (and therefore zero net entropy increase) heat transfer, the systems must exchange energy via heat while at the same temperature. So, if the reservoir changes temperature, the system must as well. But then, you no longer have a Carnot cycle. Last question: no: as long as the heat transfer happens across $\Delta T =0$, and as long as all systems (including reservoirs) are undergoing quasi-static, reversible processes, the net entropy increase will be zero. $\endgroup$
    – march
    Jan 15, 2016 at 18:05
  • $\begingroup$ Thanks. But on the last question, could you have no change in temperature to a finite-sized reservoir if there's a net transfer of heat in or out, and no external work being done on it to change its volume? I'm wondering if you can avoid having an increase in the entropy of an isolated system that includes heat engine similar to a Carnot engine but with finite reservoirs. $\endgroup$
    – Hypnosifl
    Jan 15, 2016 at 21:49
  • $\begingroup$ @Hypnosifl. By "isolated", I assume you mean thermally isolated but not mechanically isolated. $S$ of a thermally isolated system is constant as long as its subsystems undergo quasi-static processes, there is no friction between subsystems, and all heat transfer occurs across infinitesimal temperature differences. The increase in $S$ of the cold reservoir is then always exactly balanced by the decrease in $S$ of the hot reservoir, whether or not they change their temperatures during the process. $\endgroup$
    – march
    Jan 15, 2016 at 22:10
  • $\begingroup$ Sorry, I was fuzzy on the details if the Carnot cycle, reviewing it I see the steps where some external part must do mechanical work on the working fluid, and have mechanical work done on it--I assume that's why you made the point about it not being mechanically isolated? But then let's consider the combined system that includes the device D that gives and recieves mechanical work along with the reservoirs and working fluid. If this combined system is totally isolated in every way, can we say that as net work is done on D (increasing some stored potential energy in D perhaps), $\endgroup$
    – Hypnosifl
    Jan 16, 2016 at 6:25

Hot reservoir is giving away its energy to the gas in the form of heat. The problem of any heat engine is that not all of the heat that is absorbed can be transformed in a macroscopic kinetic energy of a piston. It is a simple consequnce of a molecular chaos which heat actually is. The idea is to maximize the heat transfered into mentioned macroscopic kinetic energy in order to increase machines effectivness. It seems strange to try to lower the Q hot in order to do this because this simply decreases the heat flow. Idea in the text quoted is to try to turn as much heat into work and in that way minimize entropy.Another thing is, if you always keep gas temperature that is in contact with the hot reservoir very high and the gas which is in contact with the cold reservoir very low, you get the most out of your machine. Saying that temperature of the gas is the same as the temperature of the hot reservoir is the statement of perfect flow of heat to the gas from the hot reservoir. So i believe that if you try to lower the entropy produced because of heating the gas, you wont get a thing. It is almost as if you would lower the entropy by means of seting the heat to zero. You must produce heat and the entropy but you must try to turn as much of it into work done by a piston. So yes, high T hot would produce huge entropy but also would give much energy at the same time. The point is to turn as much of this heat into macroscopic work.


If you read the quote again, you will see that the only statement about entropy is, that, because it is a function of state, after a cycle it must take the same value, thus what was gained is the same that what lost. The conclusion that this implies that entropy production must be minimized is unwarranted. There is no net entropy production in a reversible cycle. To see this you can see Carnot's cycle, the most efficient possible. The efficiency of the cycle depends on the difference in temperature of reservoirs only, you can generate as much entropy as you want and then take as much as you can, and the efficiency will not change (see for instance the graphs on https://en.wikipedia.org/wiki/Carnot_cycle, specially the T vs S).


The unique increase of entropy takes place out of the internal ignition engine, not inside the engine. Carnot combustion cycle has zero entropy resultant as entropy increase in the isothermal expansion (leg 1) is equal to the entropy decrease in the isotherm compression (leg 3) and there is no change of entropy in the adiabatic expansion (leg 2) nor in the adiabatic compression (leg 4) since there is no Q term in any adiabatic process at all. This is OK, but what makes the people perplex is the meaning of entropy increase that is in general viewed as a lost chance to use some heat energy in doing work. The reconciliation is to look at the heat dissipation and exhaust gas release. Both are delivered at relatively high temperature with respect to the cold reservoir outside the engine, and the two air-masses (the relatively hot and the relatively cold) will mix. This irreversible mixing of the two air-masses is like the mixing of hot-water mass with cold-water mass where the net change of entropy will always be positive as the cooling down up of the hot-water is accompanied by relatively small negative entropy decrease (due to the low final temperature compared to the high initial temperature of hot-water) wheres the inverse is right for the cold-water (that ends up by experiencing relatively higher final temperature in the mixture than its initial low temperature) resulting in showing relatively large entropy increase. Adding the two entropy terms for the two water bodies with different temperatures will give net entropy increase in the water mixture. The two mixing processes (the two water bodies on one hand or the air-masses on the other hand) are irreversible and each of them should result in net increase of entropy in the obtained liquid or gaseous mixture. Such entropy increase - in the two fluid cases of mixing - means that there is a lost chance of using some heat to make some work; this is called an increase of disorder or am increase of entropy in the water mixture in the fist case of mixing or in the environmental air outside the heat engine in the case of releasing the dissipated heat and gas exhaust outside the heat engine. The cycle of the heat engine has nothing to to do with any net entropy increase inside the engine since the combustion cycle has zero net change of entropy for its summed up four legs as said above. This is my modest understanding of the issue.


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