# Why does heat engine efficiency use absolute zero rather than thermal differential?

Carnot's theorem calculates efficiency of a heat engine versus absolute zero. I read the question Why can Carnot efficiency only be 100% at absolute zero? But I don't feel it addresses it precisely. One assumption of thermodynamics is that temperatures equalize. So within a system the only heat energy you actually have to work with is the difference between the heat source, and the cold sink. It seems to me the efficiency of a heat engine would be how well it handles this energy, since that is all the energy it could ever theoretically use based on other laws of thermodynamics. So basically it seems like for instance a 60% efficient engine might actually be using all of the available heat energy as well as possible with no waste. Obviously this is a ideal engine. This is all based on my understanding that a heat engine is not powered by heat, it is powered by the flow of heat from the heat source to the cold sink.

You get confused because you start with the concept of "heat" instead of entropy that is the true transported thermal "charge". So let us say your entropy reservoir transfers $$\Delta S$$ entropy to the engine. You can actually measure and verify that it has transferred $$\Delta S$$ entropy by using, say, a mercury thermometer whose expansion/contraction shows the empirical temperature, call it $$\theta$$, measured relative to some arbitrarily fixed zero point. You can calibrate a large mass calorimeter with such thermometer and declare that it absorbs or rejects $$\delta S = k\delta \theta$$ entropy for some material constant $$k$$ and then if the empirical temperature between the entropy reservoir (ie., entropy source/sink) and the engine is infinitesimally small then the entropy emitted is equal the entropy absorbed, a reversible process. During the isothermal stage of the Carnot cycle the engine absorbs/rejects infinitesimal $$\delta S = k\delta \theta$$ entropies while it works on the environment. The coupling to the reservoirs and the ensuing entropy exchange keeps the exchange almost isothermal and the process reversible by having the engine's temperature infinitesimally close to that reservoirs.

If in the higher temperature isothermal stage the engine absorbs in total $$\Delta S$$ entropy then at the lower temperature isothermal stage the same amount of entropy must expelled form the engine. As a result if the higher and lower experimental temperature difference is $$\Delta \theta =\theta_1-\theta_0$$ then the work done in the full cycle is $$W = \Delta S \Delta \theta$$ and this has nothing to do with the absolute temperature for it depends only on the experimental temperature differences.

This delivered work $$W$$ is the same for all Carnot cycles operating between the same experimental temperature differences and transporting the same $$\Delta S$$ entropy.

• Thank you. I think I got most of that, but will work on it some more. I think I see the problem and disagreement I have now. Carnot's theorem assumes the fuel provided all the heat/entropy for the engine. Which would mean the engine starts at zero entropy, or absolute zero. If the engine was warm / equalized with the cold sink when it started, then a different dedication would result. Jan 30 at 21:05
• The engine can start with an arbitrary amount of initial internal entropy, after all everything has some entropy above $T>0$, but by the end of the cycle its state must return to its initial state, it is cycle. Hence all absorbed and all internally generated entropy must be rejected before the end. During an adiabatic stage (there two in a Carnot cycle) no entropy is exchanged with the environment and if the stage is reversible then there is no internal entropy generation either what entropy is absorbed at the higher temperature is completely expelled at the lower temperature. Jan 30 at 21:20
• What makes the engine work is the drop of entropy from a higher temperature to a lower temperature reversibly and to achieve this reversible transport is the role of the engine the same way as the pale of water is pulled up by a lowered weight of almost equal mass or moving electric charges from a higher potential to a lower one can generate mechanical work. No concept of heat is needed here. You need the concept of absolute temperature when you want to know the relationship between the isothermally transported entropy and energy, specifically thermal energy, and you get $\delta Q = TdS$. Jan 30 at 21:27

The efficiency of a heat engine is defined as the amount of work exiting the engine divided by the amount of heat entering the engine. Any heat that gets rejected to the cold sink is heat that was not converted into work. The only way to obtain 100% efficiency is to reject no heat. Since heat transfer considerations prohibit rejection of heat at a lower temperature than the heat sink (heat always "flows" from hot to cold), that heat sink must necessarily be at a temperature of 0 K in order to have a chance to achieve 100% efficiency.

• thank you. I understand that. I see that it is a valid measure. Since only the difference between hot and cold is theoretical useful, it seems to me a more useful, not more correct, measure would be based off of that. Jan 30 at 18:26

So basically it seems like for instance a 60% efficient engine might actually be using all of the available heat energy as well as possible with no waste.

For a heat engine cycle to produce net work, there always has to be waste, i.e., heat transfer to a cold reservoir. Otherwise, it would violate the Kelvin-Planck statement of the second law:

No heat engine can operate in a cycle while transferring heat with a single heat reservoir

Although mathematically a Carnot heat engine cycle can be 100% efficient if the cold reservoir is 0 K, the problem is 0 K is not achievable. To quote from the following link:

https://blogs.scientificamerican.com/observations/racing-toward-absolute-zero/

"There’s a catch, though: absolute zero is impossible to reach. The reason has to do with the amount of work necessary to remove heat from a substance, which increases substantially the colder you try to go. To reach zero kelvins, you would require an infinite amount of work. And even if you could get there, quantum mechanics dictates that the atoms and molecules would still have some irreducible motion"

Hope this helps.

• Thank you. I know any real engine will be less than 100%. I was more bothered by the idea, or quantum, that a theoretically perfect heat engine could only be 100% efficient if the cold sink was at absolute zero. Jan 30 at 21:09
• Well, it looks like got the answer you were looking for Jan 31 at 0:33

So within a system the only heat energy you actually have to work with is the difference between the heat source, and the cold sink. It seems to me the efficiency of a heat engine would be how well it handles this energy, since that is all the energy it could ever theoretically use based on other laws of thermodynamics.

That kind of makes sense in some way, maybe, I'm not sure.

But:

If we increase the lower temperature by 10 degrees and also increase the upper temperature by 10 degrees, the efficiency decreases. Efficiency of an ideal engine. More thermal energy goes from warm side to cool side and less mechanical energy is produced.

So it seems that absolute temperature matters. Actually it is a fact that it matters.

It isn't completely efficient because it is not using all the energy which is provided to it. This energy is supposed to come from us, we provide it through some fuel etc. Yes, definitely the process of transfer of energy itself is 100% efficient but the energy difference between cold reservoir and hot reservoir is less than our input if the cold reservoir is not at absolute zero. Since that difference determines our output and it is less than the input, efficiency cannot be 100%. Remember that we had provided that hot reservoir with a specific amount of energy, but it has not converted all of that to work. Also, remember the definition of efficiency: $$\text{efficiency} = \frac{\text{output energy}}{\text{input energy}}*100%$$

• Thank you. It's the fact that the theoretical max is low that bothers me, not that real world engines will always lose energy. Jan 30 at 21:12
• I think you didn't get my point. Even if the real world was 100% ideal, it would not be using all the energy provided by us unless the sink was at absolute zero. We provided 'x' amount of energy and it only used 'x/2' for work, not because it "lost" it due to inefficiency but due to a design problem. @KenHorne Feb 5 at 10:19