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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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: $$ \text{efficiency} = \frac{\text{output energy}}{\text{input energy}}*100% $$