# Carnot Engine - Maximum Efficiency Proof?

As an introductory physics student (independently studying before my return to official classes this summer), I am given a "proof" that the Carnot engine has the maximum efficiency possible by way of a diagram in which, assuming a more efficient engine whose work is fed into a reverse Carnot engine, we arrive at the contradiction of a net result of 100% efficient work because it is assumed the engine feeds, and the refrigerator (i.e. Carnot engine in reverse) takes, equal amounts into the cold reservoir. Having done research online, I have seen similar proofs in which the net effect is a heat transfer from a colder to a warmer body, but the main point is that, if there were a more efficient engine, we would violate the Second Law of Thermodynamics by feeding one into the other.

However, all of these "proofs" work equally well for ANY engine and refrigerator hooked up in tandem, one being more efficient than the other. I have looked everywhere for some sort of address on what seems to be an obvious oversight, yet cannot find anything except the occasional hand-waving.

I can "see" that a Carnot engine would be the most efficient based on complete reversibility, but I struggle with this manifestation of the proof (unfortunately, the only one I am advanced enough to understand, more than likely) because of what seems to be underlying assumptions/knowledge not addressed in any explicit statement I can find.

Can someone please clarify this proof for me, specifically addressing why it would not also apply to any engine/refrigerator in tandem where one is more efficient than the other? As I said, I'm sure there is a more advanced proof that improves the rigor, but I might not be ready for it, and I am hoping to get clarification on this specific proof.

Any help is MUCH appreciated. Thank you!

## 1 Answer

The key here is that the work derived from the engine is fed into a "reverse Carnot engine"—to use your phrase—or to focus on the important point into a reversible engine being run in refrigerator/heat-pump mode.

Not every heat engine has the property that running it "backward" generates a heat pump with the same ratios of $$\frac{|Q_H|}{|W|} \quad \text{and} \quad \frac{|Q_C|}{|W|}$$ as when you run it as an engine.1 That property could be taken as defining a "reversible" engine.

If you start with a non-reversible engine you're going to have to examine the details of the cycle in order to figure out what those ratios should be when running as a heat pump or refrigerator. And the answers you get are going to be less performant than if the starting engine had been reversible.

The end result of all that is that to arrive at a contradiction of the Clausius statement of the second law (i.e. to move heat "uphill" without work) the engine under test must be considerably more efficient than the one you're using as a heat-pump.2

1 I've written the ratios in the forms that represent the coefficients of performance for the reversed machine when regarded as a $$\text{heat pump} \quad \text{or} \quad \text{refrigerator}$$ respectively.

2 Whereas, if you start with a reversible engine the test engine need only be infinitesimally more efficient to arrive at a violation of Clausius's form of the second law.

• As an aside, the historical development of classical thermodynamics is chock full of details like this: things that are "obvious" if you have totally mastered the subject but frustratingly easy to miss when you're learning it. I only began to master the subject (a process still under way, I assure you) when I worked back into the historical approach after learning the statistical approach on a later pass. Commented Feb 18, 2019 at 6:12