A common method to show the equivalence of the Clausius and Kelvin statements of the 2nd law is to show that breaking Kelvin implies breaking Clausius and vice versa. I understand the logic for breaking Kelvin --> Clausius (take an ideal engine --> breaks Kelvin --> attach it to a fridge --> fridge also breaks Clausius). However, I have doubts about the following construction for the other direction:

We assume that a heat pump/refrigerator (on the left) pumps out heat from $T_L$ into $T_H$ without any work input (breaks Clausius). A non-ideal engine, which dumps $Q_L$ into the low temperature reservoir, is connected across the same temperatures.


The argument is that, as shown in the schematic below, the net system takes $Q = Q_H2 - Q_H1$ to output work $W$ without any heat output into the low temperature reservoir, which violates the Kelvin statement. Doesn't this only work when the heat pumped out by the fridge equals the heat dumped by the engine (two $Q_L$s are equal)? Would that not make the argument non-universal as it would not apply when the two $Q$ values are not equal?


Images from: https://www.ohio.edu/mechanical/thermo/Intro/Chapt.1_6/Chapter5.html

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    $\begingroup$ Suppose we have a refrigerator which takes $Q_L$ amount of heat from low temperature reservoir. The work which is done by the engine is completely in our hand. Like it depends on us how much work we need. So accordingly the engine takes a particular amount of heat from high temperature reservoir. Suppose efficiency of heat engine is $\eta=W/Q_H$, so we can adjust our engine such that $Q_H-W=Q_L$. So, it dumps $Q_L$ amount of heat. $\endgroup$
    – Iti
    Apr 4, 2021 at 3:55

1 Answer 1


Doesn't this only work when the heat pumped out by the fridge equals the heat dumped by the engine (two 𝑄𝐿s are equal)?

Yes, indeed!

Would that not make the argument non-universal as it would not apply when the two 𝑄 values are not equal?

The point is that assuming Clausius' statement is violated, you eventually get to extract some amount of work out of the heat. We do not need require perfect simultaneity or balance in running the Clausius violator and the heat engine.

Think about it like this. There is a little demon inside the combined system controlling both the Clausius violator and the nonideal engine. The demon runs the heat engine for some time and extracts some amount of useful work $W$ from the hot reservoir and dumps $Q_L$ into the cold reservoir. Then before the hot reservoir gets too cold it runs the Clausius violator to return it to its original temperature, rinse and repeat! The demon is effectively turning the heat from the hot reservoir into work.

It is true that it is not completely symetrical to the other implication as it is usually stated but that is not for a deep reason. In the other implcation, you start with zero useful work and you need to wait to run the Kelvin violator for some time until you can use that to power a refrigerator. You could have assumed that you already had a strictly positive amount of work available in the beginning to start running the refrigerator from the get go (just like we had a strictly positive amount of thermal energy to run the engine at the beginning of the previous paragraph). The point is sooner or later you are making heat flow from the cold reservoir to the hot reservoir.

  • $\begingroup$ That makes a lot more sense when thinking about it as an ongoing process rather than something at a given time as shown on the diagrams. I think my major source of confusion was that the diagrams made me think that those Q and T values hold at all times and that they do not change $\endgroup$
    – Ethiopius
    Apr 4, 2021 at 1:19

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