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If I consider a heat engine $\mathrm{H}$ and a refrigerator $\mathrm{R}$ such that: enter image description here

enter image description here

where $Q_1<Q'_1$ and $Q_2<Q'_2$.

Now if we were to connect the engine and the refrigerator in the following way :

enter image description here

We notice that :

(a) $Q_1=W+Q_2$

(b) $Q'_1=W+Q'_2$

(by the conservation of energy)

From some manipulations of these equations, we arrive at :

$$Q'_1-Q_1=Q'_2-Q_2$$

Now if we were to draw a box around the refrigerator and the engine and in turn call it the new refrigerator, we see that :

enter image description here

Where I have defined $Q=Q'_1-Q_1=Q'_2-Q_2$. The resulting refrigerator that we get has the ability to transfer heat from the hot body to the cold body without the help of any external work, which is a violation of the second law of thermodynamics.

(When I asked the above question in other places, the response was that just like perpetual machines, such a configuration, although might seem to work on paper, cannot exist in real life. Is this what the answer should be?)

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  • $\begingroup$ Your first and third diagrams have $Q_1$ and $Q_2$ reversed. $\endgroup$
    – Bob D
    Feb 23, 2022 at 14:49
  • $\begingroup$ Also your second and third diagrams for the refrigerator have heats reversed. Get these straightened out or I will vote to close. $\endgroup$
    – Bob D
    Feb 23, 2022 at 17:40
  • $\begingroup$ Thanks for pointing it out. I have made the necessary corrections $\endgroup$
    – xyz1234
    Feb 28, 2022 at 7:32
  • $\begingroup$ In the real world, the amount of work required to drive the refrigerator is greater than the amount of work provided by the heat engine. $\endgroup$ Feb 28, 2022 at 16:46

2 Answers 2

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Assuming reversible heat engines, the engines are Carnot engines (since they operate between fixed temperatures). The efficiency of a Carnot engine is given by

$$\eta = \frac{W}{Q_H}$$.

So if the work $W$ is the same in both cycles the $Q_i$ can't be different either.

On the other hand, the proof that all Carnot engines have the same efficiency (given by the above equation) are based on that if they didn't they would violate the second law of thermodynamics. So maybe the correct answer to your questions is that heat engines you have proposed can't exist because they would violate the second law of thermodynamics.

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First of all your first and second diagrams are not consistent with the third diagram because you have reversed the heat transfers between the engine/refrigerator and the reservoirs. However, since your subsequent conservation of energy equations are consistent with the third diagram, I am assuming the third diagram is the basis of your analysis.

That said, your error is in defining $Q$ as you indicated and saying that is the input and output to the "new" refrigerator with no work input. I see no technical rationale for it and it leads to contradictions in your third figure.

Your energy conservation equations were based on manipulating the right sides of the following two equations, where we assume $W\gt0$ (otherwise the heat engine cannot operate the refrigerator in the third figure):

$$W=Q_{2}-Q_1$$ $$W=Q'_{2}-Q'_1$$

to get the equation

$$Q'_{1}-Q_{1}=Q'_{2}-Q_1$$

But when you defined $Q$ as equal to the left and right sides of the above equation and set it to the heat transferred in the fourth figure, you get, where the input and outputs to the refrigerator in the third figure are $Q'_1$ and $Q'_2$:

$$Q=Q'_{1}-Q_{1}=Q'_{1}$$ $$Q=Q'_{2}-Q_{2}=Q'_{2}$$

Which gives:

$$Q_{1}=0$$

$$Q_{2}=0$$

Which results in $W=0$ in the third figure, in contradiction to your basis of establishing the conservation of energy equations.

Hope this helps.

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