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In a Carnot engine, represented above, it is said that work is done by the system, but in reality , work is done by the system only in the isothermal and adiabatic expansions (from 1 to 3 in the picture above) then work is done on the system, as it contracts in an isothermal and adiabatic process (3 to 1). So why do we say that work is done only by the system in this engine, as the above pictures suggests, does it refer only to the first 2 processes (the expansions)? And the same goes to the reverse process, the refrigerator engine, where it is said that work is done on the system, and obviously in that case i still don't understand why it is said that work is done on the system.

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    $\begingroup$ It refers to a full cycle, so, globally, it does work, altough it's only doing work in two periods. $\endgroup$
    – FGSUZ
    Jan 11, 2019 at 21:22
  • $\begingroup$ The net work from all four segments of the cycle is not zero. It is equal to the area inside the "parallelogram." $\endgroup$ Jan 11, 2019 at 21:29
  • $\begingroup$ If you calculate the individual works in all 4 processes, add them all together, you'll get a total work $ > 0$ (i.e., out of the system). $\endgroup$ Jan 12, 2019 at 0:04

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I think you need to look at the “system” as the working fluid (e.g. an ideal gas) undergoing the Carnot cycle, and that the work done by the system is the net work for the entire cycle (the area enclosed by the PV diagram for the cycle). That work is positive, meaning net work is done by the system over the entire cycle, even though for two of the processes (isothermal and adiabatic compression) work is done on the system in order to complete the cycle.

Hope this helps.

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  • $\begingroup$ When you derive Carnot efficiency, I presume you therefore also consider the net work for the entire cycle such that efficiency n is calculated as n = net work / Q1? $\endgroup$
    – slew123
    Mar 26, 2021 at 2:46
  • $\begingroup$ Yes of course. What’s your point? $\endgroup$
    – Bob D
    Mar 26, 2021 at 7:49
  • $\begingroup$ What do you mean what is my point? I wanted to double check that I was using the formula correctly. A lot of sources don't actually explicitly state it is net work. It may be obvious to you or other teachers but not always obvious to others. $\endgroup$
    – slew123
    Mar 26, 2021 at 16:14
  • $\begingroup$ @slew123 I had no way of knowing you were trying to determine if you were" using the formula" correctly, or what formula you were using, for that matter. In any case, $\eta$ = net work/Q1 applies to any heat engine, including the Carnot engine. More importantly, unique to the Carnot engine is $$\eta=1-\frac{T_2}{T_1}$$ $\endgroup$
    – Bob D
    Mar 26, 2021 at 16:26
  • $\begingroup$ That formula isn't unique to the Carnot cycle; it applies to all reversible cycles operating between high temperature $T_1$ and low temperature $T_2$. What's unique about the Carnot cycle is it doesn't require additional reservoirs besides these two. Perhaps this was the intended meaning, but it's important point to clarify to avoid the misconception that the Carnot cycle is the only reversible cycle. $\endgroup$ Jun 26 at 21:44

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