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I was thinking what will happen to the standard Carnot cycle, if instead of - say - the usual cold reservoir, we adopt a smaller refrigerator, whose temperature in fact changes after the heat release from the piston. My opinion is that in this case the cycle is no longer reversible, am I wrong?

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  • $\begingroup$ Sorry, your question is not clear to me. In the Carnot cycle, the cold reservoir heats up. What "changes" in your setup, except that now it heats up faster, being smaller in volume. Thanks $\endgroup$ – user154420 Jun 27 '17 at 6:06
  • $\begingroup$ AFAIK (but obviously I could be wrong) the two reservoirs are described as being so large that the addition/subtraction of heat doesn't alter their temperature... $\endgroup$ – Lo Scrondo Jun 27 '17 at 6:12
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    $\begingroup$ No you are correct in what you say, but I worded my comment badly, sorry. The idealised Carnot engine with a smaller reservoir will result in an increasing tendancy towards irreversabilty, as you go from essentially infinite volume to finite volume. en.m.wikipedia.org/wiki/Carnot_cycle $\endgroup$ – user154420 Jun 27 '17 at 6:18
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    $\begingroup$ Thank you! I think it's the usual problem with idealizations - e.g., from Wikipedia, the reservoirs "are so large that their temperatures are practically unaffected by a single cycle" - if we apply literally this concept, even the first cycle would not be perfectly reversible $\endgroup$ – Lo Scrondo Jun 27 '17 at 6:51
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If the temperature of the cold reservoir (the heat sink) increases with each cycle, the efficiency of the cycle will decrease with each iteration of the cycle. Eventually, the temperature of the cold reservoir will equal the temperature of the hot reservoir and the cycle will stop.

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