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anonymous
 one year ago
What is the basis for a "reversible heat engine"? I know that it is theoretical and can't exist. But even mathematically it doesn't seem to be possible. Specifically, a reversible heat engine needs two temperature bodies to operate. Any heat transfer would automatically result in entropy generation (dS=dQ/T). Any kind of heat transfer from the high temperature body would result in entropy generation, which makes the process irreversible?
anonymous
 one year ago
What is the basis for a "reversible heat engine"? I know that it is theoretical and can't exist. But even mathematically it doesn't seem to be possible. Specifically, a reversible heat engine needs two temperature bodies to operate. Any heat transfer would automatically result in entropy generation (dS=dQ/T). Any kind of heat transfer from the high temperature body would result in entropy generation, which makes the process irreversible?

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2All thermodynamic transformations which occur in nature are irreversible, now for the Carnot's machine we suppose that the two involved thermodynamic transformations, verify this condition: \[\Large \frac{{{Q_1}}}{{{T_1}}} + \frac{{{Q_2}}}{{{T_2}}} = 0\] and that is why the efficiency of a Carnot's machine is the highest among the one of all of heat machines

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see, so does this mean that for a Carnot heat engine, the ratio of entropy lost by the high temperature body is equal to the entropy gained by the lower temperature body, hence the net is zero and therefore there is no net entropy generation?
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