Answer: Is My Logic Correct: Reversible Processes & Heat Transfer?

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SUMMARY

The discussion confirms that for two reversible processes with identical internal energy changes, if the work done (W) is maximized in one process, the heat transfer (Q) must be minimized, and vice versa. This conclusion is derived from the First Law of Thermodynamics, expressed as U2 - U1 = Q - W. The relationship between internal energy change, work, and heat transfer is maintained, affirming the logic presented in the query.

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Rohit93
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If for two reversible processes we have the same internal energy change, if work done is maximum for one, is heat transfer minimum and vice versa?
The logic that I use is that since dU=dQ+dW, for constant dU if dW is more for one, heat transfer dQ is less and vice versa just to keep the sum constant.
Is my logic correct or wrong?
 
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Your logic is essentially correct. Using the First Law for a process between state 1 and state 2,
U2-U1 = Q-W,

which leads to the same conclusion.​
 
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