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

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For two reversible processes with the same internal energy change, if the work done is maximized in one process, the heat transfer must be minimized, and vice versa. This is derived from the First Law of Thermodynamics, which states that the change in internal energy (dU) is equal to the heat transfer (dQ) minus the work done (dW). Therefore, if dW increases while dU remains constant, dQ must decrease to maintain the equality. The logic presented aligns with the principles of thermodynamics, confirming its correctness. This relationship highlights the trade-off between work and heat transfer in reversible processes.
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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