How can both the statements "For a closed system undergoing a cycle, net heat transfer is equal to net work transfer" and "According to the first law of thermodynamics, net heat less the net work done is equal to the change in internal energy" be true simultaneously, when one statement implies q - w = 0 and the other implies q - w = ΔU?
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2$\begingroup$ For a closed system undergoing a cycle, the change in internal energy iz aero at the end of the cycle. That is, the final state is the same as the initial state in a cycle. $\endgroup$– Chet MillerCommented Apr 10 at 18:17
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2 Answers
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"For a closed system undergoing a cycle, net heat transfer is equal to net work transfer"
That statement follows directly from the first law for a closed system over a complete cycle.
$$\Delta U=Q-W$$
For a complete cycle all system properties, in including internal energy, return to their original state. Thus
$\Delta U_{cycle}=0$ and therefore $Q=W$.
"According to the first law of thermodynamics, net heat less the net work done is equal to the change in internal energy"
That applies to a process where $\Delta U\ne 0$, and not to a cycle.
There is no contradiction.
Hope this helps.
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Because closed systems are closed by-virtue-of not having any external system that they are thermally connected to, to have any heat flow to, and because the definition of a cycle is that all state variables including $U$ return to their original states.
So, the first statement is identifying a very particular case, where there were flows of work and thermal energy, but none of them outside of a certain box, and the box returned back to the state that it started: so all the energy must have basically “rebalanced itself” within that box to the same $U$. The second statement is giving a more general rule that energy is conserved.
