Why define entropy with heat instead of work?

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Andrew U
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From what I understand, in the Carnot cycle summing qi/Ti for each step results in zero, thus indicating a new state function, entropy = qrev/T. But since dE = 0 = q+w, then q = -w, and looking at the equations derived from the cycle summing wi/Ti for each step should also result in zero. So why can't one also define entropy as wrev/T?
 
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