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Why define entropy with heat instead of work?

  1. Jan 14, 2016 #1
    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?
  2. jcsd
  3. Jan 14, 2016 #2
    Suppose you heat a body without doing any work. Does its entropy change?
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