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Entropy, resistor in a temperature bath

  1. Aug 9, 2011 #1
    Consider a resistor with current running through it for some time in a constant temperature bath. I understand that the change in entropy of the resistor is zero because there is no change between the initial and final thermodynamic state. However, I am trying to come up with a reversible process in which to calculate explicitly the change in entropy as zero.

    First I initially consider the work performed on the resistor and no heat is added.

    TdS = dU - dW = 0, since the change in the energy is due solely to the work.

    Then the resistor is brought into contact with the bath and heat flows from the resistor to the bath

    TdS = -dW = -I^2*R*t.

    So I'd get a negative change in entropy for the entire process (for the resistor). Where in teh cycle have I made a mistake?
     
  2. jcsd
  3. Aug 9, 2011 #2
    I expect it has something to do with the last step not being reversible.
     
  4. Aug 9, 2011 #3
    Sorry, put this in the wrong section, please feel free to move it.
     
  5. Aug 9, 2011 #4

    Andrew Mason

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    Heat is added. If you stopped the heat flow out of the resistor, the temperature of the resistor (and, hence, its internal energy) would keep increasing.

    You are not taking into account the heat flow into the resistor in the form of electrical energy. Electricity is converted into heat in the resistor. This is thermodynamically equivalent to heat flow into the resistor from a heat reservoir.

    The heat flow into the resistor is the same as the heat flow out of the resistor and they both occur at the same temperature, so there is no change in entropy of the resistor.

    AM
     
  6. Aug 9, 2011 #5
    Thanks Andrew, cleared it right up.
     
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