Change of entropy

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Homework Statement
A 10 ohms resistor is held at temperature of 300 K. A current of 5 A is passed through the conductor for 2 mins. Ignoring changes in the source of the current, what is the change of entropy in the resistor.

I solved it like this:
[tex]\Delta S=\int dS=\int\frac{dQ}{T}=\frac{1}{T}\int dQ=\frac{\Delta Q}{T}=\frac{I^2Rt}{T}[/tex]

However, the book says the answer is 0, because temperature is held constant. Well, so is during an isothermal expansion, but there the change of entropy is Rln2. I believe the answer lies in that the isothermal expansion is reversible process, while the dissipation of heat in the conductor isn't, so there the change of entropy must be 0, but can someone explain it in detail.
 

Answers and Replies

  • #2
Mapes
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A better book answer would be that the state of the resistor---temperature, volume, mass---is the same at the beginning and the end of the process, and because entropy is a state variable (i.e., its value depends upon the state only), the entropy change must be zero. The same isn't true for a gas expansion because the volume has changed.
 
  • #3
Andrew Mason
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Just to add to what Mapes has said, looking at this the way your book does, in order to keep the resistor at a constant temperature of T=300K the heat flow out of the resistor (-Q) has to equal the heat flow into the resistor (+Q) so the net change in entropy is -Q/T + Q/T = 0. With a gas, one keeps the gas at a constant temperature by expanding the gas so there need not be heat flow out of the gas in order to maintain constant temperature. You can't do that with a resistor.

AM
 
  • #4
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thank you a lot
 

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