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A very large swimming pool filled with water of temperature equal to $20\ \mathrm{^\circ C}$ is heated by a resistor with a heating power of $500\ \mathrm{W}$ for $20$ minutes. Assuming the water in the pool is not in any contact with anything besides the resistor:

Is the change of entropy of the resistor positive, negative, or zero?

My Attempt

Entropy is given by the formula: $\mathrm dS = \frac{q_\text{rev}}{T}$

My first thought was that since the resistor is giving off heat energy to the water, $q$ must be negative and therefore the entropy change must be negative. However, does the fact that there is a continuous current of electrons passing through the resistor complicate this simple calculation? I am thinking that the current of electrons increase the entropy of the resistor and offsets the decrease in entropy due to heat loss. Am I correct?

The answer in the book is that the entropy change is $0$.

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  • $\begingroup$ Resistor is heating itself too, but as amount of water is huge there's almost stationary state here. $\endgroup$ – Mithoron May 21 '16 at 13:37
  • $\begingroup$ @Mithoron What do you mean by 'stationary state'? $\endgroup$ – Nanoputian May 21 '16 at 13:45
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    $\begingroup$ Passing current produces 500W of heat which is passed on to water completely so it's temp doesn't change - process is stationery $\endgroup$ – Mithoron May 21 '16 at 16:07
  • $\begingroup$ @Mithoron Oh, okay. So q of the resistor is $0$ then, right? $\endgroup$ – Nanoputian May 22 '16 at 1:34
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Here entropy change will be zero because resistor convert the electrical energy into heating so it take as work not heat so entropy change will be zero

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