# Entropy Change of Resistor

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

We know that 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 $$0$$.

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