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 $0$.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.