A 10 Ω resistor is held at a temperature of 300 K. A current of 5 A is passed through the resistor for 2 minutes. Ignoring changes in the source of the current, what is the change of entropy in (a) the resistor and (b) the Universe?

My attempt:

$\Delta Q=I^2 R t=5^2\times 10\times 2\times 60=30000\ J$

$\displaystyle\Delta S=\frac{\Delta Q}{T}=\frac{30000}{300}=100\ JK^{-1}$

Won't $\Delta S_\text{univ}$ assume the same value of $100\ JK^{-1}$?

  • $\begingroup$ Sequentially address: What is the temperature of the surroundings? What is the heat flow to the surroundings? What is the entropy change of the surroundings? $\endgroup$ – Jeffrey J Weimer Jun 24 at 15:50
  • $\begingroup$ @JeffreyJWeimer Room temperature. -30,000 J. So I just divide the latter by the former? $\endgroup$ – Thomas Jun 24 at 16:10

Since the resistor is kept at $T=300K$(which is wierd since its heating up), the entropy change of the resistor is
$$\Delta S_{resistor}=\frac{-\Delta Q}{T}$$ (heat flow out of the system taken -ve)
Since the surroundings stay at $T$(since no other temp is given, assuming that), $$\Delta S_{surroundings}=\frac{\Delta Q}{T}$$ therefore $$\Delta S_{universe}=\Delta S_{resistor}+\Delta S_{surroundings}=0$$

To be clear,you already had the answer.

Note: Joule heating is irreversible therefore $\Delta S_{universe}>0$. The reason that doesn't happen here is because the additional system keeping the resistor at surroundings' temp isn't considered here.

  • 1
    $\begingroup$ This answer is incorrect. Entropy is a function of state, and the state of the resistor does not change; it is 300 K to begin with, and 300 K at the end. So the change in entropy of the resistor is zero. Any entropy that is generated within the resistor during this irreversible process is transferred from the resistor to the surroundings. So the entropy of the surroundings increases by 100 J/K. And the increase in the entropy of the universe is also 100 J/k. $\endgroup$ – Chet Miller Jun 24 at 23:08
  • $\begingroup$ @chet miller are you implying this example is just a heat source in surroundings $\endgroup$ – lineage Jun 24 at 23:14
  • $\begingroup$ The system is the resistor, and the surroundings is everything else. If the resistor's state does not change, its entropy does not change. $\endgroup$ – Chet Miller Jun 24 at 23:15
  • $\begingroup$ There is electrical work being done on the resistor, and since its state is not changing, an equal amount of heat is being transferred to the surroundings. So there is no change in entropy of the resistor. All the entropy generated in the resistor from the dissipated work is transferred to the surroundings. $\endgroup$ – Chet Miller Jun 24 at 23:27
  • $\begingroup$ Is the resistor at the same temperature in the end as at the beginning? Irrespective of the process, what does this mean about its entropy change? $\endgroup$ – Chet Miller Jun 24 at 23:30

In this problem, it would be incorrect to say that the change in entropy of the resistor between its initial and final states is anything other than zero. Entropy is a function of state, and the initial state of the resistor (300K) is exactly the same as its final state (300K).

The fallacy in applying the expression $\int{dq/T}$ to determine the entropy change of the resistor in this problem would be that the process the resistor experiences is irreversible, and this expression can only be used to determine the entropy change for a reversible processes. For an irreversible process, one must first devise an alternate reversible path between the same initial and final states, and then determine the value of the integral for that path. If we follow this procedure for a solid, like our resistor, we find that $$\Delta S=MC\ln{(T_2/T_1)}$$where M is the mass of the solid, C is its heat capacity, $T_1$ is the temperature in the initial state, and $T_2$ is the temperature in the final state. And if, as in our problem, $T_2=T_1=300K$, $\Delta S =0$.

Here is the mechanistic explanation of what takes place: In the present irreversible process, our system, the resistor, receives work W from its surroundings (electrial work) and it dissipates this work irreversibly, returning an equal amount of heat Q to its surroundings. This dissipation of electrical work within the resistor translates into generation of entropy. But, the generated entropy does not stay in the resistor. If it transferred via the heat flow Q to the surroundings. So the net effect is no entropy change for the resistor.

With the entropy change of the resistor being zero and the entropy change for the surroundings being 100 J/kg,the entropy change for the universe is $$\Delta S_{universe}=\Delta S_{syst}+\Delta S_{surr}=0+100=100\ J/K$$ As expected, for this irreversible process, the change in entropy of the universe is positive.


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