What is the change of entropy for a resistor at constant temperature? 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 $100\ JK^{-1}$?
 A: 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.
A: 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. 
