Thermodynamics: What is the entropy change of an electrical resistance taken as system? Consider a system consisting only of an electrical resistor through which a constant current is flowing. Simultaneously, it is rejecting heat such that its temperature is constant throughout the process. With this much of information, what can we conclusively say about the entropy change of the system?
Thanks in advance. 
 A: The differential equation for steady state heat conduction within the resistor (neglecting axial conduction) is given by:
$$k\frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right)+\frac{I^2R}{\pi r_0^2L}=0\tag{1}$$where k is the thermal conductivity of the resistor material, $r_0$ is the outer radius of the resistor (assumed cylindrical), T is the temperature (a function of r), r is the radial location, and L is the length of the resistor.  The solution to this equation for the temperature profile within the resistor, subject to the condition that the mass average temperature is $T_{ave}$ is given by:  $$T=T_{ave}+\frac{I^2R}{8\pi k L}\left[1-2\left(\frac{r}{r_0}\right)^2\right]\tag{2}$$
If the current to the resistor were suddenly shut off and the resistor were suddenly insulated, transient heat conduction would cause the temperature of the resistor to eventually equilibrate to $T_{ave}$.  Since this process would take place spontaneously and adiabatically, we would expect the entropy of the resistor in this final equilibrium state to be higher than the entropy with the current flowing and heat being rejected (given that the average temperatures are the same in both cases).
The entropy of the resistor operating at steady state with the current flowing (and heat being rejected) minus the entropy of the resistor at the final uniform  average temperature is given by:$$\Delta S=\rho C L\int_0^{r_0}{2\pi r \ln{(T/T_{ave})}}dr\tag{3}$$where $\rho$ is the material density and C is its heat capacity.
If we substitute Eqn. 2 into Eqn. 3 and carry out the indicated integration, we obtain (if I integrated correctly):
$$\Delta S=mC\left[\ln{(1-\beta)-1}+\frac{(1+\beta)}{2\beta}\ln{\left(\frac{1+\beta}{1-\beta}\right)}\right]\tag{4}$$where m is the mass of the resistor and $$\beta=\frac{I^2R}{8\pi k L T_{ave}}\tag{5}$$As we indicated above, we expect this entropy difference to be less than zero (i.e., the entropy of the resistor at steady state with the current flowing and rejecting heat to the surroundings should be less than the entropy of the resistor in the final state at the same average temperature $T_{ave}$.  Therefore, the expression involving $\beta$ in Eqn. 4 should always be negative.  If we use Taylor series expansions to evaluate the expression involving $\beta$ in Eqn. 4 in the limit of small values of $\beta$, be obtain:
$$\Delta S=-\frac{mC\beta^2}{6}\tag{6}$$  As expected this entropy difference is negative.
A: (I am assuming that temperature is constant in the resistor. If you want, think about it as a quasi-1D object)
If we assume the system to be stationary, then all the thermodynamic parameters in the resistance will be constant. Therefore, since entropy is a state function, the entropy change in any given time interval will be $0$: the system is increasing its entropy by absorbing energy from the battery, but since we are assuming that $T$ is constant it means that it is simultaneously releasing the same quantity of energy in the environment, thus decreasing its entropy.
The medium to which the resistance is giving off its heat, on the other hand, will heat up, and its entropy will increase. Using the formula for Joule heating, we have
$$dS = \frac {\delta Q} T = \frac{I^2 R dt}{T} \to \frac{dS}{dt}= \frac{I^2 R} T$$
