HEAT TRANSFER IN A CONDUCTING BAR

Very nice problem. As current flows through the slab, the heat generated per unit time in the whole volume of the slab is:

$P = \frac{RL}{S}I^2$

Therefore, the power generated per unit time, per unit volume is:

$p = \frac{P}{LS}$

Now, consider an elementary volume of cross-section area $S$ and length $\delta x$ at a distance $x$ from the left end of the slab. The hear generated per unit time in this elementary volume is:

$P_v = p S \ \delta x$

Now, consider this elementary volume again. The heat per unit time flowing into the left facet of this element is $Q_{in}$ and heat per unit time flowing out of this slab through its right facet is $Q_{out}$ . Applying the first law of thermodynamics for this elementary volume gives:

$Q_{in} + P_v = Q_{out} \ \dots (1)$

Now, the rate of hear flow varies with $x$ . Since $\delta x$ is small, one can use a Taylor series first order approximation (since $\delta x$ is small) to relate $Q_{in}$ and $Q_{out}$ as such:

$Q_{out} = Q_{in} + \frac{dQ_{in}}{dx}\delta x$ $\implies Q_{out} - Q_{in} = \frac{dQ_{in}}{dx}\delta x$

Now, using Fourier's law, we know that:

$Q_{in} = -KS \frac{dT}{dx}$ $\implies \frac{dQ_{in}}{dx} = -KS \frac{d^2T}{dx^2}$ $\implies Q_{out} - Q_{in} = -KS \frac{d^2T}{dx^2} \delta x$

Substituting the above result in (1) gives:

$P_v = -KS \frac{d^2T}{dx^2}$ $\implies \boxed{\frac{d^2T}{dx^2}=-\frac{RI^2}{KS^2}}$

We know that $T(0) = T_1$ and $T(L) = T_2$ .

Double integrating the above equation and substituting all boundary conditions and values gives the temperature distribution as a function of $x$ as follows. Simplifications have been left out in this solution.

$T(x) = -\frac{8739\,x^2}{21800}-\frac{1626261\,x}{10900}+500$

From the above equation, $T(L/2)$ can be conveniently computed. The result is $\boxed{T(L/2) \approx 350.4 \ K}$ .