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Q:
Why is the solution to the Heat Equation infinite at a steady temperature?
I know that a steady state solution to the heat equation is given by:
$$T(r,t) = T_0 + \int_{r’r’} \kappa(r-r’) \cdot \frac{dr’}{\kappa}$$
Now if we integrate this from an infinite wall to a 1cm slab of constant temperature, the first term is easily shown to be infinite, however the second term isn’t zero.
Why is this the case? I expected to find a finite answer such that the average temperature is $T_0$.
A:
Imagine you have a slab of material with constant surface area $A$, constant heat capacity $C$, and thermal conductivity $\kappa$, whose thickness is $d$. Now consider the temperature $T$ on one face of the slab, i.e., a constant temperature gradient will be applied on the slab, (again neglecting the heat capacity of the slab.)
We now want to find the heat flux per unit area $\Phi$ coming out of the slab, i.e., the amount of heat in one second $Q$, per unit area, due to the constant temperature gradient across the slab. This turns out to be given by
$$
Q = \Phi A C \kappa d
$$
or
$$
Q = \Phi
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