## Temperature Distribution Throughout a Hollow Sphere

Suppose a hollow sphere of inner radius

$a$
at temperature
$T_q$
$b$
at temperature
$T$
is amde of material of thermal conductivity
$\kappa$
.

We want to find the temperature as a function of the radius - by symmetry the temperature will depend only on the radius and the normal will be the unit radial vector

$\mathbf{n} = \mathbf{e_r}$
. The nrate of heat flow is
$H= \int_s (- \mathbf{\nabla} T) \cdot \mathbf{n} dS= \int_s - \frac{\partial T}{\partial r} \mathbf{e_r}\cdot \mathbf{e_r} dS- =-4 \pi r^2 \kappa \frac{\partial T}{\partial r}$

When the temperature is steady
$H$
will be constant. We can write
$\frac{H}{r^2} dr = -4 \pi \kappa dT \rightarrow \int^r_a \frac{H}{r^2} dr = -4 \pi \kappa \int^T_{T_1} dT$

Hence
$-H(\frac{1}{r} - \frac{1}{a}) = -4 \pi \kappa (T-T_1) \rightarrow T= (-H(- \frac{1}{r} + \frac{1}{a}) +T_a$

When
$r=b, T=T_2$

$H(\frac{1}{b} - \frac{1}{a}) = 4 \pi \kappa (T_2-T_1) \rightarrow H= \frac{ 4 \pi \kappa (T_2-T_1)}{(\frac{1}{b} - \frac{1}{a})}$

Then
$-H(\frac{1}{r} - \frac{1}{a}) = -4 \pi \kappa (T-T_1) \rightarrow T= (\frac{ (T_2-T_1)}{(\frac{1}{a} \frac{1}{r})}( \frac{1}{a} - \frac{1}{b}) +T_a$