Question

Heat flows radially outward through a spherical shell of outside radius $R_2=3m$ and inner radius $R_1=1m$. The temperature of inner surface of shell is ${\theta }_1$ and that of outer is ${\theta }_2$. If the radial distance from the center of the shell at which the temperature is just half way between ${\theta }_1$ and ${\theta }_2$ is given by $\frac{X}{2}$. Then $X$ will be

Answer 1

Cut a differential shell of radius $r$ with width $dr$

$H=\frac{dQ}{dt}=\frac{\left(KA\right)d\theta }{dr}$

$d\theta$ = difference in temperature

dr=distance traveled by heat

K= Thermal conductivity of shell

A= area

$H=\left(4K\pi r^2\right)\frac{d\theta }{dr}$

$\frac{H}{4\pi K}\int \frac{1}{r^2}dr=\int d\theta .............\left(1\right)$

We will apply the equation (1) for two times,

1. Firstly for the range $R_1$ to $R_2$

2. And Secondly for $R_1$ to $r$

For $R_1$ to $R_2$

$\frac{H}{4\pi K}{\int }_{R_1}^{R_2}\frac{dr}{r^2}={\int }_{{\theta }_1}^{{\theta }_2}d\theta$

On solving

$\frac{H}{4\pi K}[\frac{1}{R_1}-\frac{1}{R_2}]={\theta }_2-{\theta }_1$

$\frac{H}{4\pi K}\frac{(R_2-R_1)}{R_1{R}_2}={\theta }_2-{\theta }_1...............\left(2\right)$

We know that heat flow at a constant rate so H is constant

Now solving the equation for the second case (i.e., from $R_1$ to $r$

$\frac{H}{4\pi K}{\int }_{R_1}^r\frac{dr}{r^2}={\int }_{{\theta }_1}^{\frac{{\theta }_1+{\theta }_2}{2}}d\theta$

$\frac{H}{4\pi K}[\frac{1}{R_1}-\frac{1}{r}]=\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_1$

$\frac{H}{4\pi K}\frac{(r-R_1)}{r{R}_1}=\frac{{\theta }_2-{\theta }_1}{2}.................\left(3\right)$

On dividing (2) by (3), we will get

$\frac{R_2-R_1}{R_2{R}_1}\times \frac{r{R}_1}{r-R_1}=2$

On solving, we will get

$r=\frac{2R_1{R}_2}{R_1+R_2}$

On putting $R_1=1m$ and $R_2=3m$, we will get

$r=\frac{2\times 1\times 3}{1+3}$

$r=\frac{3}{2}$

Hence $X=3$