Question

Two thin metallic spherical shells of radii $20\, cm$ and $30\, cm$, respectively are placed with their centres coinciding. A material of thermal conductivity $\alpha$ is filled in the space between the shells. The inner shell is maintained at $300\, K$ and the outer shell at $310 \,K .$ If the rate at which heat flows radially through the material is $40\, W$, find the value of $\alpha$ (in units of $\left.J \,s ^{-1} \,m ^{-1} \,K ^{-1}\right)$.

• A. $\frac{3}{\pi}$
• B. $\frac{4 \pi}{3}$
• C. $\frac{5}{3 \pi}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (C)

Given, radius of two thin metallic spherical shells
$r_{1}=20 \,cm =0.2 \,m$
$r_{2}=30 \,cm =0.3 \,m$
temperature of inner shell, $T_{1}=300 \,K$,
temperature of outer shell, $T_{2}=310 \, K$,
and rate of heat flow, $H=40 \, W$.
Radial rate of flow of heat through the shell in the steady state,
$H =\frac{d Q}{d t}=\alpha A \frac{d T}{d r}=\alpha\left(4 \,\pi r^{2}\right) \frac{d T}{d r} $
$\frac{d r}{r^{2}} =\frac{4 \,\pi \alpha}{H} \cdot d T$
Integrate on both sides, we get
$\Rightarrow \int\limits_{r_{1}}^{r_{2}} \frac{d r}{r^{2}}=\frac{4 \,\pi \alpha}{H} \int\limits_{T_{1}}^{T_{2}} d T$
$\frac{r_{2}-r_{1}}{r_{1} r_{2}}=\frac{4 \pi \alpha}{H}\left(T_{2}-T_{1}\right) $
$\therefore \alpha=\frac{H\left(r_{2}-r_{1}\right)}{4 \,\pi \, r_{1} \,r_{2}\left(T_{2}-T_{1}\right)}$
Putting the given values, we get
$\alpha=\frac{40(0.3-0.2)}{4 \pi(0.3)(0.2)(310-300)}=\frac{4}{2.4 \,\pi}=\frac{5}{3 \,\pi}$
Hence, the value of $\alpha=\frac{5}{3 \pi} J \,s ^{-1} \,m ^{-1} \, K ^{-1}$.