Question

A hollow metallic sphere of radius 20 cm surrounds a concentric metallic sphere of radius 5 cm. The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at ${50}^{\circ }C$ and ${10}^{\circ }C$ respectively and it is found that 100 J of heat second. Find the thermal conductivity of the material between the spheres.

Answer 1

$a=r_1=5cm=0.05m,$

$b=r_2=20cm=0.20m$

${\theta }_1={50}^{\circ }C,{\theta }_2={10}^{\circ }$

Now, considering a small strip of thickness 'dr' at a distance 'r'

A =$4\pi r_2$

H = $\frac{-4\pi r^{2}kd\theta }{dr}$

-ve because as 'r' increase 'Q' decreases

$\Rightarrow {\int }_a^b\frac{dr}{r^2}=\frac{-4\pi K}{H}{\int }_a^{b}d\theta$

On integration

H =$\frac{dQ}{dt}=K\frac{4\pi ab({\theta }_1-{\theta }_2)}{(b-a)}$

Putting the values we get

$\frac{K\times 4\times 3.14\times 5\times 20\times 40\times {10}^{-3}}{15\times {10}^{-2}}=100$

$\Rightarrow K=\frac{15}{4\times 3.14\times 4\times {10}^{-1}}$

= 2.985 = 3$W/m{-}^{\circ }C$