Question

A spherical body of radius $b=\frac{4}{\pi }$ $\text{m}$ has a concentric cavity of radius $a=\frac{2}{\pi }$ $\text{m}$ as shown in the figure. Thermal conductivity of the material is $100{\text{W/m}}^{\circ }\text{C}$. If temperature of inner surface is kept at ${70}^{\circ }\text{C}$ and of the outer surface at ${30}^{\circ }\text{C}$, then find the rate of heat flow. (in $\text{kW}$)

• A. $16$
• B. $32$
• C. $64$
• D. $40$

Answer 1

The correct option is C $64$

Given,
Radius of spherical body $\left(b\right)=\frac{4}{\pi }\text{m}$
Radius of spherical cavity $\left(a\right)=\frac{2}{\pi }\text{m}$
Thermal conductivity of material $\left(K\right)=100{\text{W/m}}^{\circ }\text{C}$
Inner surface temperature $\left(T_1\right)={70}^{\circ }\text{C}$
Outer surface temperature $\left(T_2\right)={30}^{\circ }\text{C}$
Then,
Rate of heat flow $\left(\frac{dQ}{dt}\right)=\frac{\Delta T}{R}$
Thermal resistance offered by a spherical shell, having inner radius $a$ and outer radius $b$ is given by
$\left(R\right)=\frac{1}{4\pi K}(\frac{1}{a}-\frac{1}{b})$
$R=\frac{1}{4\pi \times 100}(\frac{\pi }{2}-\frac{\pi }{4})\Rightarrow R=\frac{1}{1600}{}^{\circ }\text{C/W}$
$\because \Delta T=T_1-T_2={40}^{\circ }\text{C}$
$\Rightarrow \frac{dQ}{dt}=1600\times \left(40\right)=64\text{kW}$
Thus, option (c) is the correct answer.