Question

A copper sphere is suspended in an evacuated chamber maintained at $300\text{K}$. The sphere is maintained at a constant temperature of $500\text{K}$ by heating it electrically. A total of $300\text{W}$ of electric power is needed to do it. When half of the surface of the copper sphere is completely blackened, $600\text{W}$ is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.

• A. $e=\frac{1}{2}$
• B. $e=\frac{1}{3}$
• C. $e=\frac{1}{4}$
• D. $e=\frac{1}{5}$

Answer 1

The correct option is B $e=\frac{1}{3}$



$T_{chamber}=300\text{K},T_{sphere}=500\text{K}$$P_1=300\text{W},P_2=600\text{W}$

From Stefan-Boltzmann law,
$P=e\sigma A(T^4-T_o^4)$

Applying Stefan-Boltzmann law :
$P_1=\sigma eA(T_{sphere}^4-T_{chamber}^4)$
$300=\sigma eA[{500}^4-{300}^4]....\left(\text{I}\right)$

After half surface is blackend ,
$600=\frac{\sigma eA}{2}[{500}^4-{300}^4]+\frac{\sigma A}{2}[{500}^4-{300}^4]...\left(\text{II}\right)$
[Blackened part of the copper sphere will behave like a black body]

Dividing equation (II) by equation (I)
$2=\frac{(\frac{e}{2}+\frac{1}{2})}{e}\Rightarrow 2e=\frac{e}{2}+\frac{1}{2}$
$\Rightarrow e=\frac{1}{3}$
Hence, option (b) is the correct answer.