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 $e=\frac{1}{3}$


Applying <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Stefan-Boltzmann's Law,
$P=e\sigma A(T^4-T_s^4)$
$T_s=300\text{K},T=500\text{K}$
$\Rightarrow 300=\sigma eA[{500}^4-{300}^4]\dots \dots \left(1\right)$
After half surface is blackened,
$600=\frac{\sigma eA}{2}[{500}^4-{300}^4]+\frac{\sigma A}{2}[{500}^4-{300}^4]\dots \dots \left(2\right)$
[$e$ for black surface $=1$]
Dividing (2) by (1),
$2=\frac{(\frac{e}{2}+\frac{1}{2})}{e}\Rightarrow 2e=\frac{e}{2}+\frac{1}{2}$
or $e=\frac{1}{3}$