Question

A copper is suspended in an evacuated chamber maintained at $300 K$. The sphere is maintained at a constant temperature of $500K$ by heating it electrically. A total of $210W$ of electric power is needed to do it. When the surface of the copper sphere is completely blackened, $700W$ is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.
A. $3$
B. $0.3$
C. $30$
D. $300$

Answer 1

Use Stefan’s law to express the emissive power of the copper sphere due to increase in temperature. For the perfectly black body, the emissivity of the copper is 1. Again, express the emissive power when the power supplied is 700 W. Solve these two equations and find the value of emissivity.

Formula used:
Stefan’s law, $P = eA\sigma \left( {{T^4} - T_0^4} \right)$
Here, e is the emissivity of the material, A is the area of the material T is the final temperature and ${T_0}$ is the reference temperature and $\sigma$ is the Stefan-Boltzmann constant.

Complete step by step answer:
We have given the initial temperature of the copper sphere, ${T_0} = 300\,{\text{K}}$ and final temperature of the copper sphere is ${T_1} = 500\,{\text{K}}$ and the power required to do it is${P_1} = 210\,{\text{W}}$.

Let’s express the heat power emitted by the copper sphere when the electric power needed is 210 W,
${P_1} = eA\sigma \left( {T_1^4 - T_0^4} \right)$
Here, e is the emissivity of the material, A is the area of the copper sphere and $\sigma$ is the Stefan-Boltzmann constant.
Substituting ${P_1} = 210\,{\text{W}}$, ${T_1} = 500\,{\text{K}}$ and ${T_0} = 300\,{\text{K}}$ in the above equation, we get,
$210 = eA\sigma \left( {{{\left( {500} \right)}^4} - {{\left( {300} \right)}^4}} \right)$ …… (1)

Let’s express the heat power emitted by the copper sphere when the electric power needed is 700 W,
${P_2} = eA\sigma \left( {T_1^4 - T_0^4} \right)$
Substituting ${P_2} = 700\,{\text{W}}$, ${T_1} = 500\,{\text{K}}$ and ${T_0} = 300\,{\text{K}}$ in the above equation, we get,
$700 = eA\sigma \left( {{{\left( {500} \right)}^4} - {{\left( {300} \right)}^4}} \right)$

At this power, we have given that the copper sphere is completely blackened. We know that the emissivity of the perfectly back body is 1.

Therefore, substituting 1 for e in the above equation, we get,
$700 = A\sigma \left( {{{\left( {500} \right)}^4} - {{\left( {300} \right)}^4}} \right)$ …… (2)
Dividing equation (1) by equation (2), we get,
$\dfrac{{210}}{{700}} = \dfrac{{eA\sigma \left( {{{\left( {500} \right)}^4} - {{\left( {300} \right)}^4}} \right)}}{{A\sigma \left( {{{\left( {500} \right)}^4} - {{\left( {300} \right)}^4}} \right)}}$
$\therefore e=0.3$

Therefore, the emissivity of the copper sphere is 0.3. So, the correct answer is option B.

Note: Always remember, the emissivity of any material lies between 0 and 1. Therefore, the emissivity of the copper sphere cannot be greater than 1 as given in other options. Here, we have assumed that the sphere does not undergo expansion with increasing temperature. Otherwise, the area of the sphere would increase.