Question

A body of surface area $5{\text{c}}{{\text{m}}^2}$ and temperature $727^\circ {\text{C}}$ emits $300{\text{J}}$ of energy per minute. Find its emissivity.
A) $0.18$
B) $0.28$
C) $0.81$
D) $1$

Answer 1

Hint: Stefan-Boltzmann law gives the energy emitted per unit time by a body to be directly proportional to the surface area of the body, its emissivity and the fourth power of its temperature. The emissivity of a body essentially refers to the capability of a material to emit energy.

Formula used:
The energy emitted for the time $t$ by a body is given by, $E = e\sigma {T^4}At$ where $e$ is the emissivity of the body, $\sigma $ is the Stefan-Boltzmann constant, $T$ is the temperature of the body and $A$ is the surface area of the body.

Complete step by step answer:
Step 1: List the parameters given in the question.
The temperature of the body is given to be $T = 727^\circ {\text{C}}$ .
The surface area of the body is given to be $A = 5{\text{c}}{{\text{m}}^2}$ .
The energy emitted per minute is given to be $E = 300{\text{J}}$.
So the time is $t = 1{\text{minute}} = 60{\text{s}}$ .
Let $e$ be the emissivity of the body which is to be determined.

Step 2: Express the relation for the emissivity of the body.
The energy emitted for the time $t$ by the given body can be expressed as $E = e\sigma {T^4}At$
$ \Rightarrow e = \dfrac{E}{{t\sigma {T^4}A}}$
Thus the emissivity of the body is given by, $e = \dfrac{E}{{t\sigma {T^4}A}}$ ------- (1)
Substituting for $T = 1000{\text{K}}$, $A = 5 \times {10^{ - 4}}{{\text{m}}^2}$, $E = 300{\text{J}}$, $t = 60{\text{s}}$ and $\sigma = 5.67 \times {10^{ - 8}}{\text{W}}{{\text{m}}^{ - 2}}{{\text{K}}^{ - 4}}$ in equation (1) we get, $e = \dfrac{{300}}{{60 \times 5 \cdot 67 \times {{10}^{ - 8}}{{\left( {1000} \right)}^4} \times 5 \times {{10}^{ - 4}}}} = 0.18$
Thus we obtain the emissivity of the body as $e = 0.18$ .

So the correct option is (A).

Note: Here, the temperature $T = 727^\circ {\text{C}}$ is given in the Celsius scale so we converted it into the Kelvin scale as $T = 727^\circ {\text{C}} = 727 + 273 = 1000{\text{K}}$ before substituting in equation (1). Also, we have converted the surface area of the body as $A = 5 \times {10^{ - 4}}{{\text{m}}^2}$ before substituting in equation (1). The value of the Stefan-Boltzmann constant is known to be $\sigma = 5.67 \times {10^{ - 8}}{\text{W}}{{\text{m}}^{ - 2}}{{\text{K}}^{ - 4}}$. If the body was a perfect black body its emissivity would be one.