Question

A black rectangular surface of area ‘A’ emits ‘E’ per second at ${27^ \circ }C$. If length and breadth are reduced to $\dfrac{1}{3}rd$ of initial value and temperature is raised to ${327^ \circ }C$ the energy emitted per second becomes
A. $\dfrac{{4E}}{9}$
B. $\dfrac{{7E}}{9}$
C. $\dfrac{{10E}}{9}$
D. $\dfrac{{16E}}{9}$

Answer 1

Energy radiated per second from a body surface is given by $E = \sigma eA{T^4}$, where $\sigma$ Stefan’s constant. A is the area of the surface, $e$ is the emissivity, and $T$ is the temperature.
Stefan’s law states that the energy radiated of a black body is proportional to its absolute temperature, T raised to the fourth power. Stefan’s-Boltzmann law states the total intensity radiated over all wavelength increases as the temperature increases.

Complete step by step answer:
The initial temperature of the body $T = {27^ \circ }C = 300K$
Since the energy radiated per second from a body surface is given by $E = \sigma eA{T^4}$
Hence the energy radiated per second from a body surface at an initial temperature will be
$E = \sigma eA{\left( {300} \right)^4}$
Now the length and breadth of the rectangular surface is reduced by $\dfrac{1}{3}$, and the temperature is raised to ${327^ \circ }C$
So the new area of the body $A' = \dfrac{L}{3}.\dfrac{B}{3} = \dfrac{A}{9}$, hence we can say the area of the body reduce by $\dfrac{1}{9}$ times.
The new temperature of the body $T' = {327^ \circ }C = 600K$
Hence the energy radiated when the area reduces by $\dfrac{1}{9}$times and at new temperature $600K$will be $E' = \sigma e\dfrac{A}{9}{\left( {600} \right)^4}$
So the ratio of the radiation of energy when the area is reduced will be

$$\dfrac{{E'}}{E} = \dfrac{{\sigma e\dfrac{A}{9}{{\left( {600} \right)}^4}}}{{\sigma eA{{\left( {300} \right)}^4}}} \\\ = \dfrac{{2.2.2.2}}{9} \\\ = \dfrac{{16}}{9} \\\$$

Hence the energy emitted per second will be$E' = \dfrac{{16}}{9}E$
Option D is correct.

Note: It should be noted down here that the temperature should always be in the units of Kelvin and not in the Celsius. In the question, all the temperatures are given in Celsius only so we need to convert it to kelvin first.