Question

At SP-WB(TV)-PH-85 two black bodies at 327℃ and 627℃ are suspended in an environment at 27℃. The ratio of their emissive powers is
A. 15:8
B. 16:3
C. 3:16
D. 5:8

Answer 1

The Emissive power of a body at a given temperature is the quantity of radiant energy emitted by the body per unit time per unit surface area of the body at that temperature. Using the Stefan Boltzmann Law, we can find out the required ratio.

Complete step by step answer:
Radiation is a process of transferring heat energy from one body to another without heating the intervening medium.
A perfectly black body is the one that absorbs all the heat radiations irrespective of the wavelength, incident on it. It doesn’t reflect or transmit any incident radiation due to which it appears completely black.
Energy radiated per unit time per unit surface area of the body is called emissive power.
$E = \dfrac{Q}{{\Delta A\Delta t}}$
Now According to Stefan Boltzmann Law, the amount of radiation emitted per unit time from an area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature.
$u = sA{T^4}$
Where s is Stefan’s constant, $s = 5.67 \times {10^{ - 8}}W/{m^2}{k^4}$ and $u$ is emissive power.
For a non-black body, $u = e\sigma A{T^4}$
Where e is the emissivity which lies between 0 to 1.
With the surrounding temperature ${T_0}$, net energy radiated by an area A per unit time.
$\Delta u = u - u^\circ = e\sigma A\left[ {{T^4} - T_0^4} \right]$
The Emissive power of the black body at 327℃ or 600K in an environment at 27℃ or 300K,
$u = e\sigma A(T_1^4 - T_0^4)$
The Emissive power of the black body at 627℃ or 900K in an environment at 27℃ or 300K,
${u_2} = e\sigma A{\left( {T_2^4 - T_0^4} \right)^4}$
Dividing the above two equations we get,
$\dfrac{{{u_1}}}{{{u_2}}} = \dfrac{{T_2^4 - T_0^4}}{{T_1^4 - T_0^4}} \\\ \implies \dfrac{{{u_1}}}{{{u_2}}} = \dfrac{{{{\left( {600} \right)}^4} - {{\left( {300} \right)}^4}}}{{{{\left( {900} \right)}^4} - {{\left( {300} \right)}^4}}} \\\ \implies \dfrac{{{u_1}}}{{{u_2}}} = \dfrac{{{{(6)}^4} - {{(3)}^4}}}{{{{(9)}^4} - {{(3)}^4}}} \\\ \implies \dfrac{{{u_1}}}{{{u_2}}} = \dfrac{{{3^4}({2^4} - 1)}}{{{3^4}({3^4} - 1)}} \\\ \implies \dfrac{{{u_1}}}{{{u_2}}} = \dfrac{{16 - 1}}{{81 - 1}} = \dfrac{{15}}{{80}} = \dfrac{3}{{16}} \\\$
Thus the ratio of their emissive powers is 3:16.

So, the correct answer is “Option C”.

Additional Information:
Stefan Boltzmann Law establishes a relation between the temperature of the black body and the amount of power it emits per unit area. The law states that; The total energy emitted/radiated per unit surface area of the black body across all the wavelengths per unit is directly proportional to the fourth power of the black body’s thermodynamic temperature.

Note:
The temperature first has to be converted into kelvins before applying in the equation of Stefan Boltzmann law otherwise you may not get the correct answer. In reality, a perfect black body doesn’t exist. The lampblack and platinum black are an approximation of the black body as they absorb 99% of the incident radiation.