Question

A blackbody emits heat at the rate of 20W. When its temperature is ${227^ \circ }C$. Another black body emits heat at the rate of $15W$ when its temperature is ${277^ \circ }C$. Compare the area of the surface of the two bodies, if the surrounding is at NTP:
(A) $16:1$
(B) $1:4$
(C) $12:1$
(D) $1:12$

Answer 1

Here we are given two black bodies that emit different powers at different temperatures. We have to compare the surface area of the two black bodies when the surrounding of the black bodies is at normal temperature and pressure. We have to consider one of the laws for black body radiation to solve this problem.
Formula used
$W = e\sigma A{T^4}$
Where, $W$ stands for the power emitted by the black body, $e$ stands for the emissivity of the black body, $\sigma$ stands for the Stefan’s constant, $A$ stands for the surface area of the black body, and $T$ stands for the temperature of the black body.

Complete step by step solution:
According to the Stefan-Boltzmann law, the amount of power radiated per unit time for a unit area of the black body is directly proportional to the fourth power of the temperature.
It can be expressed as,
$W = e\sigma A{T^4}$
The radiation emitted by the first black body can be written as,
${W_1} = e\sigma {A_1}T_1^4$
Where ${T_0}$is the temperature of the surrounding.
The radiation emitted by the second black body can be written as,
${W_2} = e\sigma {A_2}T_2^4$
Comparing the powers of the two black bodies,
$\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{e\sigma {A_1}T_1^4}}{{e\sigma {A_2}T_2^4}}$
Cancelling out the same values, we get
$\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{{{A_1}T_1^4}}{{{A_2}T_2^4}}$
From this we can write the comparison of the areas as,
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{{W_1}T_2^4}}{{{W_2}T_1^4}}$
From the question, we know that
${W_1} = 20W$
${W_2} = 15W$
${T_1} = {227^ \circ }C = 500K$
${T_2} = {277^ \circ }C = 550K$
(To convert degree Celsius to Kelvin we have to add the$273K$with the given temperature.)
Substituting the values in the above equation, we get that,
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{20 \times {{550}^4}}}{{15 \times {{500}^4}}}$
After solving, we get the ratio as, $2:1$

The correct answer is not given in the option.

Note:
A blackbody will absorb all radiation incident on it without considering the frequency or the wavelength of the incident radiation. Since it absorbs all the radiations, it will appear black in color hence it is called a black body. The black body will also emit radiations of a certain frequency which are called blackbody radiations.