Question

The rectangular $8cm \times 4cm$ of a black body at a temperature of ${127^o}C$ emits energy at the rate of E per second. If the length and breadth of the surface are each reduced to half of the initial value and the temperature is raised to ${327^o}C$ , the rate of emission energy will become:
(A) $\dfrac{3}{8}E$
(B) $\dfrac{{81}}{{16}}E$
(C) $\dfrac{9}{{16}}E$
(D) $\dfrac{{81}}{{64}}E$

Answer 1

Hint
According to Stefan’s law i.e., $E = \sigma A{T^4}$
Where $\sigma$ is Stefan’s constant
A is the area of the blackbody, T is the temperature of the blackbody.
Since it is given the area is halved and the temperature is changed from ${127^o}C$ to ${327^o}C$ . We take the ratio of their Energies and find the relation between the energy emissions.

Complete step by step answer
From Stefan’s law,
$\Rightarrow \dfrac{E}{{{E'}}} = \dfrac{A}{{{A'}}}{\left( {\dfrac{T}{{{T'}}}} \right)^4}$ ……(i)
Where ${E'},{A'},{T'}$ are the new energy, area and temperature respectively?
${A'} = \dfrac{A}{4}$ Because it is given both the length and breadth are reduced by half.
Since the area of a rectangle is length $\times$ Breadth, the new area will become one-fourth of the older area.
And the temperature is given in $^oC$ and we have to convert it into K because $\sigma$ units are ${m^{ - 2}}{K^{ - 4}}$ .
So, $T = 127 + 273 = 400K$
And ${T'} = 327 + 273 = 600K$
Substituting all the values in eq. (i)
We get $\dfrac{E}{{{E'}}} = \dfrac{A}{{{A'}}}{\left( {\dfrac{T}{{{T'}}}} \right)^4} = 4{\left( {\dfrac{{400}}{{600}}} \right)^4} = 4 \times {\left( {\dfrac{2}{3}} \right)^4} = 4 \times \dfrac{{16}}{{81}} = \dfrac{{64}}{{81}}$
$\Rightarrow {E'} = \dfrac{{81}}{{64}}E$
Hence option (D) is correct.

Note
In this answer we have to convert $^0C$ into K because $\sigma$ units are in ${m^{ - 2}}{K^{ - 4}}$ and also while considering the new area it will be one-fourth of the older area because it is given that both length and breadth are halved. These are the places where the majority of the students go wrong.