Question

The rate of emission of radiation of a black body at ${273}^{\circ }C$ is $E$, then the rate of emission of radiation of this body at $0^{\circ }C$ will be

• A. $\frac{E}{16}$
• B. $\frac{E}{4}$
• C. $\frac{E}{8}$
• D. $0$

Answer 1

Answer: (A)

$\frac{E}{16}$

The ratio of energy emission will be directly proportional to fourth power of temperature ratio can be found out from given data.

$\frac{\Delta Q_2}{\Delta Q_1}=\frac{\sigma A{T}_2^4}{\sigma A{T}_1^4}$

$T_1=273+273K=546K;T_2=273K$

substituting the temperature values in the Stefan's law

$\frac{\Delta Q_2}{\Delta Q_1}=\frac{\sigma A{273}^4}{\sigma A{546}^4}=\frac{1}{2^4}$

$\Rightarrow \Delta Q_2=\frac{\Delta Q_1}{16}$

Thus the correct answer is A.