Question

A black body is heated from ${27}^{o}C$ to ${127}^{o}C$. The ratio of their energies of radiation emitted will be:

• A. $9:16$
• B. $27:64$
• C. $81:256$
• D. $3:4$

Answer 1

Answer: (C)

$81:256$

We know that
$E=\sigma T^4$
Where $E$ is rate of emission of radiation of a body at temperature $T$.
$E_1=\sigma {(27+273)}^2$
$E_2=\sigma {(127+273)}^2$
$\frac{E_1}{E_2}=\frac{{\left(300\right)}^4}{{\left(400\right)}^4}=\frac{81}{256}$