Question

The temperature of a black body is increased by $50\%$ . Then the percentage of increase of radiation is approximately

• A. 100%
• B. 25%
• C. 400%
• D. 500%

Answer 1

The correct option is C 400%

By $Stefa{n}^{\prime }s$ $Law$,

Power = $\sigma A{T}^4$ for a black body

$\sigma$ is known as Stefan's constant

$\frac{P_1}{P_2}$ = $\frac{T_1^4}{T_2^4}$

$T_2$ = $\frac{3T_1}{2}$

$\frac{T_1^4}{T_2^4}$= $\frac{16}{81}$

By the above equations we get

$P_2$=$\frac{81P_1}{16}$

Percentage increase in radiation = $\frac{P_2-P_1}{P_1}$ x $100$ = $406.25$%