Question

If the temperature of a black body increases from $7^{o}C$ to ${287}^{o}C$ then the rate of energy radiation increases by

• A. ${\left(\frac{287}{7}\right)}^4$
• B. $16$
• C. $4$
• D. $2$

Answer 1

The correct option is B $16$

By Stefan's law, energy radiated per sec by a black body is given by $E=A\sigma T^4$
Where $A=$ area of black body, $\sigma =$ Stefan's consant. For a black body at temperature $T_1,E_1=A\sigma T_1^4$ at $T_2,E_2=A\sigma T^4$
(Since $A,\sigma$ all same)
$\frac{E_2}{E_1}=\frac{T_2^4}{T_1^4}$
$\Rightarrow$ $E_2={\left(\frac{T_2}{T_1}\right)}^4{E}_1$
$T_2={287}^{o}C=287+273=560K$
$T_1=7^{o}C=7+273=280K$,
$\therefore$ $E_2={\left(\frac{560}{280}\right)}^4{E}_1=2^4{E}_1=16E_1$
$\therefore$ Rate of energy radiated increases by $16$ times