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If the temperature of a black body increases from $7\,^oC$ to $287\,^oC$ then the rate of energy radiation increases by
${\left( {\frac{{287}}{7}} \right)^4}$
$16$
$4$
$2$
Solution
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 constant. 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)
$\therefore \frac{E_{2}}{E_{1}}=\frac{T_{2}^{4}}{T_{1}^{4}}$
$\Rightarrow \mathrm{E}_{2}=\left(\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}\right)^{4} \mathrm{E}_{1}$
$\mathrm{T}_{2}=287^{\circ} \mathrm{C}=287+273=560 \mathrm{K},$
$\mathrm{T}_{1}^{2}=7^{\circ} \mathrm{C}=7+273=280 \mathrm{K},$
$\therefore \mathrm{E}_{2}=\left(\frac{560}{280}\right)^{4} \mathrm{E}_{1}=2^{4} \mathrm{E}_{1}=16 \mathrm{E}_{1}$
Rate of energy radiated increases by
$16$ times.
