# Find the temperature at which a black body radiates energy at the rate of $\;5.67\;W/cm^{2}\;$

$(a)\;10^{2}K\qquad(b)\;10^{3}K\qquad(c)\;10^{4}K\qquad(d)\;10^{5}K$

Answer : $\;10^{3}K$
Explanation :
By stefan's law
energy radiated per second / unit area = $\; \sigma T^{4}$
$5.67 \times 10^{4} W/m^{2} = 5.67 \times 10^{-8} \times T^{4}$
$T^{4}=10^{12}$
$T = (10^{12})^{\large\frac{1}{4}}= 10^{3}K$