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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

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

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

1 Answer

Answer : $\;10^{3}K$
Explanation :
By stefan's law
energy radiated per second / unit area = $\; \sigma T^{4}$
$5.67 \times 10^{4} W/m^{2} = 5.67 \times 10^{-8} \times T^{4}$
$T^{4}=10^{12}$
$T = (10^{12})^{\large\frac{1}{4}}= 10^{3}K$
answered Mar 26, 2014 by yamini.v
 

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