A black body radiates energy at the rate of $E\ Wm^{- 2}$ at a high temperature$\ T\ K$. When the temperature is reduced to $\left( \frac{T}{2} \right)$ K, the radiant energy is

Question

Dr. Abubacker · Accepted Answer

(d) According to Stefan’s law,
$E \propto T^{4}$
or $\frac{E_{2}}{E_{1}} = \left( \frac{T_{2}}{T_{1}} \right)^{4}$
or $\frac{E_{2}}{E} = \left( \frac{\frac{T}{2}}{T} \right)^{4} = \left( \frac{1}{2} \right)^{4}$
or $E_{2} = \frac{E}{16}$

Q. A black body radiates energy at the rate of $E\ Wm^{- 2}$ at a high temperature$\ T\ K$. When the temperature is reduced to $\left( \frac{T}{2} \right)$ K, the radiant energy is

Explanation

Reviewed by Dr. Abubacker

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