A black body at a high temperature $T$ radiates energy at the rate of $U$ (in ${Wm}^{- 2}$). When the temperature falls to half ($ie$,$\frac{T}{2}$), the radiated energy (in ${Wm}^{- 2}$) will be

Question

Dr. Abubacker · Accepted Answer

(b) According to Stefan’s law of radiation
$U \propto T^{4}$ $(\because U\ is\ the\ energy)$
$\Rightarrow$ $\frac{U_{1}}{U_{2}} = \left( \frac{T_{1}}{T_{2}} \right)^{4}$
$\frac{U_{1}}{U_{2}} = \left( \frac{T}{\frac{T}{2}} \right)^{4}$ ($\because\ T_{2\ } = \frac{T}{2})$
Or $\frac{U_{1}}{U_{2}} = \left( \frac{2}{1} \right)^{4}$
Or $\frac{U_{1}}{U_{2}} = \left( \frac{16}{1} \right)$
Or $U_{2\ }$=$\ \frac{U_{1}}{16}$
$\Rightarrow$ $U_{2\ }$=$\ \frac{U}{16}$ ($\because\ U_{1\ } = U)$

Q. A black body at a high temperature $T$ radiates energy at the rate of $U$ (in ${Wm}^{- 2}$). When the temperature falls to half ($ie$,$\frac{T}{2}$), the radiated energy (in ${Wm}^{- 2}$) will be

Explanation

Reviewed by Dr. Abubacker

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