The rate of radiation of a black body at \(0^{\circ} \mathrm{C}\) is \(E \mathrm{Js}^{-1}\). The rate of…
The rate of radiation of a black body at \(0^{\circ} \mathrm{C}\) is \(E \mathrm{Js}^{-1}\). The rate of radiation of the black body at \(273^{\circ} \mathrm{C}\) will be
\(E \mathrm{Js}^{-1}\)
\(4 E \mathrm{Js}^{-1}\)
\(\frac{E}{2} \mathrm{Js}^{-1}\)
\(16 E \mathrm{Js}^{-1}\)
Solution
Given, \(T_1=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}\)
\(T_2=273^{\circ} \mathrm{C}=(273+273) \mathrm{K}=546 \mathrm{~K}\)
According to Stefan-Boltzmann's law, rate of radiation,
\(\begin{array}{ll}
& E \propto T^4 \\
\therefore & \frac{E_2}{E_1}=\left(\frac{T_2}{T_1}\right)^4=\left(\frac{546}{273}\right)^4=(2)^4=16 \\
\Rightarrow & E_2=16 E_1=16 E \mathrm{Js}^{-1} \quad\left[\because E_1=E \mathrm{Js}^{-1}\right]
\end{array}\)