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A metal rod at a temperature of $145^{\circ} \mathrm{C}$, radiates energy at a rate of 17 W . If its temperature is increased to $273^{\circ} \mathrm{C}$, then it will radiate at the rate of
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49.6 W
Given, initial temperature of metal $\operatorname{rod}\left(T_1\right)=145^{\circ} \mathrm{C}$ $=418 \mathrm{~K}$
Rate of radiated energy $\left(E_1\right)=17 \mathrm{~W}$
Final temperature $\left(T_2\right)=273^{\circ} \mathrm{C}=273+273=546 \mathrm{~K}$
We know, from the Stefan's law, $E \propto T^4$
$\begin{aligned} & \text { or } \quad \frac{E_1}{E_2}=\left(\frac{T_1}{T_2}\right)^4=\left(\frac{418}{546}\right)^4 \\ & \therefore \quad \frac{E_1}{E_2}=0.343 \\ & \end{aligned}$
Therefore, final radiated energy, $E_2=\frac{E_1}{0.343}$
$=\frac{17}{0.343}=49.6 \mathrm{~W}$
Rate of radiated energy $\left(E_1\right)=17 \mathrm{~W}$
Final temperature $\left(T_2\right)=273^{\circ} \mathrm{C}=273+273=546 \mathrm{~K}$
We know, from the Stefan's law, $E \propto T^4$
$\begin{aligned} & \text { or } \quad \frac{E_1}{E_2}=\left(\frac{T_1}{T_2}\right)^4=\left(\frac{418}{546}\right)^4 \\ & \therefore \quad \frac{E_1}{E_2}=0.343 \\ & \end{aligned}$
Therefore, final radiated energy, $E_2=\frac{E_1}{0.343}$
$=\frac{17}{0.343}=49.6 \mathrm{~W}$
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