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For a black body at temperature $727^\circ C$, its radiating power is 60 watt and temperature of surrounding is $227^\circ C$. If temperature of black body is changed to $1227^\circ C$ then its radiating power will be : -
A
304 W
B
320 W
C
240 W
D
120 W
Detailed Solution
Net power radiated by a black body in surroundings at $T_0$: $P \propto (T^4 - T_0^4)$ (Stefan's law).
Convert to kelvin: $T_1 = 1000$ K, $T_2 = 1500$ K, $T_0 = 500$ K.
$\dfrac{P_2}{P_1} = \dfrac{1500^4 - 500^4}{1000^4 - 500^4} = \dfrac{500^4(3^4 - 1)}{500^4(2^4 - 1)} = \dfrac{80}{15}$
$\dfrac{P_2}{60} = \dfrac{80}{15} \Rightarrow P_2 = 320$ W
So the radiating power becomes 320 W.
Convert to kelvin: $T_1 = 1000$ K, $T_2 = 1500$ K, $T_0 = 500$ K.
$\dfrac{P_2}{P_1} = \dfrac{1500^4 - 500^4}{1000^4 - 500^4} = \dfrac{500^4(3^4 - 1)}{500^4(2^4 - 1)} = \dfrac{80}{15}$
$\dfrac{P_2}{60} = \dfrac{80}{15} \Rightarrow P_2 = 320$ W
So the radiating power becomes 320 W.