For a black body at temperature 727^{circ} C, | vedclass

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(B) According to Stefan-Boltzmann Law,the net power radiated by a black body is given by $P = \sigma A (T^4 - T_0^4)$,where $T$ is the temperature of

Vedclass · Accepted Answer

$320$

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(B) According to Stefan-Boltzmann Law,the net power radiated by a black body is given by $P = \sigma A (T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.Given:$T_1 = 727^{\circ} C = 727 + 273 = 1000 \; K$$T_0 = 227^{\circ} C = 227 + 273 = 500 \; K$$P_1 = 60 \; W$$P_1 = k(T_1^4 - T_0^4) \Rightarrow 60 = k(1000^4 - 500^4) \quad \dots(1)$Now,$T_2 = 1227^{\circ} C = 1227 + 273 = 1500 \; K$$P_2 = k(T_2^4 - T_0^4) \Rightarrow P_2 = k(1500^4 - 500^4) \quad \dots(2)$Dividing equation $(2)$ by $(1)$:$\frac{P_2}{60} = \frac{1500^4 - 500^4}{1000^4 - 500^4} = \frac{500^4 (3^4 - 1^4)}{500^4 (2^4 - 1^4)}$$\frac{P_2}{60} = \frac{81 - 1}{16 - 1} = \frac{80}{15} = \frac{16}{3}$$P_2 = 60 	imes \frac{16}{3} = 20 	imes 16 = 320 \; W$.

For a black body at temperature $727^{\circ} C$,its radiating power is $60\; W$ and the temperature of the surrounding is $227^{\circ} C$. If the temperature of the black body is changed to $1227^{\circ} C$,then its radiating power will be ..... $W$.

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