A black body radiates 20,W at temperature 227 | vedclass

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(C) According to Stefan-Boltzmann Law,the power radiated by a black body is given by $P = A\sigma T^4$,where $A$ is the surface area,$\sigma$ is the S

Vedclass · Accepted Answer

$320$

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(C) According to Stefan-Boltzmann Law,the power radiated by a black body is given by $P = A\sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature in Kelvin.Given:$P_1 = 20\,W$$T_1 = 227^{\circ}C = 227 + 273 = 500\,K$$T_2 = 727^{\circ}C = 727 + 273 = 1000\,K$Since $P \propto T^4$,we have:$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} ight)^4$Substituting the values:$\frac{P_2}{20} = \left( \frac{1000}{500} ight)^4$$\frac{P_2}{20} = (2)^4$$\frac{P_2}{20} = 16$$P_2 = 16 	imes 20 = 320\,W$Therefore,the radiating power will be $320\,W$.

$A$ black body radiates $20\,W$ at temperature $227^{\circ}C$. If the temperature of the black body is changed to $727^{\circ}C$,then its radiating power will be ..... $W$.

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