Parallel rays of light of intensity `I=912 WM^-2` are incident on a spherical black body kept in surroundings of temperature 300K. Take Stefan-Boltzmann constant `sigma=5.7xx10^-8`
`Wm^-2K^-4` and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to

Options

A. 330 K
B. 660 K
C. 990 K
D. 1550 K

Answer

A

Explanation

Rate of radiation energy lost by the sphere = rate of radiation energy incident
`=sigma(4pir^(2))[T^(4)-(300)^(4)]=912xxpir^(2)`
`sigmaxxT^(4)=sigma(300)^(4)+(912)/(4)`
`T^(4)=(300)^(4)+(912)/((4xx5.7xx10^(-8)))=(300)^(4)+(912)/(22.8)xx10^(8)`
`=(300)^(4)+(40xx10^(8))=(81+40)xx10^(8)`
`=121xx10^(8)thereforeT=330K`