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

Answer

A

Explanation

When parallel rays are incident on sphere, then energy is effectively absorbed through the projected area perpendicular to the beam and that is `pi r^2` , where r is radius of sphere.
Rate of heat input: `U_1 = pi r^2 I = 912 pi r^2`
Note that radiation and absorption is through complete surface area `(4pi r^2)` . If T is temperature of sphere, then
Net rate of heat loss: `U_2 = sigma (4pi r^2) [T^4 - (300)^4 ]`
For steady state
`U_1 = U_2`
` 912 pi r^2 = sigma (4pi r^2) [T^4 (300)^4]`
` rArr T = [ (300)^4 + (912)/(4 xx 5.7 xx 10^(-8) )]^(1//4) = 330 K`