JEE Advanced | 2014Heat TransferMedium

Question

Parallel rays of light of intensity I = 912 Wm-2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Boltzmann constant σ = 5.7 × 10-8 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

Solution

σ × 4πR2.(T4 - T04) = 912 × πR2
T4 - T04 = = 40 × 108
T4 = 40 × 108 + (300)4 = (40 + 81) × 108
T = 330 K

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