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-Bolzmann constant $σ = 5 .7 \times 10^{- 8}$ 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

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330 K

1550 K

660 K

990 K

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Detailed Solution

Rate of radiation energy lost by the sphere = rate of radiation energy incident

$σ (4 {πr}^{2}) [T^{4} - (300)^{4}] = 912 \times {πr}^{2}$
$\begin{array}{l} Now, σ \times T^{4} = σ (300)^{4} + \frac{912}{4} \\ T^{4} = (300)^{4} + \frac{912}{(4 \times 5 .7 \times 10^{- 8})} = (300)^{4} + \frac{912}{22 .8} \times 10^{8} = (300)^{4} + (40 \times 10^{8}) = (81 + 40) \times 10^{8} \\ = 121 \times 10^{8} \end{array}$ $∴ T = 330 K$