Engineering
Physics
Radiation
Question

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

330 K

660 K

990 K

1550 K

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Solution

I × πR2 = 4πR2 σ (T4 – 3004)

 9124×5.7 × 109 + 3004 = T4

⇒ 4 × 109 + 8.1 × 109 = T4

 121 × 108 = T4

 11×102=T

T = 330 K