Q.

Parallel rays of light of intensity I=912Wm2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take Stefan-Bolzmann constant σ=5.7×108 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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a

330 K

b

1550 K

c

660 K

d

990 K

answer is A.

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Detailed Solution

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

σ4πr2T4(300)4=912×πr2
 Now, σ×T4=σ(300)4+9124T4=(300)4+9124×5.7×108=(300)4+91222.8×108=(300)4+40×108=(81+40)×108=121×108 T=330K

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