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Question

Physics Question on radiation

Parallel rays of light of intensity I=912 Wm2^{-2} are incident on a spherical black body kept in surroundings of temperature 300 K . Take Stefan constant σ=5.7×18Wm2K4\sigma =5.7 \times 1^{-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

A

330 K

B

660 K

C

990 K

D

1550 K

Answer

330 K

Explanation

Solution

In steady state Energy incident per second = Energy radiated per second IπR2=σ(T4T04)4πR2i=σ(T4T04)4\therefore \, \, I \pi R^2 =\sigma (T^4 -T_0^4) 4\pi R^2 \, \, \Rightarrow \, \, \, i = \sigma (T^4 -T_0^4)4 T4T04=40×108T481×108=40×18\Rightarrow \, \, \, \, \, \, T^4 -T_0^4 =40 \times 10^8 \, \, \Rightarrow \, \, \, T^4 -81 \times 10^8 =40 \times 1^8 T4=121×108T=330K\Rightarrow \, \, \, \, \, T^4 = 121 \times 10^8 \, \, \Rightarrow \, \, \, T=330 K