Question
Physics Question on radiation
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 constant σ=5.7×1−8Wm−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
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(T4−T04)4πR2⇒i=σ(T4−T04)4 ⇒T4−T04=40×108⇒T4−81×108=40×18 ⇒T4=121×108⇒T=330K