If the temperature of a black body increases from 7∘C to 287∘C, then the rate of emission of radiation energy is:
A
8 times
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
16 times
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2 times
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4 times
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B16 times For black body radiation E=σT4 [E is energy radiated per unit time per unit area, T is temperature of the body] E2E1=(T2T1)4⇒E2E1=(273+287273+7)4 =(560280)4=161⇒E2=16E1