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Q. Energy is being emitted from the surface of a black body at 127C the rate of 1.0×106J/sm2. The temperature of black body at which the rate of energy emission is 16.0×106J/sm2 will be

JIPMERJIPMER 2003

Solution:

E2E1=(T2T1)4
T2T1=(E2E1)1/4
=[16.0×1061.0×106]1/4=2
T2=2T1=2×T1
=2×400
=800K=527C