Rate of radiation by a black body is 'R' at temperature 'T'. Another body has same area but emissivity is 0.2 and temperature 3T. Its rate of radiation is

Step-by-step solution

Rate of radiation for a black body is given by Stefan-Boltzmann law: R = A T^4 For the second body, the rate of radiation is: R' = e A (T')^4 Given e = 0.2 and T' = 3T , substituting these values: R' = 0.2 A (3T)^4 R' = 0.2 81 A T^4 R' = 16.2 R Answer: 16.2R