Two bodies A and B having temperatlires 327^ C and 427^ C are radiating heat to the surrounding. The surrounding temperature is 27^ C . The ratio of rates of heat radiation of A to that of B is

Step-by-step solution

If temperature of surrounding is considered, then net loss of energy of a body by radiation Q= Ae (T^4-T₀^4 ) Q (T^4-T₀^4 ) Q₁ Q₂ = T₁^4-T₀^4 T₂^4-T₀^4 = (273+327)^4-(273+27)^4 (273+427)^4-(273+27)^4 = (600)^4-(300)^4 (700)^4-(300)^4 =0.52