Area of surface of a person is "1.9 m"^(2) and its body temp. is 37^(@)C and room temp. is 22^(@)C. Temperature of skin is 28^(@)C, then find the rate of emission of heat. Emissivity of skin is 0.97.

Area of surface of a person is "1.9 m"^(2) and its body temp. is 37^(@

1.	Area of surface of a person is "1.9 m"^(2) and its body temp. is 37^(@)C and room temp. is 22^(@)C. Temperature of skin is 28^(@)C, then find the rate of emission of heat. Emissivity of skin is 0.97.
Answer» Solution :Rate of energy emission of body, `H=Aesigma(T^(4)-T_(S)^(" 4"))""T=273+28=301K` `""T_(S)=273+22=295K` `1.9xx0.97xx5.67xx10^(-8)[(301)^(4)-(295)^(4)]` `=10.44981xx[8208541201-7573350625]xx10^(-8)` `=10.44981xx635190576xx10^(-8)` `=66.37620832` `~~66.4W` This rate is greater than half of rate of energy produced on body (120 W) under thermal steady state. To decrease this LOSS, in modern arctic clothes, there is a small smooth excessive METAL layer which reflects the emitter energy from body hence loss DECREASES.