Question

A naked person with surface area of 1.8 m^2 and skin temperature 33 degrees C is in a room at 10 degrees C. (assume e=1) a) at what rate does the person radiate energy? b) what is the person's net rate of energy loss due to radiation? (Answers are: a) 895 W, b) 240 W)

Answer 1

Given
surface area of the person A = 1.8 m2, temperature T1= 33 0C= 306 k, emmisivity e = 1, room temperature T2 = 10 0C = 283 k

   from stefan's boltzman law Q = A*e * (sigma) T^4

where sigma is stefan's constant = 5.6703*10^-8 Watt/m2k^4

       Q = 1.8*1*5.6703*10^-8 *306^4 W
       Q = 894.87886 W
       Q = 895 W
  

now the rate of energy loss due to radiation per sec is

       Q = A*e*sigma(T1^4 - T2^4)

       Q = 1.8*1*5.6703*10^-8 (306^4-283^4)
       Q = 240.2060
       Q = 240 W