Question

Overall 80 % of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation, evaporation of sweat (2430 kJ/kg), evaporation from the lungs (38 kJ/h), conduction, and convection.
A person working out in a gym has a metabolic rate of 2500 kJ/h. His body temperature is 37 C, and the outside temperature 24 C. Assume the skin has an area of 2.0 m2 and emissivity of 0.97. (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates 0.40 kg of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) At what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

Can someone help and show how to do it?

Answer 1

(a) Radiative heat loss rate =
(emissivity)(sigma)*(body area)*[T2^4 - T1^4]
T2 is the body temp in Kelvin and T1 is the average temp of surrounding walls.
Sigma is the Stefan-Boltzmann constant. Look it up. Emissivity = 0.97
(b) Multiply the perspiration rate by the heat of skin evaporation
(c) Do the same using the rate of lung moisture evaporation.
(d) Subtract the sum of the previous three mechanisms from 80% of the metabolic rate, which would be 2000 kJ/h.
Express all four in units of kJ/h, for easy comparison.

Answer 2

183.06

Answer 3

I think that's right