Overall 80% of the energy used by the body (by metabolism) must be eliminated as
ID: 2047748 • Letter: O
Question
Overall 80% of the energy used by the body (by metabolism) must be eliminated as excess thermal energy and needs to be dissipated. the mechanism of elimination are radiation, evaporation, of swear (2430 kJ/kg), evaporation from the lungs (38kJ/h) conduction, and convection.
A person working out in the gym has a metabolic rate of 2500 kJ/h. His body temperature is 37 degrees C, and the outside temperature is 24 degrees celcius. Assume the skin has an area of 2 m2 and emissitivity of 0.97. In addition, he eliminates 0.40 kg of perspiration by evaporation during that hour. At what rate (in watts) must the remaining excess energy be eliminated through conduction and convection?
Explanation / Answer
(a) P = s A e (T4 - T4) = 5.6696 x 10-8 * 2.0 * 0.97 *(3104 - 2994) = 136.7 W . (b) P = energy / time = 0.34 kg* 2430000 J/kg / 3600 s = 229.5W . (c) P = 38 kJ/hr = 38000 J /3600 s = 10.56 W . (d) Total power to eliminate 80% of energy = 0.80* 2500 kJ/hr = 2000000 J / 3600 s = 555.56 W . power remaining = 555.56 - 136.7 - 229.5 - 10.56 = 178.8W
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.