(a) At what rate is his excess thermal energy dissipated by radiation? (Enter yo
ID: 694734 • Letter: #
Question
(a) At what rate is his excess thermal energy dissipated by radiation? (Enter your answer to at least one decimal place.)
W
(b) If he eliminates 0.41 kg of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (Enter your answer to at least one decimal place.)
W
(c) At what rate is energy eliminated by evaporation from the lungs? (Enter your answer to at least one decimal place.)
W
(d) At what rate must the remaining excess energy be eliminated through conduction and convection?
W
Explanation / Answer
a)
Radiative heat loss rate = (emissivity)()*(body area)*[T_2^4 - T_1^4]
T2 is the body temp in Kelvin and T1 is the average temp of surrounding
= 0.97*5.6696*10^-8*2.0*[310.15^4-294.15^4]
=194.31W
b)
given prespiration = 0.34
evaporation of sweat = 2430
rate is thermal energy dissipated by evaporation of sweat is
= prespiration*evaporation of sweat
= 0.34*2430
=826.2Kj
c)
evaporation from the lungs = (38 kJ/h)
38*0.34 =12.92
d)
80% of the metabolic rate = 2500*80/100 =2000
2000 - 194.31+826.2+12.92 =966.57
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