How much water evaporates from jogger’s skin each hour when heat energy is at a rate of 878.2W?
A jogger generates heat energy at a rate of 878.2 W. If all this energy is removed by sweating, how much water (in kg) must evaporate from the jogger’s skin each hour?
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Water’s latent heat of evaporation is 2,270 kJ/kg. A watt is a J/s . That’s all you need.
Let m kg water evaporate in 1 sec
=>Q(total) = Q(37*C to 100*C) + Q(water to vapor)
=>Q(t) = m x c x delta t*C + m x L
=>878.2 = m[4187 x (100-37) + 334000]
=>878.2 = 597781m
=>m = 14.69 x 10^-4 kg
=>m for 1 hour = 14.69 x 10^-4 x 3600 = 5.29 kg