Question

During exercise, a person may give off 125 kcal of heat in 17 min by evaporation of water (at 20∘C) from the skin. The heat of vaporization for water at 20∘C is 585 kcal/kg. How much water has been lost?

Answer 1

(Assume a temperature of 20°C.) The human body is at 37 Celsius.
The heat is involved is used evaporating the water at a constant temperature of 37 Celsius.
If the human body needed to raise its sweat to 100 C before boiling it off, we'd be dead after any workout not in cold weather.

Q = m*h_vap(T)

h_vap is a function of the temperature. Looking it up in steam tables (assuming the salt content doesn't change it)

The heat of vaporization for water at 20∘C is 585 kcal/kg = 585x4.18 J/Kg = 2.445x10^6 J/kg

h_vap = 2.445x10^6 J/kg
Solve for m:
m = Q/h_vap
Converting Q from kcal to Joules:
Q= 125 k cal = 522500 J

SO, m=522500/2.445X10^6 kilograms = 0.2137 kg = 213.7 g
of water loss as sweat