That's 10 liters of air per minute. If the partial pressure of exhaled H2O is 24 torr, the number of moles of water (at 20 C and 1 atm) is
(10/24)*24/760 = 1.32*10^-2 moles
which has a mass of 0.237 g. Multiply that by 60 for the mass loss per hour. It's about 14.2 g.
For heat removal, multiply that by the heat of vaporization of H2O
A person takes about 20 breaths per minute with 0.5 litre of air in each breath. How much heat is removed per hour by the moisture in the exhaled breath if the incoming air is dry and the exhaled breath is fully saturated? Assume that the water vapour pressure in the saturated exhaled air is 24 torr.
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