Posted by Victor on Friday, May 21, 2010 at 1:53pm.
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
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