Answer:
There is 4.88 kg of water evaporated.
Step-by-step explanation:
Step 1: Data given
5500 kJ of energy during one hour of strenuous exercise.
Δh = Heat of vaporization of water is 40.6 kJ/mol
Step 2: Calculate the mass of water evaporated
Q = 2* 5500 kJ = 11000 kJ
mass = [( Q ) / ( Δh ) ] [ Molar mass of water]
mass = [ ( 11,000 kj / 40.6 kJ/mol) * 18.02 g/mol )
mass = 4882.27 grams ≈ 4.88 kg of water
There is 4.88 kg of water evaporated.