Answer:
There is 4.88 kg of water evaporated.
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.
Answer:
4,874 g
Explanation:
The body can generate 5,500 kJ during 1 hour of strenuous exercise. The heat generated during 2 hours of exercise is:
2 h × (5,500 kJ/1 h) = 11,000 kJ
This heat is used for the vaporization of water. We can calculate the heat (Q) using the following expression.
Q = ΔH°vap × m
where,
ΔH°vap is the enthalpy of vaporization of water
m is the mass of water
m = Q/ΔH°vap
m = (11,000 × 10³ J) / (2,257 J/g)
m = 4,874 g
