Sweaty Biker and Energy Consumption

[yoshiko]

In a hot day’s race, a bicyclist consumes 8L water over 4 hrs.
Assume that 80% of his energy goes into evaporating this water as sweat. (This isn't a bad approximation, b/c the mechanical efficiency of a bicycle rider is only ~20%, w/ the rest of the energy consumed going into heat.)
How much total energy, in kcal, did the rider use during the race?

I know that 8L is evaporated by heat (body). I can obtain the mass of water, giving me Q.

I need to use the equation Lxm= Q, from this get 80% of total --> total
Latent heat and vaporization

You can get the mass from knowing 1L= 1000cm^3
Because there are 8L, you have 8000cm^3
Convert to grams knowing the density of water: 1g/cm^3
(8000cm^3)(1g/cm^3)= 8000g or 8 kg water.

I am stuck from here please help!

 

[Shooting Star]

Energy to convert 8 Litres of water into vapour = mass*latent heat = (8000 gm)*(540 cal/gm). This is 80% of total energy. What's the problem now?