- #1
yoshiko
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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!
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!
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