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Sweaty Biker and Energy Consumption

  1. Dec 3, 2007 #1
    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!
    Last edited: Dec 3, 2007
  2. jcsd
  3. Dec 5, 2007 #2

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    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?
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