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First Law Thermodynamics Problem

  1. Feb 1, 2014 #1
    1. The problem statement, all variables and given/known data
    For an hour , an athlete with 50kg of weight produces a work of 75kcal pedaling on the gym bike. During this period of time he consumes 60L of oxygen.

    Calculate:

    a) The power produced by the athlete
    b) The variation of the internal energy of the athlete knowing that for each liter of oxygen consumed, 5kcal of energy are spent.
    c) The heat transfered to the envolving environment.


    2. Relevant equations
    First law of thermodynamics E2-E1=Q-W


    3. The attempt at a solution

    For the first one I converted the 75kcal that he produces in an hour to Joule/s giving me the power. Now, the other two I'm having trouble knowing what to do. If you could help me I would appreciate it.
    Thank you,
     
    Last edited: Feb 1, 2014
  2. jcsd
  3. Feb 1, 2014 #2

    SteamKing

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    It's not cool to make duplicate posts of the same HW problem.
     
  4. Feb 1, 2014 #3

    DrClaude

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    b) Can you calculate how much energy was produced knowing how much oxygen was consumed?

    c) I guess the athlete didn't develop a fever. So where did the energy go?
     
  5. Feb 1, 2014 #4
    in b) I know that 300kcal are spent (60L*5kcal) but that isn't the internal energy, I think,

    in c) no, I guess he didn't :p and I know that he releases heat, but I'm not getting on the way to calculate the heat that he releases.
     
  6. Feb 1, 2014 #5

    DrClaude

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    Indeed no, that is the total energy. You have to use the first law (conservation of energy) here.

    Solve b first, then c should be clear.
     
    Last edited: Feb 1, 2014
  7. Feb 1, 2014 #6
    I know for c) I'll have to use the formula ΔE=Q-W, because I'm going to have the variation of internal energy and the work, but in b) When you say 300kcal is the total energy you mean it is ΔE=Ekinetic+Epotential+U?
     
  8. Feb 1, 2014 #7

    DrClaude

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    Use ΔE=Q-W for b).
     
  9. Feb 1, 2014 #8
    ΔE are the 300kcal and W are the 75kcal?
     
  10. Feb 1, 2014 #9

    DrClaude

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    Correct. Actually, you only need the equation ΔE=Q-W, so sorry for the misdirection.
     
  11. Feb 1, 2014 #10
    No problem, yes, the heat is 225, in this case -225. And the variation is 300kcal, converting to Joules is 1255,2. Thank you! I'll try to find the variation now.
     
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