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Hiker climbs top of mountain; energy conversion problem

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A 60 kg hiker wishes to climb to the summit of Mt. Ogden , an ascent of 1500 m.

    a) Assuming that she is 25 % efficient at converting chemical energy from food into mechanical work, and that essentially all the mechanical work is used to climb vertically , roughly how many bowls of corn flakes(standard serving 1 ounce, 100 kilocalories) should the hiker eat before setting out?

    b) As the hiker climbs the mountain , 3-quarters of the energy from the corn flakes is converted to thermal energy. If there were no way to dissipate this energy , how many degrees would her body temeperature increase?

    c) In fact, the extra energy does not warm the hiker's body significantly; instead, it goes(mostly) into evaporating water from her skin. How many liters of water should she drink during the hike to replace the lost fluids?(At 298 K, a reasonable temperature toassume , the latent heat of vaporization of water is 580 cal/g, 8 % more than at 373 K)


    2. Relevant equations

    mgh=U
    Q=delta(T)*C
    PV=RT
    delta(H)=delta(U)+Pdelta(V)

    3. The attempt at a solution

    a) mgh=(60)(9.8)(1500) =882000 joules=882 kJ

    If she works at 25 % efficiency, I should considered only .75*mgh=662 kJ

    4.184 kJ=1 kilocalorie => 662 kJ=158 kilocalories==> 1 .58 ounces or 1.58 bowls of corn flakes

    b) delta(H)=Q+W_other. Does no dissipation mean Q=0? If so then delta(T) = 0

    c) Q=Lm=(580 cal/g)(18 g)= 10440 g; PV=RT; T=298 K, R=8.31 J/K, P=1.01e5 , why would they give me the latent heat of vaporization when I can just T, R and P to find the volume?
     
  2. jcsd
  3. Feb 15, 2009 #2

    mgb_phys

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    Looks good

    Nope, if the hiker is only 25% efficent you need 100/25 = 4x as much energy.

    You are evaporating liquid water into vapour
     
  4. Feb 15, 2009 #3

    Borek

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    Staff: Mentor

    b - no dissipation measn no heat is lost to the surroundings, all 75% heats her body

    c - you need latent heat to calculate amount of water that have to be evaporated to dissipate the heat
     
  5. Feb 15, 2009 #4
    so Q=0 and therefore delta(T)=0 since Q=delta(T)*C

    What quantity represents the amount of water? Q? I don't need the relation PV=RT, to find V?
     
  6. Feb 15, 2009 #5

    Borek

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    No, Q is not 0.

    I think you are on the right track.
     
  7. Feb 15, 2009 #6
    Wait, I thought no dissipation means no heat? should I used the fact that delta(U)=f/2*N*k*delta(T)?
     
  8. Feb 15, 2009 #7

    Borek

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    No dissipation means no heat exchange, that's not the same as no heat produced.
     
  9. Feb 15, 2009 #8
    delta(U)=f/2*N*k*Delta(T),f=5 since hiker is evaporating H2O

    delta(U)=.75*mgh

    therefore I can now find delta(T), right?
     
  10. Feb 15, 2009 #9

    Borek

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    You are overcomplicating. You know hikers mass, assume specific heat of water.
     
  11. Feb 15, 2009 #10
    Ahh, of course, delta(U)-Q and I can find delta(T) from this relation since mechaniccal energy goes into 75 % thermal energy
     
  12. Feb 16, 2009 #11
    my attempt

    a) mgh=(60)(9.8)(1500) =882000 joules=882 kJ

    If she works at 25 % efficiency, she must do 4 times the work, or 4*mgh= 3528 kJ

    4.184 kJ=1 kilocalorie => 3528 kJ= 843 kilocalories==> 8.43 ounces or 8.43 bowls of corn flakes


    3/4 energy from the corn flakes is 3*882 = 2646 kJ

    It takes 4.186 J to heat 1 kg water 1 degree.

    (2646 kJ/60 kg)*1K*kg/4.186kJ = 10.5 degree increase


    If 580 cal of energy will evaporate 1 g of water, the thermal energy she gives off 2464 kJ or 632 kcal will evaporate 632/.58 g = 1090 g = 1.09 L of water that she should drink

    Or?
     
  13. Feb 16, 2009 #12

    Borek

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    AROD, what you just did was you have done someone else's homerwork. Please don't.
     
    Last edited: Feb 16, 2009
  14. Feb 17, 2009 #13
    gotcha, my bad.

    physics rocks.
     
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