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Efficiency of a heat engine

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    A 52-kg mountain climber, starting from rest, climbs a vertical distance of 730 m. At the top, she is again at rest. In the process, her body generates 4.1 × 10^10 J of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by e = |W|/|QH|, where |W| is the magnitude of the work she does and |QH| is the magnitude of the input heat. Find her efficiency as a heat engine.

    2. Relevant equations

    e = |W|/|Q_H|

    PE = mgh

    Q_C = Q_H - W

    3. The attempt at a solution

    I calculated the potential energy gained by climbing the mountain. I called this Q, and called the 4.1 X 10 ^10 J the Work. However, this did not give me the correct efficiency. Could someone guide me toward what I am missing? Thanks so much for your time
  2. jcsd
  3. Dec 7, 2011 #2


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    Gold Member

    For mgh I got about 370,000 J. This is no where near the number you give 4.1 × 10^10 J.

    Humans are pretty efficient so that number 4.1 × 10^10 J should be closer to 370,000 J.

    The energy content of a gallon of gasoline is about 2.2E7 J. See,

    http://www.phy.syr.edu/courses/modules/ENERGY/ENERGY_POLICY/tables.html [Broken]

    From the same link, the energy content of a candy bar is about a million J. The climber would have to eat more then 10,000 candy bars to make the required energy.
    Last edited by a moderator: May 5, 2017
  4. Dec 8, 2011 #3

    Andrew Mason

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    Homework Helper

    You have it reversed. Qh is the heat flow = 4.1e10 J. W, the work done, is the climbing = mgh. I agree with Spinner that Qh seems extremely high.

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