# Homework Help: Efficiency of a heat engine

1. Dec 7, 2011

### darw

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. Dec 7, 2011

### Spinnor

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
3. Dec 8, 2011

### Andrew Mason

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.

AM