A 52 g mountain climber, start from rest, climbs a vertical distance of 730 m. At top, she is again at rest. In the process, her body generates 4.1*10^6 J of energy via metabolic processes. Her body acts like a heat engine with efficiency ( W/Q_h.) W is the magnitude of work done by her, and Q_h is magnitude of the input heat. So.. can you tell me her effiency as a heat engine?

Question

drwls · Accepted Answer

They tell you the formula to use for the efficiency:
W/Q_h

They also tell you that
Q_h = 4.1*10^6 J

The formula for work done (W) is M g H, where H = 730 m

You only have to do the indicated calculation. I am quite sure the climber's weight is 52 kg, not 52 g.

I get 9.1% for the efficiency

david · Answer

wow. that was simple. thank you.. i should read the question more carefully next time :D