Question

A 59-kg mountain climber, starting from rest, climbs a vertical distance of 706 m. At the top, she is again at rest. In the process, her body generates 4.2 × 106 J of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as 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.

Answer 1

weight of the woman = 59 kg

distance = 706 m

work done = force * displacement

                 = mg * displacement

                 = 59*9.8*706

                 = 408209.2 J

efficiency of heat engine e = W/QH

where QH = 4.2 *10^6 J

now   e = 408209.2 /4.2*10^6

             = 9.72*10^-2 J