Question

A 63-kg mountain climber, starting from rest, climbs a vertical distance of 705 m. At the top, she is again at rest. In the process, her body generates 4.5 × 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

come riven, Climber weight virnical DistanceH= = 63 kg 705m. alro Energy They tell giun by generated by climber body = &n=4.5X105 us the formula to ure for efficiency e = wl The formula for work done (W) = mgh where g = g. 8 m/2 n = 705m (Ginen) W = = = efficiency Now MgH 63 x 9.8 X 705 4-353 Xios J e = Iwl = Ton) = 4.353810 x 100% 4.58106 0.096 X 100% i % e = Tee 0.0967 x 100 9.67%