A 58 kg mountain climber, starting from rest, climbs a vertical distance of 770

Question

A 58 kg mountain climber, starting from rest, climbs a vertical distance of 770 m. At the top, she is again at rest. In the process, her body generates 4.4 x 10^6 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. Use g=9.8 m/s^2 A 58 kg mountain climber, starting from rest, climbs a vertical distance of 770 m. At the top, she is again at rest. In the process, her body generates 4.4 x 10^6 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. Use g=9.8 m/s^2

Explanation / Answer

The energy or out put energy is,

U = mgh = 58*9.8*770 = 437668 J.
The input energy = 4.4 x 10^6 J

Effficiency is calculated as follows:

e = Output energy / Input energy = 437668 J / 4.4 x 10^6 J = 0.09947 = 9.947%

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