A 55-kg mountain climber, starting from rest, climbs a vertical distance of 701 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.

Question

Physics

Hugh Montoya

26 October, 17:25

A 55-kg mountain climber, starting from rest, climbs a vertical distance of 701 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.

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Answers (1)

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Stephen Hess · Answer 1

7.75%

Explanation:

mass of climber, m = 55 kg

height, h = 701 m

Qc = 4.5 x 10^6 J (heat exhaused by the body)

Work = m x g x h

W = 55 x 9.8 x 701

W = 377839 J

W = QH - Qc

Where, QH is the heat input

QH = 377839 + 4.5 x 10^6

QH = 4877839 J

So, the efficiency

e = W / QH

e = 377839 / 4877839

e = 0.0774 = 7.75 %

Thus, the efficiency of the body is 7.75 %.