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4 June, 10:31

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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  1. 4 June, 10:39
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    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 %.
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