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10 June, 07:31

A firebox is at 650 K, and the ambient temperature is 250 K. The efficiency of a Carnot engine doing 146 J of work as it transports energy between these constant-temperature baths is 61.5%. The Carnot engine must take in energy 146 J/0.62 = 237.3 J from the hot reservoir and must put out 91.3 J of energy by heat into the environment. To follow Carnot's reasoning, suppose some other heat engine S could have efficiency 70.0%. (a) Find the energy input and exhaust energy output of engine S as it does 146 J of work

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  1. 10 June, 10:18
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    Qce = 91.25 J

    Explanation:

    Given

    W = 146 J, Tc = 250 k, Th = 650 K

    Qhe = W / [ 1 - (Tc / Th) ]

    Tc / Th = 250 / 650 = 0.384615384

    Qhe = 146 J / 0.615384615

    Qhe = 237.25 J

    So replacing to find the energy exhaust energy

    Qce = Qhe - W

    Qce = 237.25 J - 146 J

    Qce = 91.25 J
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