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28 July, 13:17

If a heat engine takes in 4565 kJ and gives up 2955 kJ during one cycle, what is the engine's efficiency?

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  1. 28 July, 13:41
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    The engine efficiency is 64.73 %

    Explanation:

    Given data

    To find the engine's efficiency we have the formula,

    Energy input - 4565 KJ

    Energy output - 2955KJ

    Efficiency = energy output / energy input * 100%

    =2955/4565

    =0.6473 * 100

    η = 64.73 %

    The engine efficiency is 64.73 %
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