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22 December, 05:24

You/'ve designed a new heat engine that uses two containers held at different temperatures. One container is at 353 K, while the other is kept at 494 K. What is the maximum efficiency you could hope for with your new engine?

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  1. 22 December, 07:35
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    The maximum possible efficiency, i. e the efficiency of a Carnot engine, is give by the ratio of the absolute temperatures of hot and cold reservoir.

    η_max = 1 - (T_c/T_h)

    For this engine:

    η_max = 1 - [ (20 + 273) K / (600 + 273) K ] = 0.66 = 66%

    The actual efficiency of the engine is 30%, i. e.

    η = 0.3 ∙ 0.664 = 0.20 = 20 %

    On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir:

    η = W/Q_h

    So the heat required from hot reservoir is:

    Q_h = W/η = 1000J / 0.20 = 5000J
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