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25 July, 03:10

An ideal heat-engine is to be used in an environment where the ambient temperature is 23.5 °C. What should be the minimum temperature of the hot heat reservoir in order to reach at least 41.2 percent efficiency with the heat-engine? (Give your answer in Celsius.)

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  1. 25 July, 03:26
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    The minimum temperature of the hot reservoir = 231.25 °C

    Explanation:

    η₁ = (1 - T (c) / T (h)) * 100 ... equation 1

    Where η₁ = efficiency of the heat engine in percentage, T (c) = Temperature at which the low temperature reservoir operates, T (h)) = Temperature at which the high temperature reservoir operates.

    Making T (h) the subject of the equation in equation 1

    T (h) = T (c) / {1 - (η₁ / 100) }

    where η₁ = 41.2%, T (c) = 23.5 °C = 23.5 + 273 = 296.5 K

    ∴T (h) = 296.5/{1 - (41.2/100) }

    T (h) = 296.5 / (1 - 0.412)

    T (h) = 296.5/0.588

    T (h) = 504.25 K

    Convert to temperature in Celsius

    T (h) = 504.25 - 273 = 231.25 °C

    The minimum temperature of the hot reservoir = 231.25 °C
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