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A Carnot engine, operating between hot and cold reservoirs whose temperatures are 800.0 and 400.0 K, respectively, performs a certain amount of work W1. The temperature of the cold reservoir is then lowered to 200.0 K (see the drawing), and the work done by the engine is W2. Assuming that the input heat QH is the same in both cases, what is the ratio W2/W1?

  • physics -

    The W2/W1 ratio equals the efficiency ratio, since the heat input is constant.

    Initially, the Carnot efficiency is
    1 - (Tc/Th) = 1 - 400/800 = 0.50

    After lowering Tc, the efficiency becomes
    1 - (Tc/Th) = 1 - 200/800 = 0.75

    W2/W1 = 0.75/0.50 = 1.5

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