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18 February, 05:09

A certain heat engine extracts 1.30 kJ of heat from a hot temperature reservoir and discharges 0.70 kJ of heat to a cold temperature reservoir. What is the efficiency of this engine?

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Answers (2)
  1. 18 February, 05:11
    0
    46% (0.46)

    Explanation:

    temperature of hot reservoir (Th) = 1.3 kJ

    temperature of COLD reservoir (Tc) = 0.7 kJ

    Efficiency = 1 - (Tc/Th)

    Efficiency = 1 - (0.7/1.3) = 0.46 = 0.46 x 100 = 46 %
  2. 18 February, 05:28
    0
    46.2 %

    Explanation:

    heat in (Q_in) = 1.3kJ

    heat out (Q_out) = 0.7kJ

    work out = heat in - heat out = 1.3 - 0.7 = 0.6 kJ

    efficiency = work out / heat in * 100%

    = 0.6/1.3 * 100 % = 46.2 %
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