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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?

Respuesta :

usmanbiu

Answer:

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 %

Answer:

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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