Question

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?

Answer 1

46.2 %

Step-by-step 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 %

Answer 2

46% (0.46)

Step-by-step 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 %