The heating temperature of the heat engine is +127 degrees, and the temperature of the refrigerator is +31 degrees.

The heating temperature of the heat engine is +127 degrees, and the temperature of the refrigerator is +31 degrees. During one cycle, the working fluid of the engine transferred the amount of heat to the refrigerator equal to 1.14 kJ. determine: 1) the efficiency of the heat engine 2) the amount of heat received by the refrigerator from the working fluid 3) the work performed by the heat engine in one cycle

1)
efficiency = (T1-T2) / T1 = (400-304) / 400 = 0.24 or 24%
2)
efficiency = (Q1-Q2) / Q1
efficiency * Q1 = Q1-Q2
Q1 (1-efficiency) = Q2
Q1 = Q2 / 1-efficiency = 1.14 * 10 ^ 3/1-.024 = 1500J
3)
q = c * u
q = 5 * 10 (-6) * 600 = 0.003 (cl)
q-charge
c-capacitance
u-voltage