A heat engine accepts 500 kJ of energy from a thermal reserv

A heat engine accepts 500 kJ of energy from a thermal reservoir at 700 K and rejects 100 kJ to a thermal reservoir at 300 K. Determine the amount of work produced and the thermal efficiency.

Solution

Let Q_in denotes the amount of heat accepted. Q_in=500kJ at T_h=700K

Let Q_out denotes the amount of heat rejected. Q_out=100kJ at T_h=300K

Now from energy balance of work, W=Q_in - Qout
or W=500-100 kJ
W=400kJ

Thermal Efficiency=W/Qin

or E=400/500
Thermal Efficiency is 0.8 or 80%