A heat engine receives 50 kcal of heat from the source per cycle, and operates with an efficiency of 20%. The heat rejected by engine to the sink per cycle is

Given Q1 = 50 kcal and η = 20%Using η = WQ1, we have W = η ×Q1i.e., work obtained per cycle W = 20% × 50 kcal      = 10 kcal = 42kJSince, Q1 = Q2+W so, Q2 = Q-Wi.e., heat rejected to the sink per cycle Q2     = 50 kcal -10 kcal = 40 kcal