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 ${\mathrm{Q}}_1 = 50 \mathrm{kcal} \mathrm{and} \mathrm{\eta } = 20\%$

Using $\mathrm{\eta } = \frac{\mathrm{W}}{{\mathrm{Q}}_1}, \mathrm{we} \mathrm{have} \mathrm{W} = \mathrm{\eta } \times {\mathrm{Q}}_1$

i.e., work obtained per cycle W = $20\% \times 50 \mathrm{kcal}$

      = 10 kcal = 42kJ

Since, ${\mathrm{Q}}_1 = {\mathrm{Q}}_2+\mathrm{W} \mathrm{so}, {\mathrm{Q}}_2 = \mathrm{Q}-\mathrm{W}$

i.e., heat rejected to the sink per cycle ${\mathrm{Q}}_2$

     = 50 kcal -10 kcal = 40 kcal