Chapter 21 - Heat Engines and Refrigerators - Exercises and Problems - Page 595: 25

a) $\eta = 0.4 = 40\%$ b)$T_H = 215.33^oC$

We have $W=Q_H -Q_C$ $ 10J = Q_H - 15J$ $Q_H = 25J$ a) Efficiency of engine is $\eta = \frac{W}{Q_H} = \frac{10J}{25J}$ $\eta = 0.4 = 40\%$ b)We have $T_C = 20^oC= 293K$ Also, $\frac{Q_H}{Q_C} =\frac{T_H}{T_C}$ $\therefore \frac{25J}{15J} =\frac{T_H}{293K}$ $ T_H = 488.33K$ $T_H = 215.33^oC$