Answer to Question #285456 in Physics for haasggdfnsd

Question

 The heat exhausted to the heat sink by a heat engine is twice the mechanical work done. What is its efficiency?

Answer 1

By definition, the efficiency of an engine $\eta$ is equal to the ratio of the heat energy $Q_h>0$ taken from the high temperature heat source and $W>0$ mechanical work done in one cycle:

$$\eta = \dfrac{W}{Q_h}\times 100\%$$

According to the energy conservation law:

$$Q_h = W+Q_s$$

where $Q_s>0$ is the the heat exhausted to the heat sink. Since by the task $Q_s = 2W$, find:

$$Q_h = W+2W = 3W\\ \eta = \dfrac{W}{3W}\times 100\% \approx 33\%$$

Answer. 33%.