Question

a carnot engine absorbs heat $q_h$ from a high - temperature reservoir and exhausts heat $q_c$ equal to two - thirds of $q_c=(2q_h)/3$. what is the efficiency of this engine?

a. 1/4
b. 1/2
c. 1/3
d. 1/6
e. 2/5

Explanation:

Step1: Recall efficiency formula

The efficiency $\eta$ of a heat - engine is given by $\eta = 1-\frac{Q_c}{Q_h}$, where $Q_h$ is the heat absorbed from the high - temperature reservoir and $Q_c$ is the heat exhausted to the low - temperature reservoir.

Step2: Substitute given value

We are given that $Q_c=\frac{2Q_h}{3}$. Substituting this into the efficiency formula, we get $\eta = 1-\frac{\frac{2Q_h}{3}}{Q_h}$.

Step3: Simplify the expression

Since $\frac{\frac{2Q_h}{3}}{Q_h}=\frac{2}{3}$ (the $Q_h$ terms cancel out), then $\eta = 1 - \frac{2}{3}=\frac{1}{3}$.

Answer:

C. $1/3$