Question

during each cycle, a reversible engine absorbs 2500 joules of heat from a high - temperature reservoir and performs 2200 joules of work. what is the ratio (t_{h}/t_{c}) of the reservoir temperatures?

a. 1.6 m
b. 8.3 m
c. 5.2 m
d. 7.7 m
e. 9.1 m

Explanation:

Step1: Calculate heat rejected

The heat absorbed $Q_H = 2500$ J and work done $W=2200$ J. By the first - law of thermodynamics $Q_H=W + Q_C$, so $Q_C=Q_H - W$.
$Q_C=2500 - 2200=300$ J

Step2: Use the efficiency formula for a reversible engine

The efficiency of a reversible engine is $\eta = 1-\frac{Q_C}{Q_H}=1-\frac{T_C}{T_H}$. Rearranging for $\frac{T_H}{T_C}$, we get $\frac{T_H}{T_C}=\frac{Q_H}{Q_C}$.
$\frac{T_H}{T_C}=\frac{2500}{300}=\frac{25}{3}\approx8.3$

Answer:

b. 8.3