An ideal Carnot engine operates between a warm reservoir at 233 K and a colder reservoir.During each cycle,this engine extracts $15.0 \mathrm {~J}$ of energy from the warm reservoir and does $4.0 \mathrm {~J}$ of work.What is the temperature of the colder reservoir?

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by $ \eta = 1 - \frac{T_c}{T_h} $, where $T_c$ is the temperature of the colder reservoir and $T_h$ is the temperature of the warmer reservoir. The work done $W$ is the difference between the heat extracted from the warm reservoir $Q_h$ and the heat rejected to the cold reservoir $Q_c$, so $W = Q_h - Q_c$. Given $Q_h = 15.0 \mathrm{J}$ and $W = 4.0 \mathrm{J}$, we find $Q_c = 11.0 \mathrm{J}$. The efficiency can also be expressed as $ \eta = \frac{W}{Q_h} = \frac{4.0}{15.0} = \frac{4}{15}$. Substituting the efficiency expression into the Carnot efficiency formula gives $ \frac{4}{15} = 1 - \frac{T_c}{233} $, solving for $T_c$ gives $T_c = 171 \mathrm{K}$.

An Ideal Carnot Engine Operates Between a Warm Reservoir at 233