An ideal Carnot engine operates between a warm reservoir at $233 \mathrm {~K}$ and a colder reservoir. During each cycle, this engine extracts $15.0 \mathrm {~J}$ of heat from the warm reservoir and does $4.0 \mathrm {~J}$ of work. What is the temperature of the colder reservoir?
A) $47 \mathrm {~K}$
B) $67 \mathrm {~K}$
C) $62 \mathrm {~K}$
D) $171 \mathrm {~K}$

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 cold reservoir, $T_h$ is the temperature of the hot reservoir, and $\eta$ is the efficiency. The work done $W$ is related to the heat extracted $Q_h$ and the efficiency by $W = \eta Q_h$. Given $W = 4.0 \, \mathrm{J}$ and $Q_h = 15.0 \, \mathrm{J}$, we find $\eta = \frac{4.0}{15.0} = \frac{4}{15}$. Substituting the given $T_h = 233 \, \mathrm{K}$ and solving for $T_c$, we get $T_c = T_h(1 - \eta) = 233 \times (1 - \frac{4}{15}) = 233 \times \frac{11}{15} = 171 \, \mathrm{K}$.

An Ideal Carnot Engine Operates Between a Warm Reservoir At 233