An ideal Carnot thermodynamic engine has an efficiency of 0.600.If it operates between a deep lake with a constant temperature of $294.0 \mathrm {~K}$ and a hot reservoir,what is the temperature of the hot 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 cold reservoir temperature and $T_h$ is the hot reservoir temperature. Rearranging for $T_h$, we get $T_h = \frac{T_c}{1 - \eta}$. Substituting $\eta = 0.600$ and $T_c = 294.0 \, \mathrm{K}$, we find $T_h = \frac{294.0}{1 - 0.600} = 735 \, \mathrm{K}$.

An Ideal Carnot Thermodynamic Engine Has an Efficiency of 0 294.0 K294.0 \mathrm {~K}294.0 K

An Ideal Carnot Thermodynamic Engine Has an Efficiency of 0 $294.0 \mathrm {~K}$