An ideal Carnot heat 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?
A)735 K
B)490 K
C)470 K
D)784 K

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by $ \eta = 1 - \frac{T_C}{T_H} $. Given $\eta = 0.600$ and $T_C = 294.0 \, \text{K}$, solve for $T_H$: $0.600 = 1 - \frac{294.0}{T_H}$, which gives $T_H = 735 \, \text{K}$.

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

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