An ideal Carnot heat engine operates between $285 \mathrm {~K}$ and $460 \mathrm {~K}$. What is its efficiency?
A) $0.62$
B) $1.61$
C) $0.38$
D) $0.61$

Question

Quizplus · Accepted Answer

The efficiency ($\eta$) of an ideal 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, both in Kelvin. Thus, $\eta = 1 - \frac{285}{460} = 1 - 0.6196 = 0.3804$, which rounds to $0.38$.

An Ideal Carnot Heat Engine Operates Between 285 K285 \mathrm {~K}285 K And 460 K460 \mathrm {~K}460 K

An Ideal Carnot Heat Engine Operates Between $285 \mathrm {~K}$ And $460 \mathrm {~K}$