An ideal-gas heat engine operates in a Carnot cycle between temperatures of 266 $\circ$ C and 115 $\circ$ C.It absorbs 4250 cal per cycle at 266 $\circ$ C.What is its efficiency in %?

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$, where $T_C$ and $T_H$ are the absolute temperatures (in Kelvin). Converting the given temperatures to Kelvin: $T_H = 266 + 273 = 539K$ and $T_C = 115 + 273 = 388K$. Thus, $\eta = 1 - \frac{388}{539} \approx 0.28$ or 28%.

An Ideal-Gas Heat Engine Operates in a Carnot Cycle Between ∘\circ∘

An Ideal-Gas Heat Engine Operates in a Carnot Cycle Between $\circ$