An engine delivers 40.0 J of work while it takes in energy from a 400 K thermal source. If the engine discharges energy into a thermal reservoir at 200 K, then what is the least amount of heat needed to run the engine?
A) 55.0 J
B) 65.0 J
C) 80.0 J
D) 110 J

Question

Quizplus · Accepted Answer

The efficiency of a heat engine operating between two thermal reservoirs is given by $1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$, where $T_{\text{cold}}$ and $T_{\text{hot}}$ are the temperatures of the cold and hot reservoirs, respectively. In this case, $T_{\text{cold}} = 200 \, \text{K}$ and $T_{\text{hot}} = 400 \, \text{K}$, so the maximum efficiency is $1 - \frac{200}{400} = 0.5$ or 50%. Since the engine delivers 40.0 J of work, the least amount of heat needed to run the engine, given its maximum efficiency, would be the work output divided by the efficiency: $40.0 \, \text{J} / 0.5 = 80.0 \, \text{J}$.

An Engine Delivers 40