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 K and a hot reservoir, what is the temperature of the hot reservoir?
A) 735 K
B) 470 K
C) 784 K
D) 490 K

Question

Quizplus · Accepted Answer

The efficiency of a Carnot heat engine is given by:$$e = 1 - \frac{T_c}{T_h}$$where $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir.We are given that $e = 0.600$ and $T_c = 294.0 K$. Solving for $T_h$, we get:$$T_h = \frac{T_c}{1 - e} = \frac{294.0 K}{1 - 0.600} = 735 K$$Therefore, the temperature of the hot reservoir is 735 K.

An Ideal Carnot Heat Engine Has an Efficiency of 0