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 \, \mathrm{K} $, we solve for $ T_h $:  $ 0.600 = 1 - \frac{294.0}{T_h} $  $ 0.400 = \frac{294.0}{T_h} $  $ T_h = \frac{294.0}{0.400} = 735 \, \mathrm{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 $294.0 \mathrm {~K}$