An engine is designed to obtain energy from the temperature gradient of the ocean. What is the thermodynamic efficiency of such an engine if the temperature of the surface of the water is 59°F (15°C) and the temperature well below the surface is 41°F (5°C)?
A) 3.5%
B) 67%
C) 31%
D) 17%
E) 96%

Question

Quizplus · Accepted Answer

The efficiency of a heat engine operating between two temperatures is given by the Carnot efficiency formula: $\eta = 1 - \frac{T_c}{T_h}$, where $T_c$ and $T_h$ are the cold and hot reservoir temperatures, respectively, in Kelvin. Converting the given temperatures to Kelvin: 15°C + 273.15 = 288.15K and 5°C + 273.15 = 278.15K. Thus, $\eta = 1 - \frac{278.15}{288.15} = 0.0347$ or 3.47%, which rounds to 3.5%.

An Engine Is Designed to Obtain Energy from the Temperature