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 15$\degree$C and the temperature well below the surface is 5$\degree$C?
A)3.5%
B)67%
C)31%
D)17%
E)96%

Question

Quizplus · Accepted Answer

The thermodynamic efficiency of an engine operating between two temperatures is given by the Carnot efficiency formula: $\eta = 1 - \frac{T_c}{T_h}$, where $T_h$ and $T_c$ are the absolute temperatures (in Kelvin) of the hot and cold reservoirs, respectively. Converting the given temperatures to Kelvin: $T_h = 15 + 273 = 288$ K and $T_c = 5 + 273 = 278$ K. Thus, $\eta = 1 - \frac{278}{288} = 0.0347$ or 3.5%.

An Engine Is Designed to Obtain Energy from the Temperature °\degree°

An Engine Is Designed to Obtain Energy from the Temperature $\degree$