Question

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)

Answer 1

0.035 (3.5 %)

Step-by-step explanation:

The thermodynamic efficiency is given by:

`\eta = 1 - (T_C)/(T_H)`

where

`T_C` is the cold temperature

`T_H` is the hot temperature

In this problem we have

`T_C = 5 ^(\circ)C+ 273 = 278 KT_H = 15^(\circ)C+273 = 288 K`

So the efficiency is

`\eta = 1 - (278 K)/(288 K)=0.035`