# Max efficiency of heat engine

1. Dec 6, 2006

### mikefitz

1. The problem statement, all variables and given/known data

Calculate the maximum possible efficiency of a heat engine that uses surface lake water at 18.5°C as a source of heat and rejects waste heat to the water 0.100 km below the surface where the temperature is 4.3°C.

2. Relevant equations

e=1-Qc/Qh
Q=cm(deltaT)

3. The attempt at a solution

I order to calculate the max efficiency I need the heat capacity of water at 18.5°C and 4.3°C. My book only provides 15C and 0C - how am I to solve this problem?

2. Dec 6, 2006

### NoTime

What is different about water at 0C?
Pressure might make some difference, but they don't seem to be asking that.

http://en.wikipedia.org/wiki/Specific_heat_capacity
Are residual temperature effects important when you round your answer to significant digits?

3. Dec 7, 2006

### mikefitz

from my book

Qc/Qh = W/Qh - 1

262.54999/294.15 = (W/294.15)-1

1.89=W/294.15

W=556.699 J

Why is this wrong?

4. Dec 7, 2006

### OlderDan

Partly because you were not asked to calculate work. You were asked to calculate efficiency. Furthermore, you are not changing the temperature of an object by absorbing or emitting heat. Your Q = cmΔT equation is not what you need here. You are extracting heat form a "hot" reservior, using some of the energy to do work, and delivering the remaining heat to a "cold" reservoir. The maximum efficiency with which this can be done (most work done) is by the Carnot Cycle. That efficiency can be expressed in terms of the reservoir temperatures. You do not need anything else.

Last edited: Dec 7, 2006