What Is the Maximum Efficiency of a Heat Engine Using Lake Water Temperatures?

AI Thread Summary
The discussion focuses on calculating the maximum efficiency of a heat engine using lake water at 18.5°C as the heat source and rejecting heat to water at 4.3°C. The efficiency can be determined using the Carnot efficiency formula, which relies solely on the temperatures of the hot and cold reservoirs. The user initially attempted to use specific heat capacity and work calculations, but was advised that these were unnecessary for determining efficiency. The key takeaway is that the maximum efficiency is derived from the temperature difference between the reservoirs, without needing to calculate work or specific heat capacities. Understanding the Carnot Cycle is essential for solving this problem correctly.
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Homework Statement



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.

Homework Equations



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

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

Here is some more info
http://en.wikipedia.org/wiki/Specific_heat_capacity
Are residual temperature effects important when you round your answer to significant digits?
 
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?
 
mikefitz said:
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?

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.
 
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