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

In summary, the maximum possible efficiency of a heat engine using surface lake water at 18.5°C as a heat source and rejecting waste heat to water 0.100 km below the surface at 4.3°C can be calculated using the formula e=1-Qc/Qh, where Qc is the heat absorbed from the "hot" reservoir and Qh is the heat delivered to the "cold" reservoir. The heat capacity of water at 18.5°C and 4.3°C is needed for this calculation, which can be found from external sources. The Carnot Cycle can be used to determine the maximum efficiency of the heat engine, without needing to calculate work or consider residual temperature effects.
  • #1
mikefitz
155
0

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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  • #2
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?
 
  • #3
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
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.
 
Last edited:

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

1. What is the definition of "max efficiency of heat engine"?

The max efficiency of a heat engine is the maximum possible efficiency at which a heat engine can convert heat energy into mechanical work. It is calculated by dividing the work output by the heat input.

2. How is the max efficiency of a heat engine calculated?

The max efficiency of a heat engine can be calculated using the formula: efficiency = 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.

3. What factors affect the max efficiency of a heat engine?

The max efficiency of a heat engine is affected by the temperature difference between the hot and cold reservoirs, the type of working fluid used, and the design of the heat engine.

4. What is the Carnot cycle and how does it relate to the max efficiency of a heat engine?

The Carnot cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat energy into work. The max efficiency of a heat engine is equal to the efficiency of a Carnot cycle, which is determined by the temperature difference between the hot and cold reservoirs.

5. Can the max efficiency of a heat engine ever be 100%?

No, according to the second law of thermodynamics, it is impossible for any heat engine to have a 100% efficiency. The efficiency of a heat engine can approach 100%, but it can never reach it.

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