# Max Work From Heat Engine Reservoirs

• SevenTacos
In summary, the maximum work that can be extracted from an engine operating between an infinite heat reservoir with temperature Th and a finite cool reservoir with initial temperature Tc0 and heat capacity C is given by W = C (Th ln (Th/Tco)). This formula can be derived by integrating the work equation, dw = C ((Th/Tc) - 1) dTc, from Tc0 to Th.
SevenTacos

## Homework Statement

You have an infinite heat reservoir with temperature Th. But you’ve only got a finite cool reservoir, with initial temperature Tc0 and heat capacity C. Find an expression for the maximum work you can extract if you operate an engine between these two reservoirs.

## The Attempt at a Solution

I've been stumped at this for hours. I know I simply need to integrate work as dW= dQc + dQh. So i need to define Qc and Qh. Qc = (Cc)(dTc) and Qh= (Ch)(dTh) correct? I am completely at a loss on defining dTc and dTh however. I'm essentially confused because Tc depends on how much energy (q) the engine has deposited into the cold resevoir. When I try to create an equation, I come up with: Tc= Tc0 + (Qc)(C) But to define q, you need Tc. How do you define these variables when they rely on each other? Please Help!

Tc is going to be changing, while Th will be staying the same. So as Tc rises, a smaller fraction of the heat transferred from Th is able to be converted to work. If Tc and Th are the instantaneous values of the temperatures, and dq is transferred to Tc, what is the amount of heat dQ transferred from Th (in terms of Tc, Th, and dq)? How much work dw is done? What is the change in Tc, dTc? What is dw/dTc?

Chet

I appreciate the reply, but I'm still pretty confused. Using the statement dQh/dQc = Th / Tc, it's clear dQc = (Tc / Th) dQh. I can't really think of anywhere else to go though. I know I can define Qh and Qc in terms of nRTln(v1/v2) but those are the wrong variables here.

SevenTacos said:
I appreciate the reply, but I'm still pretty confused. Using the statement dQh/dQc = Th / Tc, it's clear dQc = (Tc / Th) dQh. I can't really think of anywhere else to go though. I know I can define Qh and Qc in terms of nRTln(v1/v2) but those are the wrong variables here.
This is good so far. dw = dQ-dq, so what do you get for that?

Chet

You would get dw = dQh- dQh(Tc / Th). But I still need to define Tc as a function of Th correct? It's 1 am here, will return in the morning!

SevenTacos said:
You would get dw = dQh- dQh(Tc / Th). But I still need to define Tc as a function of Th correct? It's 1 am here, will return in the morning!
If I take your results, and express dw in terms of dQc, I get:
$$dw=\left(\frac{T_{h0}}{T_c}-1\right)dQ_c$$
Now how is dQc related to the heat capacity C and dTc?
If you substitute this into the equation for dw, what do you get?

Chet

Chestermiller said:
If I take your results, and express dw in terms of dQc, I get:
$$dw=\left(\frac{T_{h0}}{T_c}-1\right)dQ_c$$
Now how is dQc related to the heat capacity C and dTc?
If you substitute this into the equation for dw, what do you get?

Chet
dQc = (C)(dTc), so dw = (Tho/Tc -1) (C)( dTc)

Did some more thought on this: If dQh/dQc = Th / Tc, then (Ch)(dTh)/(Cc)(dTc) = Th / Tc. If this is correct then you can cross multiply and take the integral, looking like Cc/Ch ∫ dTc/ Tc = ∫ dTh / Th. The bounds for the first integral should be Tc to Th, because the engine will stop when Tc = Th. The bounds for the second I guess would be Tho to Th? After integration you get (Cc/Ch) ln (Th/ Tc) = ln (Th/Tho), rearranging you get this final factor for Tc:

Tc = Th / (( Th/ Tho) ^ (Ch/Cc))

I feel like this should be right, but it also feels horribly wrong

SevenTacos said:
dQc = (C)(dTc), so dw = (Tho/Tc -1) (C)( dTc)

This last equation is correct, and it's all you need to finish solving this problem.

Did some more thought on this: If dQh/dQc = Th / Tc, then (Ch)(dTh)/(Cc)(dTc) = Th / Tc. If this is correct then you can cross multiply and take the integral, looking like Cc/Ch ∫ dTc/ Tc = ∫ dTh / Th. The bounds for the first integral should be Tc to Th, because the engine will stop when Tc = Th. The bounds for the second I guess would be Tho to Th? After integration you get (Cc/Ch) ln (Th/ Tc) = ln (Th/Tho), rearranging you get this final factor for Tc:

Tc = Th / (( Th/ Tho) ^ (Ch/Cc))

I feel like this should be right, but it also feels horribly wrong

Yes. It's horribly wrong. Here's what's wrong:

The problem statement says that you have an infinite heat reservoir at temperature Th. That means that Ch is infinite, and that Th never changes from its initial value Th0. So the rest of the analysis is incorrect.

Getting back to your correct equation for dw, you have:

$$dw=C\left(\frac{T_{h0}}{T_C}-1\right)dT_C$$

This can be integrated immediately from ##T_C=T_{C0}## to ##T_C=T_{h0}## to give you w. Do you know how to integrate this?

Chet

Chestermiller said:
This last equation is correct, and it's all you need to finish solving this problem.

Getting back to your correct equation for dw, you have:

$$dw=C\left(\frac{T_{h0}}{T_C}-1\right)dT_C$$

This can be integrated immediately from ##T_C=T_{C0}## to ##T_C=T_{h0}## to give you w. Do you know how to integrate this?

Chet

Yes, the integral comes out to W = C ( Th ln (Tc) - Tc ) which yields a work of C (Th ln (Th) - Th) - C (Th ln(Tco) -Tco )

So W = C (Th ln (Th) - Th) - C (Th ln(Tco) -Tco ) as a final answer?

SevenTacos said:
Yes, the integral comes out to W = C ( Th ln (Tc) - Tc ) which yields a work of C (Th ln (Th) - Th) - C (Th ln(Tco) -Tco )

So W = C (Th ln (Th) - Th) - C (Th ln(Tco) -Tco ) as a final answer?
Yes. This is an acceptable final answer to your homework problem. However, consider what happens if we manipulate your final answer mathematically into the following equivalent form:
$$W=CT_h\ln \left(\frac{T_h}{T_{co}}\right)-C(T_h-T_{co})$$
Can you show that this equation is mathematically equivalent to your answer? Can you identify what the second term on the right hand side of this equation represents physically? With that realization, what does the first term on the right hand side represent physically?

Chet

Chestermiller said:
Yes. This is an acceptable final answer to your homework problem. However, consider what happens if we manipulate your final answer mathematically into the following equivalent form:
$$W=CT_h\ln \left(\frac{T_h}{T_{co}}\right)-C(T_h-T_{co})$$
Can you show that this equation is mathematically equivalent to your answer? Can you identify what the second term on the right hand side of this equation represents physically? With that realization, what does the first term on the right hand side represent physically?

Chet
The two terms represent the heat put into the system Qh, and the heat deposited into the cold reservoir Qc respectively, after all, W = Qh - Qc.
And yes, I understand the algebra to clean up that equation.

Just have to say Thank you Chet, this problem was much easier than I made it out to be; I severely overthought this. Thanks for taking the time to put me through the process I needed to go through

SevenTacos said:
The two terms represent the heat put into the system Qh, and the heat deposited into the cold reservoir Qc respectively, after all, W = Qh - Qc.
And yes, I understand the algebra to clean up that equation.

Just have to say Thank you Chet, this problem was much easier than I made it out to be; I severely overthought this. Thanks for taking the time to put me through the process I needed to go through
My pleasure. It's great to work with someone as focused and determined as you are.

Chet

## 1. What is the concept of "Max Work From Heat Engine Reservoirs"?

The concept of "Max Work From Heat Engine Reservoirs" is the maximum amount of work that can be obtained from a heat engine by using a given amount of heat from a hot reservoir and transferring it to a cold reservoir. This concept is based on the Second Law of Thermodynamics and is used to determine the efficiency of a heat engine.

## 2. How is the "Max Work From Heat Engine Reservoirs" calculated?

The "Max Work From Heat Engine Reservoirs" is calculated using the Carnot efficiency formula, which is the ratio of the temperature difference between the hot and cold reservoirs to the temperature of the hot reservoir. This formula is expressed as: efficiency = (Th - Tc)/Th, where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir.

## 3. Why is the "Max Work From Heat Engine Reservoirs" important in thermodynamics?

The "Max Work From Heat Engine Reservoirs" is important in thermodynamics because it helps us understand the limitations of heat engines and the maximum amount of work that can be obtained from a given amount of heat. This concept is also used in the design and optimization of heat engines to improve their efficiency and performance.

## 4. Can the "Max Work From Heat Engine Reservoirs" be exceeded?

No, the "Max Work From Heat Engine Reservoirs" cannot be exceeded. This is due to the Second Law of Thermodynamics, which states that it is impossible to create a perpetual motion machine that produces work without any energy input. Therefore, the maximum work that can be obtained from a heat engine is limited by the Carnot efficiency.

## 5. How does the "Max Work From Heat Engine Reservoirs" relate to real-world applications?

The "Max Work From Heat Engine Reservoirs" is an important concept in real-world applications such as power plants and car engines. It helps engineers and scientists understand the efficiency of these systems and how to improve them. By maximizing the work obtained from heat engines, we can reduce energy waste and improve the overall performance of these systems.

• Introductory Physics Homework Help
Replies
1
Views
610
• Introductory Physics Homework Help
Replies
2
Views
991
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
3K
• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
23
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Thermodynamics
Replies
1
Views
851