Maximizing Heat Pump Efficiency with Constant Entropy

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

Let the specific heat capacity of each object be $C$, then
C(T-200)=C(400-T)+C(400-T), where T is the final temperature.
But solving this gives T=333.33 K which is wrong.

Any help is appreciated. Thanks!

 

[gneill]

Contemplate upon the second law of thermodynamics.

 

[Saitama]

Contemplate upon the second law of thermodynamics.

Still lost. Any hints about which equations to start with?

I am not too much familiar with second law of thermodynamics and I did not expect that this question would deal with it. Do I have to deal with entropy here?

 

[Mute]

By what kind of process are the objects exchanging heat?

 

[Saitama]

By what kind of process are the objects exchanging heat?

By radiating the heat?

 

[gneill]

By radiating the heat?

The problem says nothing about how they exchange heat (radiation, conduction, convection,...). But it does say that they form an isolated system. That means no heat escape, no external forces or energies.

Can you quote a statement of the second law of thermodynamics? Hint: If heat moves between the given objects, will the heat flow spontaneously from an object of lower temperature to one of higher temperature?

 

[Mute]

By radiating the heat?

If the objects are touching, they would be transferring heat by conduction, no? The problem says the objects are isolated from the rest of the universe, so I doubt they're radiating heat.

So, your objects are isolated. I guess what I was wondering was, are the objects allowed to expand/contract when heated?

I'm not sure if these questions are pointing you in the right direction - I just find it odd that the question asks for the "highest possible temperature one of them can reach". To me that suggests there may be a process of heat exchange different than the one you calculated in which the boxes can achieve a higher temperature than 333.33 K.

e: I'll let gneill handle this. Seems to have a better idea of how to solve the problem than I do, and I won't have much time to think about it anyways.

 

[Saitama]

Can you quote a statement of the second law of thermodynamics?

From Wikipedia: "The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium—the state of maximum entropy."

Hint: If heat moves between the given objects, will the heat flow spontaneously from an object of lower temperature to one of higher temperature?

No, the process is not spontaneous. But I am still unsure about what equations to start with.

 

[gneill]

From Wikipedia: "The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium—the state of maximum entropy."

Note the phrase, "isolated systems spontaneously evolve towards thermodynamic equilibrium". That means heat, if it moves, moves from things of higher temperature to things of lower temperature. When everything has the same temperature then the system is in equilibrium.

Here we have three objects with starting temperatures. Which of the objects can achieve a higher temperature if heat always moves from higher temp to lower temp? Can the highest available temp increase?

No, the process is not spontaneous. But I am still unsure about what equations to start with.

No equations are required! Just the second law and its consequences.

 

[haruspex]

Note the phrase, "isolated systems spontaneously evolve towards thermodynamic equilibrium". That means heat, if it moves, moves from things of higher temperature to things of lower temperature. When everything has the same temperature then the system is in equilibrium.

Here we have three objects with starting temperatures. Which of the objects can achieve a higher temperature if heat always moves from higher temp to lower temp? Can the highest available temp increase?

I agree with Mute, there's something fishy about this question. On the face of it, the answer is trivial (yes?), but I suspect the question intends that the objects may form a heat engine. Thus, conservation of energy and non-reduction of entropy are the only two constraints.

 

[Saitama]

Here we have three objects with starting temperatures. Which of the objects can achieve a higher temperature if heat always moves from higher temp to lower temp? Can the highest available temp increase?

The highest available temperature is 400 K and it increases only if the heat flows from lower temperature to higher but that is not possible (or spontaneous).

I agree with Mute, there's something fishy about this question. On the face of it, the answer is trivial (yes?), but I suspect the question intends that the objects may form a heat engine. Thus, conservation of energy and non-reduction of entropy are the only two constraints.

How does the system forms a heat engine?

 

[haruspex]

How does the system forms a heat engine?

In principle, the flow of heat from one of the hotter bodies to the cooler body can be used to drive some heat up the gradient to the other hot body. I know it says there are just these three bodies, nothing else, but perhaps they are in the form of a heat engine. The details don't matter.
If we assume that the only constraints are conservation of energy and non-reduction of total entropy it will be possible to calculate a max temp (and it will be > 400K). I gather that you will then be able to check if it's the accepted answer.

 

[Saitama]

If we assume that the only constraints are conservation of energy and non-reduction of total entropy it will be possible to calculate a max temp (and it will be > 400K). I gather that you will then be able to check if it's the accepted answer.

Non-reduction of entropy?

Something like initial entropy=final entropy? If so, how am I supposed to calculate the initial and final entropy?

 

[haruspex]

Let the bodies be A, B (each 400K) and C. Suppose we arrange a small quantity of heat ΔQ to flow from A to C and use that to pump a further ΔR from A to B. Assume some arbitrary specific heat for all three. Write down the equations for no net change in entropy. This should lead to a differential equation.

Edit: For the purposes of the differential equation, you can treat one body in isolation to determine the change in entropy as its temperature changes from T1 to T2. Or you may already know a formula for that.

 

[Saitama]

Edit: For the purposes of the differential equation, you can treat one body in isolation to determine the change in entropy as its temperature changes from T1 to T2. Or you may already know a formula for that.

I don't think I still get it but here is what I get:
dS=\frac{dQ}{T}
$dQ=CdT$
\Delta S=C\ln\left(\frac{T_2}{T_1}\right)
Are you talking about this? I still don't see how this would help me.

 

[haruspex]

\Delta S=C\ln\left(\frac{T_2}{T_1}\right)
Are you talking about this?

Yes. Now, if the three temperatures change from T_A0, T_B0, T_C0 to T_A1, T_B1, T_C1, what is the total change in entropy? What other equation can you write relating those 6 variables (since the bodies have the same S.H and mass)?

 

[Saitama]

Yes. Now, if the three temperatures change from T_A0, T_B0, T_C0 to T_A1, T_B1, T_C1, what is the total change in entropy? What other equation can you write relating those 6 variables (since the bodies have the same S.H and mass)?

Total change in entropy:
\Delta S_{total}=C\ln\left(\frac{T_{A1}}{T_{A0}}\right)+C\ln\left(\frac{T_{B1}}{T_{B0}}\right)+C\ln\left(\frac{T_{C1}}{T_{C0}}\right)

I have no idea about the other equations.

Sorry if I am missing something obvious. I do know the second law of thermodynamics but this section did not get much attention in my course, we only touched it during the Carnot cycle. Sorry again. :(

 

[haruspex]

Total change in entropy:
\Delta S_{total}=C\ln\left(\frac{T_{A1}}{T_{A0}}\right)+C\ln\left(\frac{T_{B1}}{T_{B0}}\right)+C\ln\left(\frac{T_{C1}}{T_{C0}}\right)

Good. Now, for max final temp, you can assume entropy stays constant. That will greatly simplify the above equation. You also know that total energy is constant. Express that using the same six variables. You know the initial temperatures, so it leaves you with three unknowns and two equations. We need one more equation.
Suppose that we start with A and B at 400K and C at 200K, and we use A to heat B and C. When we can pump no more heat into B, what can you say about the temperatures of A and C?

 

[Saitama]

Good. Now, for max final temp, you can assume entropy stays constant. That will greatly simplify the above equation. You also know that total energy is constant. Express that using the same six variables. You know the initial temperatures, so it leaves you with three unknowns and two equations. We need one more equation.
Suppose that we start with A and B at 400K and C at 200K, and we use A to heat B and C. When we can pump no more heat into B, what can you say about the temperatures of A and C?

As the entropy stays constant
T_{A1}\cdot T_{B1}\cdot T_{C1}=T_{A0}\cdot T_{B0}\cdot T_{C0}

For energy conservation, I don't think that the following equation is correct,
C(T_{A1}-T_{A0}+C(T_{B1}-T_{B0})+C(T_{C1}-T_{C0})=0
I really have no idea about the third equation. When A heats B and C, the temperature of A decreases and that of C increases.

 

[haruspex]

As the entropy stays constant
T_{A1}\cdot T_{B1}\cdot T_{C1}=T_{A0}\cdot T_{B0}\cdot T_{C0}

For energy conservation, I don't think that the following equation is correct,
C(T_{A1}-T_{A0}+C(T_{B1}-T_{B0})+C(T_{C1}-T_{C0})=0

Apart from a missing parenthesis in the second, both equations look right to me. You can cancel the C in the second and rearrange more like the first equation.

I really have no idea about the third equation. When A heats B and C, the temperature of A decreases and that of C increases.

Yes, but remember its the down gradient flow from A to C that we're using to pump heat into B. That can only continue until... ?

 

[Saitama]

Yes, but remember its the down gradient flow from A to C that we're using to pump heat into B. That can only continue until... ?

I really have no idea.

 

[haruspex]

... until there's no gradient, which will mean the temperatures of A and C will be...?

 

[Saitama]

... until there's no gradient, which will mean the temperatures of A and C will be...?

When there is no gradient, there is no flow of heat. Before that, the temperature of A decreases and that of C increases.

 

[haruspex]

When there is no gradient, there is no flow of heat. Before that, the temperature of A decreases and that of C increases.

Right, but what does that tell you about the difference between the temperatures of A and C at that point?

 

[collinsmark]

Pranav-Arora,

Keep up the conversation with haruspex; that's good stuff.

But I want to make sure there is no confusion here in the problem statement. It's not asking for the final temperature of the objects.

The temperature of each object is a function of time. At time t = 0, the temperatures of each object is specified to be:

T_1(0) = 200 \ \mathrm{K}, \ \ T_2(0) = 400 \ \mathrm{K}, \ \ T_3(0) = 400 \ \mathrm{K}.

A long time later (Edit: assuming the heat exchange lasts forever*),

T_1(\infty) = T_2(\infty) = T_3(\infty) = \mathrm{Some \ other \ temperature}.

But the problem statement is not [well, technically not necessarily] asking you what that final temperature is. Rather it's asking you "what is the maximum possible temperature anyone of them can have from the time t = 0 to the time t = \infty."

*(If the heat exchange doesn't last forever, it means the final temperature of each of the three objects will be different. But there is still something that can be said about the maximum temperature that anyone of them can have from the time 0 < t < \infty, given the second law of thermodynamics.)

 

[Saitama]

Right, but what does that tell you about the difference between the temperatures of A and C at that point?

That $T_{C1}-T_{A1}$ is positive?

 

[haruspex]

That $T_{C1}-T_{A1}$ is positive?

No, I mean the point at which there is no gradient. When no more heat can be pumped into B, the temperatures of A and C will be equal, right? That gives you your third equation.
From these you can extract a cubic in one unknown. Now, solving a cubic is a lot easier if you happen to know one root. Here's where the problem setter has been kind to you - you do already have one solution to it. (Hint)

 

[collinsmark]

I'm sorry to interject here again, but I really want to make sure we're all on the right page with the original problem statement.

The original problem said, "Find the highest possible temperature one of them can reach in Kelvin."

We know object 2 and object 3 started out at 400 K.

So the final answer (to the original problem statement) must be greater than or equal to 400 K.

Can it be greater than 400 K? The first law of thermodynamics does not prohibit that. But what about the second law? How does the second law fit in here?

 

[haruspex]

How does the second law fit in here?

On the basis that the max temp achievable will correspond to no change in entropy, that produced the equation "product of absolute temperatures = constant".

 

[collinsmark]

On the basis that the max temp achievable will correspond to no change in entropy, that produced the equation "product of absolute temperatures = constant".

I agree!

But like gneill pointed out in an earlier post, one doesn't really need any equations whatsoever to solve the original problem statement. Rather it's a matter of qualitative reflection on the implications of the second law of thermodynamics.

Hypothetically, suppose object 1 and object 3 transferred some or all their heat into object 2. In that case, object 2's temperature would be greater than 400 K. That is not prohibited by the first law of thermodynamics. But would that violate the second law?

Or in a different hypothetical situation, if the entropy of the system is kept constant, what does that say about the amount of spontaneous heat transfer that takes place?

-------------

Edit:

Oh, wait, I get it. When two of the three objects (at different temperatures) transfer heat between each other, there is some work that can be extracted. That work being extracted can be used to increase the temperature of the third object.

Nevermind. I apologize if I confused anybody. Sorry about that. I agree now that equations are needed.

 

[Saitama]

No, I mean the point at which there is no gradient. When no more heat can be pumped into B, the temperatures of A and C will be equal, right? That gives you your third equation.
From these you can extract a cubic in one unknown. Now, solving a cubic is a lot easier if you happen to know one root. Here's where the problem setter has been kind to you - you do already have one solution to it. (Hint)

I got a cubic in one unknown,
2T_{A1}^3-1000T_{A1}^2+32\times 10^6=0
Honestly, I can not figure out that one root and had to use a calculator. The calculator gave me 2 positive roots, 256.16 and 400. The second root indicates that no heat is extracted from A, I don't see how this could have been obvious to me. :(

Corresponding to these two values of $T_{A1}$, I get $T_{B1}$ equal to 200 and 487.67. Is 487.67 the correct answer?

...the max temp achievable will correspond to no change in entropy..

I haven't been able to understand why the total change in entropy is constant? How did you even bring this relation? Have I missed out anything in my notes? If so, please give me a link which explains this.

 

[haruspex]

I got a cubic in one unknown,
2T_{A1}^3-1000T_{A1}^2+32\times 10^6=0
Honestly, I can not figure out that one root and had to use a calculator. The calculator gave me 2 positive roots, 256.16 and 400. The second root indicates that no heat is extracted from A, I don't see how this could have been obvious to me. :(

You had three equations for the three temperatures: known product, known sum, and two of them the same. The initial conditions met all these criteria, so must be a solution.

Corresponding to these two values of $T_{A1}$, I get $T_{B1}$ equal to 200 and 487.67. Is 487.67 the correct answer?

That's what I got, as I recall. (Did it in my head while cycling.)

I haven't been able to understand why the total change in entropy is constant?

You were asked for the max temp that could be achieved (in principle). If entropy increases that will reduce your ability to pump heat up the gradient, so to get the max take entropy as constant.

This was a pretty problem. I hope you enjoyed the journey as much as I did.

 

[Saitama]

Thank you haruspex for your input and patience. It was fun discussing the problem here, got to learn something new.

If entropy increases that will reduce your ability to pump heat up the gradient, so to get the max take entropy as constant.

This cleared all my doubts, thank you again.