Thermodynamics Problem: Heat Engine Between Two Blocks

[Master1022]

Homework Statement

A block of metal (mass m, specific heat c) is originally at temperature T1. A reversible heat engine is run between the block and the surroundings at T0. Find an expression for the work extracted from the engine (considering both cases of T1 > T0 and T0 < T1)

Relevant Equations

1st Law: Q - W = U

Hi,

I am quite confused about how to approach this problem. I have seen variations of this problem where there is a heat engine between two blocks, but in this case the surroundings are massless, so I don't believe that approach will work here.

Method:
I have first started with the case that $T_1 > T_0$. I have noted that $Q - W = \Delta U$.

I have first written down $dU = mc dT_1$ and $dQ = -mc dT_1$. However, substituting those into the 1st law above and integrating, does not yield the correct answer.

Does anyone have any tips on where I should begin?

Thanks in advance

 

[Chestermiller]

Homework Statement:: A block of metal (mass m, specific heat c) is originally at temperature T1. A reversible heat engine is run between the block and the surroundings at T0. Find an expression for the work extracted from the engine (considering both cases of T1 > T0 and T0 < T1)
Relevant Equations:: 1st Law: Q - W = U

Hi,

I am quite confused about how to approach this problem. I have seen variations of this problem where there is a heat engine between two blocks, but in this case the surroundings are massless, so I don't believe that approach will work here.

The surroundings are not massless. They have infinite mass and infinite heat capacity. So all the heat transfer to or from the surroundings takes place at To.

Method:
I have first started with the case that $T_1 > T_0$. I have noted that $Q - W = \Delta U$.

I have first written down $dU = mc dT_1$ and $dQ = -mc dT_1$. However, substituting those into the 1st law above and integrating, does not yield the correct answer.

Does anyone have any tips on where I should begin?

Thanks in advance

The engine is going to be operating in cycles, so its internal energy and its entropy do not change.

If T1 >To, you are going to be transferring heat from the block to the engine, doing work, and rejecting a smaller amount of heat from the engine to the surroundings. You will get maximum work from this combination if the operation is reversible, so that the sum of the entropy changes of the block, the surroundings, and the reservoir are zero.[/QUOTE]

 

[Master1022]

Thank you very much for your response. Does the assumption that we don't change the internal energy or entropy arise from the fact that we should end up (on a graph) where we start off? I believe your suggestion has led me to the correct answer, but can you confirm if my logic is correct?

Okay, so if the process is reversible, then: $ds = \frac{dq}{T}$. Therefore, I have $$dS_{block} = \frac{dQ}{T_1} = \frac{-mc d T_1}{T_1}$$. Then, for the surroundings, we have: $$dS_{surroundings} = \frac{mcdT_1 - dW}{T_0}$$
Therefore, $\Delta S_{system} = \Delta S_{block} + \Delta S_{surroundings} = 0$. Subsituting into this equation we have:
$$\frac{-mc d T_1}{T_1} + \frac{mcdT_1 - dW}{T_0} = 0$$
$$\frac{-mc d T_1}{T_1} + \frac{mcdT_1}{T_0} = \frac{dW}{T_0}$$
integrating the tempertature for the block between $T_0$ and $T_1$ and the work between $0$ and $W$ yields:
$$- mc \ln \left( \frac{T_1}{T_0} \right) + \frac{mc}{T_0} \left( T_1 - T_0 \right) = \frac{W}{T_0}$$

Rearranging yields: $W = - mc T_0 \ln \left( \frac{T_1}{T_0} \right) + mc \left( T_1 - T_0 \right)$

However, one more question I have about this process is (and I believe this comes from my own lack of true understanding of limits within calculus): 'Why do the limits go from T0 to T1'? Is it because we have already put the -ve sign there to indicate that the entropy change of the block is < 0 and swapping the limits would effectively counter that change?

Thank you for your help.

 

[Chestermiller]

Thank you very much for your response. Does the assumption that we don't change the internal energy or entropy arise from the fact that we should end up (on a graph) where we start off?

Correct. Yes, the engine operates in a cycle.

Okay, so if the process is reversible, then: $ds = \frac{dq}{T}$. Therefore, I have $$dS_{block} = \frac{dQ}{T_1} = \frac{-mc d T_1}{T_1}$$. Then, for the surroundings, we have: $$dS_{surroundings} = \frac{mcdT_1 - dW}{T_0}$$
Therefore, $\Delta S_{system} = \Delta S_{block} + \Delta S_{surroundings} = 0$. Subsituting into this equation we have:
$$\frac{-mc d T_1}{T_1} + \frac{mcdT_1 - dW}{T_0} = 0$$
$$\frac{-mc d T_1}{T_1} + \frac{mcdT_1}{T_0} = \frac{dW}{T_0}$$
integrating the tempertature for the block between $T_0$ and $T_1$ and the work between $0$ and $W$ yields:
$$- mc \ln \left( \frac{T_1}{T_0} \right) + \frac{mc}{T_0} \left( T_1 - T_0 \right) = \frac{W}{T_0}$$

Rearranging yields: $W = - mc T_0 \ln \left( \frac{T_1}{T_0} \right) + mc \left( T_1 - T_0 \right)$

However, one more question I have about this process is (and I believe this comes from my own lack of true understanding of limits within calculus): 'Why do the limits go from T0 to T1'? Is it because we have already put the -ve sign there to indicate that the entropy change of the block is < 0 and swapping the limits would effectively counter that change?

Thank you for your help.

Yes. Nicely done.

 

[Master1022]

I am sorry to bring up this old thread. I have been revisiting this sort of problem and have used a (very) slightly different method to get to the same answer. Why do we use $\Delta S_{system}$ rather than $\Delta S_{engine}$? Is that just by convention? For example, we know that the process is reversible and therefore $ds = \sum \frac{dQ}{T} = 0$. If we define positive Q as flowing into the engine, we can write:
$$\frac{dQ_1}{T_1} - \frac{dQ_2}{T_2} = 0$$
$$\frac{mc dT_1}{T_1} - \frac{(mc dT_1 - dW)}{T_0} = 0$$
where $dT_1 > 0$ by my definition. We can continue as the post above does and end up with the same answer. This is basically just multiplying the equation in post #3 by -1. I know that this doesn't make any difference here, but am still keen to know about the process.

Thanks.

 

[Chestermiller]

I am sorry to bring up this old thread. I have been revisiting this sort of problem and have used a (very) slightly different method to get to the same answer. Why do we use $\Delta S_{system}$ rather than $\Delta S_{engine}$? Is that just by convention? For example, we know that the process is reversible and therefore $ds = \sum \frac{dQ}{T} = 0$. If we define positive Q as flowing into the engine, we can write:
$$\frac{dQ_1}{T_1} - \frac{dQ_2}{T_2} = 0$$
$$\frac{mc dT_1}{T_1} - \frac{(mc dT_1 - dW)}{T_0} = 0$$
where $dT_1 > 0$ by my definition. We can continue as the post above does and end up with the same answer. This is basically just multiplying the equation in post #3 by -1. I know that this doesn't make any difference here, but am still keen to know about the process.

Thanks.

One method treats the entire combination (bodies plus engine) as the system, and the other approach treats just the engine as the system. It doesn't really matter,, since the change in entropy of the engine is zero either way.

 

[etotheipi]

Therefore, $\Delta S_{system} = \Delta S_{block} + \Delta S_{surroundings} = 0$. Subsituting into this equation we have:
$$\frac{-mc d T_1}{T_1} + \frac{mcdT_1 - dW}{T_0} = 0$$
$$\frac{-mc d T_1}{T_1} + \frac{mcdT_1}{T_0} = \frac{dW}{T_0}$$
integrating the tempertature for the block between $T_0$ and $T_1$ and the work between $0$ and $W$ yields:
$$- mc \ln \left( \frac{T_1}{T_0} \right) + \frac{mc}{T_0} \left( T_1 - T_0 \right) = \frac{W}{T_0}$$

Rearranging yields: $W = - mc T_0 \ln \left( \frac{T_1}{T_0} \right) + mc \left( T_1 - T_0 \right)$

However, one more question I have about this process is (and I believe this comes from my own lack of true understanding of limits within calculus): 'Why do the limits go from T0 to T1'? Is it because we have already put the -ve sign there to indicate that the entropy change of the block is < 0 and swapping the limits would effectively counter that change?

I believe there is an inconsistency in your signs somewhere, and that is why you had to integrate the wrong way around to get the right answer.

Since $T_1$ is a constant, I would define the temperature of the block to be $T_b$ and write
$$-\frac{dQ_b}{T_b} + \frac{dQ_s}{T_0} = 0$$Since $dQ_{b} = dW + dQ_{s}$ and $-dQ_{b} = mcdT_b$, $$\frac{mcdT_b}{T_b} + \frac{-mcdT_b - dW}{T_0} = 0 \implies \int_{T_1}^{T_0} \left(\frac{1}{T_b} - \frac{1}{T_0} \right)mcdT_b = \int_0^W\frac{1}{T_0}dW$$ $$mc\ln \left(\frac{T_0}{T_1} \right) - \frac{mc(T_0 - T_1)}{T_0} = \frac{W}{T_0}$$

 

[Master1022]

I believe there is an inconsistency in your signs somewhere, and that is why you had to integrate the wrong way around to get the right answer.

Since $T_1$ is a constant, I would define the temperature of the block to be $T_b$ and write
$$-\frac{dQ_b}{T_b} + \frac{dQ_s}{T_0} = 0$$Since $dQ_{b} = dW + dQ_{s}$ and $-dQ_{b} = mcdT_b$, $$\frac{mcdT_b}{T_b} + \frac{-mcdT_b - dW}{T_0} = 0 \implies \int_{T_1}^{T_0} \left(\frac{1}{T_b} - \frac{1}{T_0} \right)mcdT_b = \int_0^W\frac{1}{T_0}dW$$ $$mc\ln \left(\frac{T_0}{T_1} \right) - \frac{mc(T_0 - T_1)}{T_0} = \frac{W}{T_0}$$

Thanks for your reply. So in my working I defined $dT_1$ (or $dT_b$) to be positive and therefore the limits needed to move in the same direction. Perhaps that is the issue?

I know the latter point is more of a calculus problem. In an earlier thread, I had a similar queries about how to define the direction of the $dx$ when integrating. Although I was dealing with vector quantities, I think the advice was to define $dx$ in the direction of $x$ increasing, regardless of which way the integral was going.

Also, just so I can follow your work, are you considering $\Delta S_{system}$ or $\Delta S_{engine}$? They lead to the same answer as advised by @Chestermiller , but there is a -ve sign difference based on what you are considering flow to/from.

Please do let me know if I am wrong.

EDIT: after a second review, I realized you are considering $\Delta S_{system}$. The difference is in the definition of the $dT_b$, which you define to be negative, thus requiring the introduction of the extra -ve signs. Once again, please do correct me if I have misunderstood

 

[etotheipi]

A differential $\text{d}q$ of any quantity $q$ can carry a sign, i.e. it can be positive or negative. That is why it is never meaningful to consider a differential in isolation, since we must also work with a pair (ratio) of differentials. E.g. if $\text{d}a = 3\text{d}b$, then either both differentials are positive or both are negative.

Thanks for your reply. So in my working I defined $dT_1$ (or $dT_b$) to be positive and therefore the limits needed to move in the same direction. Perhaps that is the issue?

This is the problem; the sign of a differential is not something that can be defined. If the block's temperature is decreasing, then you will have effectively $dT_b < 0$. But I emphasise that this should not matter, you do not need to know whether the block's temperature is increasing or decreasing in order to write the relationship between two quantities. See below!

I know the latter point is more of a calculus problem. In an earlier thread, I had a similar queries about how to define the direction of the $dx$. Although I was dealing with vector quantities, I think the advice was to define $dx$ in the direction of $x$ increasing, regardless of which way the integral was going.

If you are working with two differentials, as you must always do, the signs always work out. For instance, from mechanics we might say $dU = -F_x dx$. If $F_x$ is positive then if $dx > 0$ then $dU < 0$, and vice versa.

EDIT: after a second review, I realized you are considering $\Delta S_{system}$. The difference is in the definition of the $dT_b$, which you define to be negative, thus requiring the introduction of the extra -ve signs. Once again, please do correct me if I have misunderstood

I do consider the system, yes. But as I mention above I have not defined any sign of $dT_b$, as you cannot do that!

The method is always to determine a relationship between differentials, then construct the limits from "initial state" to "final state".

 

[Master1022]

A differential $\text{d}q$ of any quantity $q$ can carry a sign, i.e. it can be positive or negative. That is why it is never meaningful to consider a differential in isolation, since we must also work with a pair (ratio) of differentials. E.g. if $\text{d}a = 3\text{d}b$, then either both differentials are positive or both are negative.
This is the problem; the sign of a differential is not something that can be defined. If the block's temperature is decreasing, then you will have effectively $dT_b < 0$. But I emphasise that this should not matter, you do not need to know whether the block's temperature is increasing or decreasing in order to write the relationship between two quantities. See below!
If you are working with two differentials, as you must always do, the signs always work out. For instance, from mechanics we might say $dU = -F_x dx$. If $F_x$ is positive then if $dx > 0$ then $dU < 0$, and vice versa.
I do consider the system, yes. But as I mention above I have not defined any sign of $dT_b$, as you cannot do that!

The method is always to determine a relationship between differentials, then construct the limits from "initial state" to "final state".

Thank you for taking the time to explain it. The only question I have at the moment is: doesn't saying $- \frac{dQ_b}{T_b}$ implicitly define a prior knowledge of direction from the initial to final temperature? I will take some time to make sure I understand this post before writing back if I have any more questions. I think the main reason I do that for these types of problems is to make it clear to myself which quantities are either positive or negative (e.g. we know that $\Delta S_{block} < 0$, so have the - sign there just helps out - I take your point that it isn't necessary to do it the way I do)

 

[etotheipi]

Thank you for taking the time to explain it. I will take some time to make sure I understand this post before writing back if I have any questions. I think the main reason I do that for these types of problems is to make it clear to myself which quantities are either positive or negative (e.g. we know that $\Delta S_{block} < 0$, so have the - sign there just helps out - I take your point that it isn't necessary to do it the way I do)

It's certainly useful to look at the "relative" signs of different quantities, e.g. if we define $dQ_b$ as the heat leaving the block, then we must have $dQ_b = -mcdT_b$, since for instance a negative $dT_b$ implies a positive $dQ_b$. But as soon as you have these relations, you can forget all about the signs of the differentials and just integrate normally, from the initial to final values.

 

[Master1022]

It's certainly useful to look at the "relative" signs of different quantities, e.g. if we define $dQ_b$ as the heat leaving the block, then we must have $dQ_b = -mcdT_b$, since for instance a negative $dT_b$ implies a positive $dQ_b$. But as soon as you have these relations, you can forget all about the signs of the differentials and just integrate normally, from the initial to final values.

This makes a bit more sense. I just wanted to quickly ask a follow up before I go away to think about this. Just to confirm, is a good way to proceed as follows: start with the fundamental relation $dQ_{blocktooutside} = mcdT$ and then determine the sign based on asking (as you have done) "what should be the effect of a +/- change in $dT$"?

 

[etotheipi]

This makes a bit more sense. I just wanted to quickly ask a follow up before I go away to think about this. Just to confirm, is a good way to proceed as follows: start with the fundamental relation $dQ_{blocktooutside} = mcdT$ and then determine the sign based on asking (as you have done) "what should be the effect of a +/- change in $dT$"?

Yes, that's a good way of doing it. The good news is that you usually never have to worry too much about this sort of thing, because most of the time you will be using existing formulae e.g. $dW = F_x dx$, or $dx = v_x dt$, or $dE = \gamma dA$, and so on.

The only reason that we need to think about it a little more in thermodynamics is because of all of the different conventions for heat and work, etc. Sometimes you will define $Q>0$ if heat flows into a system, or sometimes $Q < 0$. So in these cases, it is helpful to think about what effect a small change in one quantity has on another quantity in order to get the right signs!

 

[Master1022]

Hi,

I am very sorry to bring this back up once again @Chestermiller and @etotheipi. I thought I actually understood what was going on in these sorts of problems, but I seem to have run into a problem: when doing the block to block problem, we can get two different expressions for work by doing two different (but somewhat similar) methods.

So in this original problem, we have a reversible heat engine between a mass and the surroundings. However, can we not use the EXACT SAME method for an engine from block to block?

So in this question, we did:
1. Write second law of thermodynamics down for a reversible system in terms of $dQ_A$ and $dQ_B$
2. Substitute $dQ_B = dQ_A - dW$
3. Integrate and re-arrange for work

This method doesn't seem to care whether B has a mass or not. Therefore, if we are working with the block-to-block engine, then can we not just use the same method? (the temperature limits will be different though)

If we do this same method for the block to block system (A is hot block and B is cold block; blocks are identical in mass and heat capacity):
1. $dQ_A = - mc dT_A$, $dQ_b = mc dT_B$; $dQ_A = dQ_B + dW$
2. $$ ds = 0 = \frac{-dQ_A}{T_A} + \frac{dQ_B}{T_B} $$
$$ \frac{mc dT_A}{T_A} + \frac{mc dT_B}{T_B} = 0 $$
if we integrate both temperatures from a respective starting temperature $T_A$ or $T_B$ to a common equilibrium final temperature $T_f$, then we can find that $T_f = \sqrt{T_A \cdot T_B }$

Now this is where the problem is: there seem to be two different ways of getting the work and I cannot see why one is incorrect?

Method 1: same as above
$$ \frac{mc dT_A}{T_A} + \frac{mc dT_B}{T_B} = 0 $$
$$ \frac{mc dT_A}{T_A} + \frac{-mc dT_A - dW}{T_B} = 0 $$
$$ \int_0^W \frac{dW}{T_B} = mc \int_{T_A}^{T_f} \frac{dT_A}{T_A} - \frac{mc}{T_B} \int_{T_A}^{T_f} dT_A $$
$$ W = mc T_B ln \left( \frac{\sqrt{T_A T_B}}{T_A} \right) - mc ( \sqrt{T_A T_B} - T_A ) $$

we have a natural logarithm in our expression

Method 2: use first law to get the work
$$ dQ_A = dQ_B + dW $$
$$ \int_0^W dW = - mc \int_{T_A}^{T_f} dT_A - mc \int_{T_B}^{T_f} dT_b $$
$$ W = mc( T_A + T_B - 2 \sqrt{T_A T_B} ) $$

now we don't...

Can you perhaps help me see why one method is invalid? Is method 1 invalid because $T_B$ isn't a constant for the block-to-block problem but method 1 assumes it is?

 

[etotheipi]

I will be a little frank, your notation is a bit of a mess. Specifically, you're re-using the same symbols for integration variables, and limits (which in this problem, should be constants). You're also mixing and matching different parts of different scenarios. There are essentially two different questions, heat transfer between two blocks (both of which have variable temperature) via a Carnot engine, and heat transfer from one block to the surroundings (which is at constant temperature) via a Carnot engine. I'll do these two examples separately so you can see why it's so important to clearly define the problem!

First, let's consider two blocks $A$ and $B$ [with $T_A(0) > T_B(0)$], of equal mass and specific heat capacity, connected via a Carnot engine. We'll also say that $T_A(0) = T_H$, and $T_B(0) = T_C$, where $T_H$ and $T_C$ are constants, the initial temperatures. The final bit of notation is that I'll use $dQ_A$ to denote the heat moving out of block A, into the engine (i.e. $dQ_A = -mcdT_A$), and $dQ_B$ to denote the heat moving into block B, from the engine (i.e. $dQ_B = mc dT_B$). Let's use the fact that the entropy of the engine is constant, i.e.$$dS_{\text{engine}} = \frac{dQ_A}{T_A} - \frac{dQ_B}{T_B} = 0 \implies \frac{dT_A}{T_A} + \frac{dT_B}{T_B} = 0$$Let's say both blocks eventually equalise to a temperature $T_f$; then$$\int_{T_H}^{T_f} \frac{dT_A}{T_A} = - \int_{T_C}^{T_f} \frac{dT_B}{T_B} \implies \ln{\frac{T_f}{T_H}} = -\ln{\frac{T_f}{T_C}} = \ln{\frac{T_C}{T_f}}$$from which it follows that $T_f = \sqrt{T_H T_C}$. Now let's apply the first law to the engine,$$\begin{align*}dU_{\text{engine}} = dQ_A - dQ_B - dW = 0 &\implies dW = -mc (dT_A + dT_B)\\

&\implies W = -mc \left(\int_{T_H}^{\sqrt{T_H T_C}} dT_A + \int_{T_C}^{\sqrt{T_H T_C}} dT_B \right) \\

&\implies W = -mc (2\sqrt{T_H T_C} - T_H - T_C)
\end{align*}$$Secondly, let's consider a single block $A$, connected to the surroundings $S$ via a Carnot engine. Let $T_A(0) = T_H$, and also let the temperature of the surroundings be constant, at $T_S(t) = T_0, \, \forall t$. Clearly $T_A(\infty) = T_0$, so in this case let's again apply the first law$$dQ_A - dQ_S - dW = 0 \implies dW = -mc dT_A - dQ_S$$We need to find a way to eliminate $dQ_S$, which we can just do by using the constant entropy condition for the engine, specifically $dQ_S = (T_0 / T_A) dQ_A = -mc (T_0 / T_A) dT_A$. Hence$$dW = -mc dT_A + mc \left(\frac{T_0}{T_A}\right) dT_A = -mc \left( 1 - \frac{T_0}{T_A} \right) dT_A$$Doing the integration,$$ \begin{align*}W &= -mc \int_{T_H}^{T_0} \left( 1 - \frac{T_0}{T_A} \right) dT_A\\ &= -mc \left[ T_A - T_0 \ln{T_A} \right]_{T_H}^{T_0}\\ &= -mc(T_0 - T_H - T_0 \ln{\frac{T_0}{T_H}})\\ &= mc(T_H - T_0) - mcT_0 \ln{\frac{T_H}{T_0}}\end{align*}$$Hopefully you can see that these are two different problems, with two different solutions!

 

[Master1022]

I will be a little frank, your notation is a bit of a mess. Specifically, you're re-using the same symbols for integration variables, and limits (which in this problem, should be constants). You're also mixing and matching different parts of different scenarios. There are essentially two different questions, heat transfer between two blocks (both of which have variable temperature) via a Carnot engine, and heat transfer from one block to the surroundings (which is at constant temperature) via a Carnot engine. I'll do these two examples separately so you can see why it's so important to clearly define the problem!

Thanks for your reply @etotheipi. Apologies, I ought to have included a few subscript zeros in the problem - that's just laziness on my part. Thanks for clarifying the solution - I believe my working follows the same method. However, at the time, I didn't recall that (for the block to block problem) the temperature of the cold block was not a constant and therefore we couldn't use the same trick of substituting in our first law expression into the second law expression and integrating.

Also, not that it matters as we were mainly discussing the concepts, but I think there is a small sign error in your working for the second problem:

$dQ_S = (T_0 / T_A) dQ_A = -mc (T_0 / T_A) dT_A$. Hence$$dW = -mc dT_A - mc \left(\frac{T_0}{T_A}\right) dT_A = -mc \left( 1 + \frac{T_0}{T_A} \right) dT_A$$

should be:

$dQ_S = (T_0 / T_A) dQ_A = -mc (T_0 / T_A) dT_A$. Hence$$dW = -mc dT_A + mc \left(\frac{T_0}{T_A}\right) dT_A = -mc \left( 1 - \frac{T_0}{T_A} \right) dT_A$$

Once again, thanks for the clarification.

 

[etotheipi]

Ah, good catch! Corrected.