Conceptualizing entropy change of surroundings

In summary: However, this path is known to be isothermal. The entropy change (of the system) in going from A-->D is simply the sum of the individual entropy changes (of the system) in each of the above steps (since entropy is a state function and since each step is reversible). Now comes the confusing part:The entropy change of the surroundings in going from A-->D is calculated by summing up the heat transferred to the surroundings in each of the individual steps, and then divided by the temperature of the isothermal process A-->D. This confuses me greatly. I don't understand why this approach is used
  • #36
Chestermiller said:
It doesn't imply this at all. At the boundary between the system and surroundings, the rate of heat flow from the surroundings to the system is always minus the rate of heat flow from the system to the surroundings, and the temperature at the boundary is the same for both the system and the surroundings, say TB. (This assumes that there is only one location where heat is being exchanged between the system and the surroundings). This all implies that dQ/TB for the system is always equal to -dQ/TB for the surroundings.
Why would Tb always be the same for both the system and the reservoir? What if there is no direct thermal contact? (eg they are connected via a Carnot device so that all heat flow occurs isothermally at the temperature of the system or reservoir). And how do we maintain the first law if dQsystem = -dQsurr? Are you assuming that no work is being done?

The entropy change for the system is only equal to dQ/TB if the system is undergoing a reversible path, and the entropy change for the surroundings is only equal to -dQ/TB if the surroundings are undergoing a reversible path.
This cannot be true if work is being done. For example, this is not true for a reversible Carnot engine operating between the system and the surroundings. Qsys ≠ -Qsurr and Tsys ≠ TB

AM
 
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  • #37
Chestermiller said:
It doesn't imply this at all. At the boundary between the system and surroundings, the rate of heat flow from the surroundings to the system is always minus the rate of heat flow from the system to the surroundings, and the temperature at the boundary is the same for both the system and the surroundings, say TB. (This assumes that there is only one location where heat is being exchanged between the system and the surroundings). This all implies that dQ/TB for the system is always equal to -dQ/TB for the surroundings. But this definitely does not mean that the entropy change for the system has to be minus the temperature change for the surroundings. The entropy change for the system is only equal to dQ/TB if the system is undergoing a reversible path, and the entropy change for the surroundings is only equal to -dQ/TB if the surroundings are undergoing a reversible path. We can regard the system and the surroundings as really two separate systems that are mutually constrained at the boundary between them.

Now, for the system, we have that

(dS)syst = dQ/TB for a reversible path

(dS)syst > dQ/TB for an irreversible path

The second relationship represents the simplified version of the so-called Cauchy inequality.

And, for the surroundings,

(dS)surr = -dQ/TB for a reversible path

(dS)surr > -dQ/TB for an irreversible path

So if both system and surroundings undergo reversible paths, then

(dS)syst + (dS)surr = 0

If either the system or the surroundings or both undergo irreversible paths in the actual process, then

(dS)syst + (dS)surr > 0

BiP, yesterday I sent you an analysis I did of the irreversible heating of a metal bar by means of heat supplied by an isothermal reservoir (comprising the surroundings). This is a case where the system is undergoing a irreversible path, while the surroundings are experiencing a reversible path. I showed how to precisely calculate the change in entropy for both the system and the surroundings from the solution to the transient heat conduction equation, and how the positive definiteness of the sum total of the entropy changes for the system and the surroundings comes about (for this problem). I have attached that analysis to this reply for others involved in this thread to examine. This provides a specific example of how the Cauchy inequality evolved, and why the sum of the entropy changes for the system and surroundings is always greater than zero for an irreversible process.

Chet

Thanks Chet! By the way I think you mean Clausius inequality and not Cauchy inequality :tongue:

BiP
 
  • #38
Bipolarity said:
Thanks Chet! By the way I think you mean Clausius inequality and not Cauchy inequality :tongue:

BiP

Yes. You are right. I meant Clausius inequality. I realized that yesterday.

Chet
 
  • #39
Andrew Mason said:
Why would Tb always be the same for both the system and the reservoir? What if there is no direct thermal contact? (eg they are connected via a Carnot device so that all heat flow occurs isothermally at the temperature of the system or reservoir). And how do we maintain the first law if dQsystem = -dQsurr? Are you assuming that no work is being done?

This cannot be true if work is being done. For example, this is not true for a reversible Carnot engine operating between the system and the surroundings. Qsys ≠ -Qsurr and Tsys ≠ TB

AM

Hi Andrew.

I don't have time to do it today, but in the next couple of days, I'm going to address all your questions to your satisfaction. Please bear with me. The reason that people struggle so much with these concepts is that the mathematics is done in such an ambiguous and imprecise way. I am also going to write up a document which explains things in a much more precise way mathematically. (That document may take another few days).

Chet
 
  • #40
I thought we calculated both the entropy change of the system and of the surrounding for the process in question? It can also be seen that the entropy increase of the surrounding is larger than the entropy decrease of the system, so that the total entropy clearly increases as should be for an irreversible process.
The change of entropy of the system is not problematic as both the start point and the end point are equilibrium states and entropy is a state function.
The entropy change of the surrounding we calculated as Q/T is only a lower bound. E.g. we could do some work on the system (stirring or electric heating) and this would produce additional heat transferred to the surrounding.
 
  • #41
DrDu said:
I thought we calculated both the entropy change of the system and of the surrounding for the process in question? It can also be seen that the entropy increase of the surrounding is larger than the entropy decrease of the system, so that the total entropy clearly increases as should be for an irreversible process.
The change of entropy of the system is not problematic as both the start point and the end point are equilibrium states and entropy is a state function.
The entropy change of the surrounding we calculated as Q/T is only a lower bound. E.g. we could do some work on the system (stirring or electric heating) and this would produce additional heat transferred to the surrounding.
The change in entropy of the system is ∫dQ/T over the reversible path A-B-C-D. That works out to (see post #19):

[itex]\Delta S_{sys} = m(C_{water}\ln{\frac{273}{263}} - L/273 - C_{ice}\ln{\frac{273}{263}})[/itex]

The change in entropy of the surroundings is the actual heat flow to the surroundings divided by its temperature. If the actual process is freezing at 263K ie. from the supercooled water to ice at the same temperature, Q = mL (L being the latent heat of fusion) so:

[itex]\Delta S_{surr} = mL/263[/itex]

There is no work done in this process. But if a process produces work (eg via a heat engine), the heat flow into the working substance will be greater than the heat flow out of the working substance but the temperatures at which the heat flows will always be such that |Q|/Th - |Q|/Tc ≤ 0

AM
 
  • #42
Why do you ignore the C_p dependent terms in Delta S_surr but not in Delta S_sys?
 
  • #43
DrDu said:
Why do you ignore the C_p dependent terms in Delta S_surr but not in Delta S_sys?
There is no change in the state of the surroundings except the heatflow to the surroundings from the system. We assume that its pressure, volume and temperature do not change. So change in entropy of the surroundings from A to D is determined by its temperature and the heatflow from the system to the surroundings. A reversible path from A to D is an isothermal process in which total heat flow Q flows to the surroundings where Q is the mass of the water x the heat of fusion for water. The change in entropy = ∫dQ/T over that path is just Q/T = mL/T.

To determine the change in entropy of the system, however, the reversible path from A-D is more complicated. It requires going through steps A-B-C-D. So you have to calculate ∫dQ/T for that path.

AM
 
  • #44
Bipolarity said:
Now comes the confusing part:
The entropy change of the surroundings in going from A-->D is calculated by summing up the heat transferred to the surroundings in each of the individual steps, and then divided by the temperature of the isothermal process A-->D.
What they are probably saying is that the entropy being transferred into the surroundings has to be calculated by reversible processes. However, the surroundings can also gain entropy if entropy is created. The creation of entropy is an irreversible process.
I think the problem is with the phrase "heat transferred". This can mean one of two things, which you haven't specified.
"Heat transferred into the surroundings" can mean the changes in internal energy of the surroundings, This can include energy transferred by heat conduction or it can mean work.
I believe that this is what you are thinking. You are considering energy transferred into the environment by any and all means.
If the energy is transferred into the surroundings by work, then there are frictional forces involved. The energy transferred into the surroundings by friction create entropy. The creation of entropy is irreversible.
"Heat transferred into the surroundings" can mean energy transferred by "heat conduction". The energy transferred is then associated with entropy transfer, not the creation of entropy. Then that by nature is reversible.
 
  • #45
Andrew, what I meant is the following:
The heat of melting, L, is a function of temperature and in formulas for both entropies we should refer to the measured value at T=273K. You calculated the temperature dependence of L correctly already in #19, i.e.
[tex]
L(263 K)=L(273 K)+(\Delta T)(C_{water}-C_{ice})
[/tex]
where [itex]\Delta T=-10K[/itex].
Developing the logarithms, the C_p dependent terms cancel to lowest order in [itex]\Delta T[/itex] and the total entropy change becomes [itex] \Delta S\approx -mL\Delta T/T [/itex].
 
  • #46
DrDu said:
Andrew, what I meant is the following:
The heat of melting, L, is a function of temperature and in formulas for both entropies we should refer to the measured value at T=273K. You calculated the temperature dependence of L correctly already in #19, i.e.
[tex]
L(263 K)=L(273 K)+(\Delta T)(C_{water}-C_{ice})
[/tex]
where [itex]\Delta T=-10K[/itex].
This is only true if the actual process by which the supercoolled water freezes is the reversible process A-B-C-D. If the water freezes at 263K the heat released to the surroundings would not be the same as in the reversible process.
Developing the logarithms, the C_p dependent terms cancel to lowest order in [itex]\Delta T[/itex] and the total entropy change becomes [itex] \Delta S\approx -mL\Delta T/T [/itex].
To find the change in entropy of the surroundings one would have to measure the specific ΔH of supercooled water at 263K turning to ice at 263K and call that L263 so that the change in entropy is mL263/263. I don't know what the temperature dependence of L is but it has to be something that results in an increase in entropy overall.

You seem to be saying that the specific heat of fusion of supercooled water at temperature T is simply:

[itex]L(T) = L_{273} + \Delta T(C_{water} - C_{ice})[/itex]

If that were true, you would be right but I don't know that it is.

AM
 
Last edited:
  • #47
Andrew Mason said:
Why would Tb always be the same for both the system and the reservoir? What if there is no direct thermal contact? (eg they are connected via a Carnot device so that all heat flow occurs isothermally at the temperature of the system or reservoir). And how do we maintain the first law if dQsystem = -dQsurr? Are you assuming that no work is being done?

This cannot be true if work is being done. For example, this is not true for a reversible Carnot engine operating between the system and the surroundings. Qsys ≠ -Qsurr and Tsys ≠ TB

AM
In the present discussion of second law concepts, we have subdivided the world into two separate and distinct entities:

Entity 1 has been loosely referred to as "the system"

Entity 2 (which includes everything not included in Entity 1) has been loosely referred to as "the surroundings."

There is a "boundary" between the system and surroundings which does not permit mass to pass through, so that both the system and surroundings can be regarded as closed . However, the boundary does permit heat to transfer between the system and surroundings (across its interface), and the boundary can move and deform, so that work is done by the system on the surroundings, and vice versa.

The introduction to this picture of a third entity, say a Carnot device, muddies the waters a little, but only slightly. The third entity can be assimilated into either the system or the surroundings, and then we are back down to 2 entities again. In practice, it might be more convenient to assimilate the third entity into the surroundings, if the focus is more on the system.

In general, both the system and the surroundings can be undergoing irreversible processes, and the two regions will be coupled with one another at the boundary interface. What is the nature of this coupling? From our knowledge of transport phenomena and continuum mechanics, we know that certain parameters must be continuous across the interface: velocity vector, heat flux vector, temperature, and traction stress. The traction stress is equal to the stress tensor dotted with a unit normal vector to the interface. For an inviscid fluid, the traction stress is equal to the pressure times a unit normal, and the work being done by the system on the surroundings per unit area of the interface is equal to the pressure times the normal component of velocity.

Now back to our Entities. When most of us studied thermo, we learned it from a historical perspective. Early 2nd law pioneers worked with multiple entities, including heat reservoirs, Carnot engines, etc. As time progressed, and our understanding evolved, the concept of entropy developed, and the mathematics became more precise and refined. It was recognized that the number of entities could be reduced from many to just two, namely, the system and the surroundings. Then Clausius realized that he did not even need two entities to capture the essence of the 2nd law mathematically; he could do it all with just a single entity, the system. (He regarded the surroundings as just another system). Clausius' work formalized the 2nd law in a concise mathematical fashion.

Here is my attempt to paraphrase Clausius' thinking: I have a closed (constant mass) thermodynamic system that is enclosed within an interface boundary β. The system is initially in thermodynamic equilibrium state A, and I wish to bring it to thermodynamic equilibrium state B. I will consider all processes, both reversible and irreversible, that can accomplish this by applying heat transfer at various locations on β, and by moving and deforming β (i.e., doing work at the interface β). As I test these various processes, I will keep track of the local heat transfer rates I apply at the boundary of β (which can vary with position on β and time), and the corresponding temperatures at these locations along the boundary. Then I will calculate the ratios of the local heat transfer rates to the local temperatures, and sum these quantities over the entire boundary. I will then integrate this sum with respect to time, from time zero (corresponding to equilibrium state A), until the time that system has attained equilibrium state B. I will then compare the values I obtain for this integral for all the possible irreversible and reversible paths that I have imposed on the system. According to my understanding of the 2nd law, this integral will vary from path to path, but will never be greater than a certain limiting value. That limiting value is called the entropy change ΔS for the system. If I represents the value of the integral, then:

ΔS = Imax

The integral I attains this maximum value only along a reversible path. Otherwise I is less than Imax.

ΔS ≥ I

Mathematically, Clausius' representation of the second law constitutes a physical principle expressed using variational calculus.

Unfortunately, the mathematical from of the equation used by Clausius to capture the above powerful ideas is very imprecise and ambiguous, and has confused thermodynamics students over the years:

dS ≥ dQ/T

This equation does not specifically indicate where dQ and T are to be evaluated. For a system experiencing an irreversible path, the temperature (as well as pressure) will vary with spatial location and time within the system. So, what value of T should be used?
Also the heat flux within the system will be non-zero, and will vary with location and time. Our discussion above indicates that dQ and T should be evaluated exclusively at the boundary of the system. Our previous discussion also indicates that there can be multiple locations of heat flow in and corresponding temperatures at the boundary.

The continuum mechanics guys have formalized the mathematics of the Clausius inequality even more precisely by integrating the dot product of heat flux and unit normal divided by temperature over the entire boundary surface, and have used the divergence theorem to show mathematically why the change in entropy exceeds the integral by a positive definite amount. Google the key words continuum mechanics and Clausius inequality to see this.
 
  • #48
In the present discussion of second law concepts, we have subdivided the world into two separate and distinct entities:

Entity 1 has been loosely referred to as "the system"

Entity 2 (which includes everything not included in Entity 1) has been loosely referred to as "the surroundings."

There is a "boundary" between the system and surroundings which does not permit mass to pass through, so that both the system and surroundings can be regarded as closed . However, the boundary does permit heat to transfer between the system and surroundings (across its interface), and the boundary can move and deform, so that work is done by the system on the surroundings, and vice versa.

You obviously went to the same school as I did.
 
  • #49
Andrew Mason said:
You seem to be saying that the specific heat of fusion of supercooled water at temperature T is simply:

[itex]L(T) = L_{273} + \Delta T(C_{water} - C_{ice})[/itex]

If that were true, you would be right but I don't know that it is.

AM

The heat of melting at 263K is equal to the enthalpy difference between states A and D.
As H is a state function it can be calculated along the path ABCD, in close analogy to the calculation of the entropy change of the system. Going from A to B and from C to D gives the terms [itex]\pm mC_{water/ice}ΔT[/itex], while the enthalpy change from B to C is -mL.
 
  • #50
DrDu said:
The heat of melting at 263K is equal to the enthalpy difference between states A and D.
As H is a state function it can be calculated along the path ABCD, in close analogy to the calculation of the entropy change of the system. Going from A to B and from C to D gives the terms [itex]\pm mC_{water/ice}ΔT[/itex], while the enthalpy change from B to C is -mL.
Ok. That makes sense. If there is constant pressure then L(T) = L(273) - (Cwater - Cice)(273-T).

AM
 
<h2>1. What is entropy change of surroundings?</h2><p>Entropy change of surroundings is a measure of the increase or decrease in the disorder or randomness of the environment surrounding a system. It is often used to describe the change in heat transfer or energy distribution between a system and its surroundings.</p><h2>2. How is entropy change of surroundings calculated?</h2><p>The entropy change of surroundings can be calculated using the equation ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred between the system and surroundings, and T is the temperature of the surroundings.</p><h2>3. What factors affect the entropy change of surroundings?</h2><p>The entropy change of surroundings is affected by the temperature of the surroundings, the amount of heat transferred, and the nature of the system and its surroundings. It is also influenced by the type of process (reversible or irreversible) and the efficiency of the system.</p><h2>4. How does the entropy change of surroundings relate to the second law of thermodynamics?</h2><p>The second law of thermodynamics states that the total entropy of a closed system and its surroundings always increases in a spontaneous process. This means that the entropy change of the surroundings will always be positive, as the system and surroundings tend towards a state of maximum disorder.</p><h2>5. Can the entropy change of surroundings be negative?</h2><p>Yes, the entropy change of surroundings can be negative if the system is undergoing a reversible process and the heat transfer is from the surroundings to the system. In this case, the surroundings become more ordered and the entropy decreases, while the system becomes more disordered and the entropy increases.</p>

1. What is entropy change of surroundings?

Entropy change of surroundings is a measure of the increase or decrease in the disorder or randomness of the environment surrounding a system. It is often used to describe the change in heat transfer or energy distribution between a system and its surroundings.

2. How is entropy change of surroundings calculated?

The entropy change of surroundings can be calculated using the equation ΔS = q/T, where ΔS is the change in entropy, q is the heat transferred between the system and surroundings, and T is the temperature of the surroundings.

3. What factors affect the entropy change of surroundings?

The entropy change of surroundings is affected by the temperature of the surroundings, the amount of heat transferred, and the nature of the system and its surroundings. It is also influenced by the type of process (reversible or irreversible) and the efficiency of the system.

4. How does the entropy change of surroundings relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system and its surroundings always increases in a spontaneous process. This means that the entropy change of the surroundings will always be positive, as the system and surroundings tend towards a state of maximum disorder.

5. Can the entropy change of surroundings be negative?

Yes, the entropy change of surroundings can be negative if the system is undergoing a reversible process and the heat transfer is from the surroundings to the system. In this case, the surroundings become more ordered and the entropy decreases, while the system becomes more disordered and the entropy increases.

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