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Isn't it contradictory ?

Sorry for my bad english and as I am posting from mobile I can't able to use latex for mathematics so apolozise for that .

Thanx in advance.

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- Thread starter Shan K
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- #1

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Isn't it contradictory ?

Sorry for my bad english and as I am posting from mobile I can't able to use latex for mathematics so apolozise for that .

Thanx in advance.

- #2

Quantum Defect

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Isn't it contradictory ?

Sorry for my bad english and as I am posting from mobile I can't able to use latex for mathematics so apolozise for that .

Thanx in advance.

Entropy is a state function. In calculating the change in entropy for a

Delta S (of the

Two different things -- the

When you calculate the change in entropy of the

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- #4

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Then how can one define a proper entropy change for irreversible process . because it was defined as dS= dQ/T for reversible process.

I think you are confusing "quasi-static" with "reversible". Quasi-static means sufficiently slow and in that case dS=dQ/T whether the process is reversible or irreversible. On the other hand reversible means change in S is zero whereas irreversible means change in S is > 0.

- #5

Chestermiller

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I have written up some notes on this that I think may be helpful to you. Here are the notes, which are really pretty short:

FIRST LAW OF THERMODYNAMICS

Suppose that we have a closed system that at initial time t

[tex]\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W[/tex]

where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

The time variation of [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

If a process path is irreversible, then the temperature and pressure within the system are inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:

[tex]\dot{w}(t)=P_I(t)\dot{V}(t)[/tex]

where P

[itex]P_I(t)=P(t)[/itex] (reversible process path)

Therefore, [itex]\dot{w}(t)=P(t)\dot{V}(t)[/itex] (reversible process path)

Another feature of reversible process paths is that they are carried out very slowly, so that [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (t

SECOND LAW OF THERMODYNAMICS

In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate [itex]\dot{q}(t)[/itex] and the rate of doing work [itex]\dot{w}(t)[/itex] as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:

[tex]Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}[/tex]

[tex]W=\int_{t_i}^{t_f}{\dot{w}(t)dt}[/tex]

In the present section, we will be introducing a third integral of this type (involving the heat transfer rate [itex]\dot{q}(t)[/itex]) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:

[tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}[/tex]

where T

Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today

So, mathematically, we can now state the Second Law as follows:

[tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq\Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}[/tex]

where [itex]\dot{q}_{rev}(t)[/itex] is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the

- #6

Chestermiller

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This is not correct. dS=dQ/T only for a reversible path between the initial and final equilibrium states of a system.I think you are confusing "quasi-static" with "reversible". Quasi-static means sufficiently slow and in that case dS=dQ/T whether the process is reversible or irreversible.

This is true only for anOn the other hand reversible means change in S is zero whereas irreversible means change in S is > 0.

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- #8

Chestermiller

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Please tell me you don't really think that.Chestermiller, the heat transfer from a hot body to a cold body is an example of an irreversible process for which dS=dQ/T.

Chet

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Thanks for the beautiful description .the change in entropy for a system between the initial and final equilibrium states. The change in entropy for a system between the two equilibrium states is thenotmaximumvalue of the integral of dQ/T over all possible paths, both reversible and irreversible. Fortunately, this maximum value is also the same as the integral over each and everyreversiblepath between the two equilibrium states. So, to determine the change in entropy, you can't use the actual irreversible path. You need to dream up a reversible path between the same two equilibrium states (any reversible path will do), and calculate the value of the integral of dQ/T over that path.

I have written up some notes on this that I think may be helpful to you. Here are the notes, which are really pretty short:

FIRST LAW OF THERMODYNAMICS

Suppose that we have a closed system that at initial time t_{i}is in an initial equilibrium state, with internal energy U_{i}, and at a later time t_{f}, it is in a new equilibrium state with internal energy U_{f}. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let [itex]\dot{q}(t)[/itex] represent the rate of heat addition across the interface between the system and the surroundings at time t, and let [itex]\dot{w}(t)[/itex] represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),

[tex]\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W[/tex]

where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

The time variation of [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

If a process path is irreversible, then the temperature and pressure within the system are inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperatureat the interfacecan be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, T_{I}(t) and P_{I}(t) can be used to impose the process path that we desire. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex]).

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:

[tex]\dot{w}(t)=P_I(t)\dot{V}(t)[/tex]

where P_{I}(t) is the pressure at the interface and [itex]\dot{V}(t)[/itex] is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

[itex]P_I(t)=P(t)[/itex] (reversible process path)

Therefore, [itex]\dot{w}(t)=P(t)\dot{V}(t)[/itex] (reversible process path)

Another feature of reversible process paths is that they are carried out very slowly, so that [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (t_{f}-t_{i}) becomes exceedingly large. In this way, Q-W remains constant and finite.

SECOND LAW OF THERMODYNAMICS

In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate [itex]\dot{q}(t)[/itex] and the rate of doing work [itex]\dot{w}(t)[/itex] as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:

[tex]Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}[/tex]

[tex]W=\int_{t_i}^{t_f}{\dot{w}(t)dt}[/tex]

In the present section, we will be introducing a third integral of this type (involving the heat transfer rate [itex]\dot{q}(t)[/itex]) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:

[tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}[/tex]

where T_{I}(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call todaythe reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first dream up a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

So, mathematically, we can now state the Second Law as follows:

[tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq\Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}[/tex]

where [itex]\dot{q}_{rev}(t)[/itex] is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is thesystemtemperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.

So you are saying that for every process the entropy change is fixed whether we take the reversible or irrevesible one. But for reversible paths thhe integral give us just the change in entropy for two states but for the irreversible paths we get less than that of the change in entropy . So it is the mathematical notion which is different for those two paths but the changing of entropy is fixed if the states are fixed. Right ?

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Ya. But it is not the whole description . We use dS=dQ/T for calculating the entropy change . But in this process the system goes from states (say) a to b but the surrounding is not it is going to some other states thats why when we take the system from b to a ,it able to recover its initial state but as the surrounding didnt transit between states a to b as by the system it will not able to come to its initial value thats why the total eenntropy of the universe increases.Chestermiller, the heat transfer from a hot body to a cold body is an example of an irreversible process for which dS=dQ/T.

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Chestermiller

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Yes. Exactly.Thanks for the beautiful description .

So you are saying that for every process the entropy change is fixed whether we take the reversible or irrevesible one. But for reversible paths thhe integral give us just the change in entropy for two states but for the irreversible paths we get less than that of the change in entropy . So it is the mathematical notion which is different for those two paths but the changing of entropy is fixed if the states are fixed. Right ?

Chet

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Chestermiller

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Hi Useful nucleus. I think what you are saying here is that, for a process path involving the irreverisble transfer of heat from a hot body to a cold body, the integral of dQ/T along the irreversible process path gives you the correct change of entropy? If this is what you are saying, I can show you why this statement is incorrect. If you are willing to work with me, I can precisely define such a process and then, with your help, determine both the integral of dQ/T along the reversible path as well as the change in entropy. We can then compare the results. Are you willing to work with me and try this?Chestermiller, the heat transfer from a hot body to a cold body is an example of an irreversible process for which dS=dQ/T.

Chet

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Hi Useful nucleus. I think what you are saying here is that, for a process path involving the irreverisble transfer of heat from a hot body to a cold body, the integral of dQ/T along the irreversible process path gives you the correct change of entropy? If this is what you are saying, I can show you why this statement is incorrect. If you are willing to work with me, I can precisely define such a process and then, with your help, determine both the integral of dQ/T along the reversible path as well as the change in entropy. We can then compare the results. Are you willing to work with me and try this?

Chet

Hi Chestermiller, sure I'm happy to work on that with you but please allow sometime for me to get back to you in this thread. Just also to clarify, in this example I mentioned , the two bodies are perfectly isolated from the surrounding. All what is allowed is heat transfer from the hot to the cold body "quasi-statically". My claim is that the process is irreversible, and my second claim is that TdS=dQ. The second claim holds in its differential form and no need to expect that integrating two equal differentials will give different quantities. Thank you for your well to clarify!

- #14

Chestermiller

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Please call me Chet.Hi Chestermiller, sure I'm happy to work on that with you but please allow sometime for me to get back to you in this thread. Just also to clarify, in this example I mentioned , the two bodies are perfectly isolated from the surrounding. All what is allowed is heat transfer from the hot to the cold body "quasi-statically". My claim is that the process is irreversible, and my second claim is that TdS=dQ. The second claim holds in its differential form and no need to expect that integrating two equal differentials will give different quantities. Thank you for your well to clarify!

OK. Let's see where this takes us. We have two identical flat slabs of the same material, one at temperature T

Chet

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OK. Let's see where this takes us. We have two identical flat slabs of the same material, one at temperature T_{H}and the other at temperature T_{C}. At time zero, we bring the bodies into direct contact with one another and allow them to thermally equilibrate. Is this what you had in mind, or do we need to add something else to obtain your quasi-static constraint? Or, is this OK to start with?

Chet

That sounds good, Chet. Although for such thought experiments I tend to use (ideal) gases and adiabatic or diathermal walls for simplicity , but the example you brought is perfectly fine. I may add the isolation from the surrounding to focus on the two bodies.

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DrDu

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- #17

Chestermiller

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Yes. Isolated as a combination.That sounds good, Chet. Although for such thought experiments I tend to use (ideal) gases and adiabatic or diathermal walls for simplicity , but the example you brought is perfectly fine. I may add the isolation from the surrounding to focus on the two bodies.

Now, what I intend for us to be showing is that, for this irreversible process as posed, the Clausius

So let's begin. I hope that this is not too elementary for you.

What is the final equilibrium temperature of the two identical slabs?

If m is the mass of each slab and C is the heat capacity of each slab, what is the amount of heat that gets transferred from the hot slab to the cold slab?

At all times after the instant that the two identical flat slabs are brought into contact with one another, what is the temperature at the interface between the two slabs?

What is the change in entropy of the initially hot slab?

What is the change in entropy of the initially cold slab?

This is where I'll stop for now.

Chet

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Chestermiller

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Chet

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Chet

Chet, I'm definitely still interested in responding and also to see your answers, but as I mentioned above it might take me sometime to respond. Sorry if took longer than expected. But if there is interest to post your replies soon , please feel free to do so.

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I agree with this, DrDu. But I'll try to find an intuitive argument to reach to this conclusion.

- #21

Chestermiller

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No. I think it would be better if I waited for your responses. That way we would both be invested in the discussion.Chet, I'm definitely still interested in responding and also to see your answers, but as I mentioned above it might take me sometime to respond. Sorry if took longer than expected. But if there is interest to post your replies soon , please feel free to do so.

Chet

- #22

Chestermiller

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Yes. Isolated as a combination.

Now, what I intend for us to be showing is that, for this irreversible process as posed, the Clausiusinequalityapplies to the combination of the two bodies as well as to each of the bodies individually (between the initial and final equilibrium states of the system): ##ΔS>\int{\frac{dQ}{T_I}}##, where T_{I}is the temperature at the interface between the system and the surroundings.

What is the final equilibrium temperature of the two identical slabs?

The final equilibrium temperature of the two slabs = ##\frac{(T_H+T_C)}{2}##

If m is the mass of each slab and C is the heat capacity of each slab, what is the amount of heat that gets transferred from the hot slab to the cold slab?

The amount of heat transferred from the hot slab to the cold slab $$Q = mC\left(T_H-\frac{(T_H+T_C)}{2}\right)=mC\frac{(T_H-T_C)}{2}$$

At all times after the instant that the two identical flat slabs are brought into contact with one another, what is the temperature at the interface between the two slabs?

From analysis of the transient heat transfer problem, the temperature at the interface between the two slabs at all times after they are brought into contact with one another = ##\frac{(T_H+T_C)}{2}##

$$ΔS_H=mC\ln \left(\frac{T_H+T_C}{2T_H}\right)$$What is the change in entropy of the initially hot slab?

$$ΔS_C=mC\ln \left(\frac{T_H+T_C}{2T_C}\right)$$What is the change in entropy of the initially cold slab?

Now let's see how the Clausius inequality applies to the two slabs individually, and then to the two slabs as a combined system:

The integral of dq/T

$$\int {\frac{dq}{T_I}}=\frac{-Q}{T_I}=-mC\frac{T_H-T_C}{T_H+T_C}$$

This can be compared with the change in entropy of the hot slab which is calculated above by:

$$ΔS_H=-mC\ln \left(\frac{2T_H}{T_H+T_C}\right)$$

From these relationships, one can readily verify that for all T

$$ΔS_H>\int {\frac{dq}{T_I}}$$

The integral of dq/T

$$\int {\frac{dq}{T_I}}=\frac{Q}{T_I}=+mC\frac{T_H-T_C}{T_H+T_C}$$

This can be compared with the change in entropy of the cold slab which is calculated above by:

$$ΔS_C=mC\ln \left(\frac{T_H+T_C}{2T_C}\right)$$

From these relationships, one can readily verify that for all T

$$ΔS_C>\int {\frac{dq}{T_I}}$$

The integral of dq/T

$$\int {\frac{dq}{T_I}}=0$$

From the previous results, the change in entropy for the combination of the two slabs is the sum of the entropy changes for each of the slabs individually:

$$ΔS=ΔS_H+ΔS_C=mC\ln \left(\frac{T_H+T_C}{2T_H}\right)+mC\ln \left(\frac{T_H+T_C}{2T_C}\right)$$

or

$$ΔS=mC\ln \left(\frac{(T_H+T_C)^2}{4T_HT_C}\right)$$

One can readily verify that, for all values of T

$$ΔS>\int {\frac{dq}{T_I}}$$

We have shown that, for the irreversible process of transferring heat from a hot slab to a cold slab, the Cauchy inequality is satisfied irrespective of whether we consider the slabs individually, or whether we consider them in combination, and thus that the change in entropy in each case is greater than the integral of dq/T at the interface.

Chet

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Not sure what are you trying to tell. How do you get that the change in the entropy of one of the slab is mCln(Tf/t0) if it is not by integration of dQ/T?

- #24

Chestermiller

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It is equal to the integral of dQ/T

Not sure what are you trying to tell. How do you get that the change in the entropy of one of the slab is mCln(Tf/t0) if it is not by integration of dQ/T?

If you want to get a better understanding of all this, see my post #5.

Chet

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Ah, ok, I know understand what you were saying.

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The final equilibrium temperature of the two slabs = ##\frac{(T_H+T_C)}{2}##

The amount of heat transferred from the hot slab to the cold slab $$Q = mC\left(T_H-\frac{(T_H+T_C)}{2}\right)=mC\frac{(T_H-T_C)}{2}$$

From analysis of the transient heat transfer problem, the temperature at the interface between the two slabs at all times after they are brought into contact with one another = ##\frac{(T_H+T_C)}{2}##

$$ΔS_H=mC\ln \left(\frac{T_H+T_C}{2T_H}\right)$$

$$ΔS_C=mC\ln \left(\frac{T_H+T_C}{2T_C}\right)$$

Now let's see how the Clausius inequality applies to the two slabs individually, and then to the two slabs as a combined system:

Hot Slab:

The integral of dq/T_{I}for the hot slab (with T_{I}is the temperature at the interface where the heat is being transferred) is just the total heat flow divided by the (constant) interface temperature:

$$\int {\frac{dq}{T_I}}=\frac{-Q}{T_I}=-mC\frac{T_H-T_C}{T_H+T_C}$$

This can be compared with the change in entropy of the hot slab which is calculated above by:

$$ΔS_H=-mC\ln \left(\frac{2T_H}{T_H+T_C}\right)$$

From these relationships, one can readily verify that for all T_{C}<T_{H},

$$ΔS_H>\int {\frac{dq}{T_I}}$$

Cold Slab:

The integral of dq/T_{I}for the cold slab (with T_{I}is the temperature at the interface where the heat is being transferred) is just the total heat flow divided by the (constant) interface temperature:

$$\int {\frac{dq}{T_I}}=\frac{Q}{T_I}=+mC\frac{T_H-T_C}{T_H+T_C}$$

This can be compared with the change in entropy of the cold slab which is calculated above by:

$$ΔS_C=mC\ln \left(\frac{T_H+T_C}{2T_C}\right)$$

From these relationships, one can readily verify that for all T_{C}<T_{H},

$$ΔS_C>\int {\frac{dq}{T_I}}$$

Combination of the Two Slabs:

The integral of dq/T_{I}for the combination of the two slabs is zero, since the combination is insulated from the surroundings. Thus,

$$\int {\frac{dq}{T_I}}=0$$

From the previous results, the change in entropy for the combination of the two slabs is the sum of the entropy changes for each of the slabs individually:

$$ΔS=ΔS_H+ΔS_C=mC\ln \left(\frac{T_H+T_C}{2T_H}\right)+mC\ln \left(\frac{T_H+T_C}{2T_C}\right)$$

or

$$ΔS=mC\ln \left(\frac{(T_H+T_C)^2}{4T_HT_C}\right)$$

One can readily verify that, for all values of T_{C}<T_{H}, this relationship gives a positive value for ΔS. Therefore,

$$ΔS>\int {\frac{dq}{T_I}}$$

We have shown that, for the irreversible process of transferring heat from a hot slab to a cold slab, the Cauchy inequality is satisfied irrespective of whether we consider the slabs individually, or whether we consider them in combination, and thus that the change in entropy in each case is greater than the integral of dq/T at the interface.

Chet

Thanks Mr. Chet.Your explanation is really good.

can you please also prove that the cyclic integral of δQ / T is negative with the same example...

- #27

Chestermiller

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I don't understand what you are asking. Can you please be more specific about the situation you want to analyze?Thanks Mr. Chet.Your explanation is really good.

can you please also prove that the cyclic integral of δQ / T is negative with the same example...

Chet

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I don't understand what you are asking. Can you please be more specific about the situation you want to analyze?

Chet

I want a proof for Claussius Inequality theorem which is closed integral of δQ / T is less than or equal to zero.

- #29

Chestermiller

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You would like a demonstration, based on the present problem, that Clausius inequality for a cyclic process is satisfied. Is that what you are asking?I want a proof for Claussius Inequality theorem which is closed integral of δQ / T is less than or equal to zero.

Chet

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I don't understand what you are asking. Can you please be more specific about the situation you want to analyze?

Chet[/QUO

You would like a demonstration, based on the present problem, that Clausius inequality for a cyclic process is satisfied. Is that what you are asking?

Chet

Yes.. exactly...

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