Hypothesis about temperature (split from Physical mechanism how matter absorbs radiation.)

[martillo]

2 hours ago, sethoflagos said:

Enthalpy includes work previously performed on the environment by the system. Since that energy is no longer contained 'in the system' how can it be pertinent to current system temperature?

Good question. Something I also thought about at some time. What I know is that's the way Enthaply is defined anywhere.
P = F/A so F = PA and W = F.distance then PV = W makes sense but at in a perfect equilibrium both P and V are constant so no work done isn't it?
I think the right anser could come considering the following: Thermal equilibrium is a dynamical equilibrium. In the case of the term PV you can think in the environment continuously doing an infinitesimal work on the system and the system continuously reacting with a correspondent infinitesimal work. You can think the environment winning in some infinitesimal areas of the surface while the system winning in the other infinitesimal ones. I didn't arrive to the right conclusion to finalyze the answer giving the not zero term PV in the equation. Something yet needs to be reviewed and solved in that subject.

2 hours ago, sethoflagos said:

If you ignore all degrees of freedom bar translational, it is a fair approximation for the internal energy of the noble gases. It is simply the wrong expression for enthalpy.

2 hours ago, sethoflagos said:

Consideration of degrees of freedom explains almost all the differences in expressions for molar heat capacity.  

2 hours ago, sethoflagos said:

A couple of points here. For solids, N refers not to number of molecules, but number of atoms. 

Secondly, solids support shearing vibrations that gases do not, and therefore have the corresponding degrees of freedom to accommodate this.

Of both counts, you are not comparing like with like and therefore your logic has no foundation. 

Well, as I already said I could agree on "atomic and molecular vibrations" present although we could differ in what we really mean by vibrations, how energy is stored in them and how they are provoked in atoms and molecules.

2 hours ago, sethoflagos said:

Isothermally? Like your last equation. your understanding of thermodynamics needs work. 

No, radiation involves a gradient of temperature so the phenomenon is not isothermal.

 

35 minutes ago, swansont said:

No, that’s the radiation transfer.

No, radiation is not transferred, heat (Q) is transferred. Radiation is dQ/dA.

35 minutes ago, swansont said:

But that transfer rate drops dramatically once you stop touching the coffee, which is not what would happen with radiation.

You are right that when we touch the cup conductivity takes place but if not touch no conductivity ad the cup remains radiating heat which can be felt with the face as I mentioned.

35 minutes ago, swansont said:

Not that much difference. The radiation is going up from the surface.

As sethoflagos said, that’s convection. The fluid in this case is the air.

There is air flowing upside and this case involves convection but I wasn't thinking in my face up the coffee cup. I was thinking in looking at the cup horizontally. We are thinking in different experiments then. Horizontally at some proper distance I can feel the radiation of the coffee cup.

37 minutes ago, swansont said:

You can’t use your idea here, that’s circular reasoning. You need empirical evidence to support it.

What idea are you referring to? Please explain.

Edited April 5 by martillo

[swansont]

1 hour ago, martillo said:

You are right that when we touch the cup conductivity takes place but if not touch no conductivity ad the cup remains radiating heat which can be felt with the face as I mentioned.

But you’ve asserted that there is no conduction, that it’s all radiation. Being inconsistent just makes this worse.

1 hour ago, martillo said:

What idea are you referring to? Please explain

YOUR IDEA THAT ALL THERMAL ENERGY AND HEAT TRANSFER IS RADIATION. 

1 hour ago, martillo said:

Horizontally at some proper distance I can feel the radiation of the coffee cup.

No, you can’t. Not if the cup is cool on the outside.

[martillo]

You know, you are wasting our times with silly questions what shows you really don't take enough attention in what we are discussing. The discussion is becoming useless now this way.

1 hour ago, swansont said:

But you’ve asserted that there is no conduction, that it’s all radiation. Being inconsistent just makes this worse.

I never said there is no conduction. I agree in heat transferring by conduction, convection and radiation.

1 hour ago, swansont said:

YOUR IDEA THAT ALL THERMAL ENERGY AND HEAT TRANSFER IS RADIATION. 

I think your idea about what I think is wrong in something. I maintain total agreement with the mathematical definitions of those concepts.

1 hour ago, swansont said:

No, you can’t. Not if the cup is cool on the outside.

If the cup is filled with hot coffee it doesn't stay cool, it warms with T superior to the environment T. Depending on the cup considered it warms more or less but it warms and if the cup is not a so good insulator its radiation can be sensed by finger or face at some proper distance.

Edited April 5 by martillo

[sethoflagos]

1 hour ago, martillo said:

Good question. Something I also thought about at some time. What I know is that's the way Enthaply is defined anywhere.

But what you don't know is what enthalpy is and where its use is appropriate. It isn't here. Temperature is defined as the inverse partial derivative of entropy with internal energy, not enthalpy.

2 hours ago, martillo said:

P = F/A so F = PA and W = F.distance then PV = W makes sense but at in a perfect equilibrium both P and V are constant so no work done isn't it?
I think the right anser could come considering the following: Thermal equilibrium is a dynamical equilibrium. In the case of the term PV you can think in the environment continuously doing an infinitesimal work on the system and the system continuously reacting with a correspondent infinitesimal work. You can think the environment winning in some infinitesimal areas of the surface while the system winning in the other infinitesimal ones. I didn't arrive to the right conclusion to finalyze the answer giving the not zero term PV in the equation. Something yet needs to be reviewed and solved in that subject.

This is just off-topic waffle. 

2 hours ago, martillo said:

Well, as I already said I could agree on "atomic and molecular vibrations" present although we could differ in what we really mean by vibrations, how energy is stored in them and how they are provoked in atoms and molecules.

Familiarise yourself with Debye's theorem. And why that superseded Einstein's photoelectron model. That should help clarify.

[swansont]

On 3/23/2024 at 8:56 AM, martillo said:

The "heat" transferring between two systems is always through the interchange of photons which can pass through the walls separating them. Photons can pass through atoms and molecules (with scattering dispersion sometimes) carrying energy.

This is your claim - heat transfer is from radiation

On 3/15/2024 at 9:58 AM, martillo said:

the temperature T (measured by a thermometer) is more related to the energy of the internal EM radiation present in the interior

This is also your claim

[martillo]

12 minutes ago, swansont said:

These is your claim - heat transfer is from radiation

I was wrong in that statement and I admitted it well correcting the thing. I after said that conductivity is actually accomplished in solid metal conductors by the absorption with posterior emission of photons in a process that has a time delay involved and so on and we have a long discussion about. I'm wondering why you aren't you considering all that now. By the way, I know I make mistakes sometimes but always try to correct them.

Edited April 5 by martillo

[swansont]

25 minutes ago, martillo said:

I was wrong in that statement and I admitted it well correcting the thing. I after said that conductivity is actually accomplished in solid metal conductors by the absorption with posterior emission of photons in a process that has a time delay involved and so on and we have a long discussion about. I'm wondering why you aren't you considering all that now. By the way, I know I make mistakes sometimes but always try to correct them.

That’s still radiation, and I’m talking about conduction transfer from one material to another, not through a material. When two materials touch the heat transfer rate is much larger than if they are separated by a small amount (especially if that is a vacuum gap) This is inconsistent with the heat transfer being radiative.

You burn yourself when you touch the hot water. What does the S-B law say is the heat flow from touching your 1 cm^2 fingertip to the coffee?

 

[martillo]

1 hour ago, sethoflagos said:

But what you don't know is what enthalpy is and where its use is appropriate. It isn't here. Temperature is defined as the inverse partial derivative of entropy with internal energy, not enthalpy.

I'm applying enthalpy because I think we are considering systems on which we can apply it. Do you think I cannot apply it?
I have studied the temperature definition for a while. It begun to be defined by Kelvin years before the concepts of enthalpy and entropy. It was defined with the concepts of heat and the Carnot cycle. The definition in terms of entropy came long after and as it is related to enthalpy by dH = TdS, then H and T can be related. I assume both definitions are totally compatible, am I wrong in that?

1 hour ago, sethoflagos said:

Familiarise yourself with Debye's theorem. And why that superseded Einstein's photoelectron model. That should help clarify.

I took a look on Debye's approach. At Wikipedia: 
"In thermodynamics and solid-state physics, the Debye model is a method developed by Peter Debye in 1912 to estimate phonon contribution to the specific heat (heat capacity) in a solid.^[1] It treats the vibrations of the atomic lattice (heat) as phonons in a box"
So Debye's approach is to explain the atomic vibrations in atoms/molecules as applied in heat transmission. I have a concept of those vibrations without the need to involve any virtual particle but who knows I would need to replace the behavior described by phonons someway. I will go a little further to see if I would need to do that or not.

 

55 minutes ago, swansont said:

That’s still radiation, and I’m talking about conduction transfer from one material to another, not through a material. When two materials touch the heat transfer rate is much larger than if they are separated by a small amount (especially if that is a vacuum gap) This is inconsistent with the heat transfer being radiative.

I'm considering conduction as defined at Wikipedia ():
"Conduction is the process by which heat is transferred from the hotter end to the colder end of an object. The ability of the object to conduct heat is known as its thermal conductivity, and is denoted k."

This doesn't match with your concept of heat transfer "not through a material" as you said.

55 minutes ago, swansont said:

You burn yourself when you touch the hot water. What does the S-B law say is the heat flow from touching your 1 cm^2 fingertip to the coffee?

When I touch hot water conduction takes place. S-B law does not apply in conduction, it only applies in radiation, I know.

Edited April 5 by martillo

[sethoflagos]

2 hours ago, martillo said:

I was wrong in that statement and I admitted it well correcting the thing. I after said that conductivity is actually accomplished in solid metal conductors by the absorption with posterior emission of photons in a process that has a time delay involved and so on and we have a long discussion about. I'm wondering why you aren't you considering all that now. By the way, I know I make mistakes sometimes but always try to correct them.

Well there's a big mistake here.

The vast majority of absorbed photons originate from emitters at a higher temperature because of S-B's T⁴.

Therefore the vast majority of absorbed photons lead to an entropy increase in the emitter/absorber system because dU = TdS.

Therefore the reemission of photons you propose would cause an overall system entropy decrease in contravention of the 2nd Law. No amount of arm waving will rescue your hypothesis from this.

[martillo]

24 minutes ago, sethoflagos said:

Well there's a big mistake here.

The vast majority of absorbed photons originate from emitters at a higher temperature because of S-B's T⁴.

Therefore the vast majority of absorbed photons lead to an entropy increase in the emitter/absorber system because dU = TdS.

Therefore the reemission of photons you propose would cause an overall system entropy decrease in contravention of the 2nd Law. No amount of arm waving will rescue your hypothesis from this.

Heat conduction increases temperature. That is well known in electronics. Entrophy increases with temperature and so the entrophy in the conductor increases. Fine for me. What I don't get is that this "would cause an overall system entropy decrease". What "overall" system are you considering? 

Edited April 6 by martillo

[swansont]

3 hours ago, martillo said:

When I touch hot water conduction takes place. S-B law does not apply in conduction, it only applies in radiation, I know

Progress

[martillo]

6 hours ago, sethoflagos said:

But what you don't know is what enthalpy is and where its use is appropriate. It isn't here. Temperature is defined as the inverse partial derivative of entropy with internal energy, not enthalpy.

You are wrong in this. T is not defined as T = dU/dS. T is defined as T = dH/dS.
So you are also wrong in this:

3 hours ago, sethoflagos said:

Well there's a big mistake here.

The vast majority of absorbed photons originate from emitters at a higher temperature because of S-B's T⁴.

Therefore the vast majority of absorbed photons lead to an entropy increase in the emitter/absorber system because dU = TdS.

Therefore the reemission of photons you propose would cause an overall system entropy decrease in contravention of the 2nd Law. No amount of arm waving will rescue your hypothesis from this.

I'm not violating 2nd law at all.
You are applying a wrong definition.

Edited April 6 by martillo

[martillo]

12 hours ago, swansont said:

Progress

Thanks. +1
The model is progressing too. Applying some adjustments and developing it further now.
I need time for some calculations (quantification) now.
I consider all the comments and criticisms in the thread as contributing to the model so this is a collaborative model between all the participants.
If it works it could be called the SFNM model. Science Forums .Net Model.
We are going to make it work, I'm sure.

 

Edited April 6 by martillo

[Bufofrog]

1 hour ago, martillo said:

The model is progressing too. Applying some adjustments and developing it further now.
I need time for some calculations (quantification) now.
I consider all the comments and criticisms in the thread as contributing to the model so this is a collaborative model between all the participants.
If it works it could be called the SFNM model. Science Forums .Net Model.
We are going to make it work, I'm sure.

I guess it's fun to fantasize that you have come up with some new physics.  

[martillo]

26 minutes ago, Bufofrog said:

I guess it's fun to fantasize that you have come up with some new physics.  

It's your guess but you don't know what I have already done in about 30 years working in Physics' subjects.

[sethoflagos]

14 hours ago, martillo said:

You are wrong in this. T is not defined as T = dU/dS. T is defined as T = dH/dS.
So you are also wrong in this:

Actually both are correct providing dU/dS is evaluated at constant volume and dH/dS is evaluted at constant chemical potential. Look at the wikipedia pages on thermodynamic potential and Maxwell's relations.

However, since 2019 the international community has agreed to settle on the former as the official definition of thermodynamic temperature:

Quote

In particular, when the body is described by stating its internal energy U, an extensive variable, as a function of its entropy S, also an extensive variable, and other state variables V, N, with U = U (S, V, N), then the temperature is equal to the partial derivative of the internal energy with respect to the entropy:^[29][30][31]

(2)

Likewise, when the body is described by stating its entropy S as a function of its internal energy U, and other state variables V, N, with S = S (U, V, N), then the reciprocal of the temperature is equal to the partial derivative of the entropy with respect to the internal energy:^[29][31][32]

(3)

The above definition, equation (1), of the absolute temperature, is due to Kelvin. It refers to systems closed to the transfer of matter and has a special emphasis on directly experimental procedures.

One thing this equation tells us is that if an amount of energy dU is transferred from a warmer body to a cooler one, the TdS values for each must be equal in magnitude (though opposite in sign). This can only be so if the lower value of T for the cooler body is balanced by a proportionally higher value of dS. Therefore the total entropy change for the two bodies is positive, 

14 hours ago, martillo said:

I'm not violating 2nd law at all.
You are applying a wrong definition.

Howl at the moon as long as you like, your absorbed photons have been dissipated (as phonons in context) and cannot be recovered intact.

[martillo]

29 minutes ago, sethoflagos said:

Howl at the moon as long as you like, your absorbed photons have been dissipated (as phonons in context) and cannot be recovered intact.

The photons could be recovered by external heat supply only, I know.

[martillo]

Here is proposed a model for the conduction process in a solid conductor transmitting photons in one degree of freedom only which is the direction of the flux of photons.
The derivation of the model has been summarized in a very concise way this way:

The assumptions made are:
_ There is a constant flux of photons along the conductor.
_ The energy density of the flux is derived from Planck law having a non linear variation in T.

The model determines an energy density Ec/V in the conduction given by the relation: 

Ec/V = µ(f,T) = [constant]xT³ 

where [constant] = 2.π.K³/(h².c³.(e-1) = 1,01660413x10² in SI units: K^-1.kg^-1.s^-5 (precision of 8 digits) based on the constants' table below

The variation is proportional to T³ and so can compete with Debye model.

Derivation made:

Agreeing with Planck distribution of the energies of the photons in the conductor:

µ(f,T) = (8.π.h.f³/c³).(exp(hf/KT) - 1)^-1

          = (8.π.h.f³/c³).(exp(1) - 1)^-1 

Considering hF = KT and doing some algebra:

µ(f,T) = [8.π.K³/(h².c³.(e - 1))].T³ 

Finally:

µ(f,T) = [constant].T³ 

I apologize for haven't used Latex notation. I have found this a convenient notation anyway.

I ask to everybody in the thread to take the appropriated time in answering your comments and criticisms. 

I need some time to have some rest now.

 

PS:
I must comment the used relation hf =KT (where f is the mean statistical average value of the frequencies of the photons in the considered flux in the conductor) as described at Wikipedia Planck's law on the section "Trying to find a physical explanation of the law":

Referring to a new universal constant of nature, h,^[108] Planck supposed that, in the several oscillators of each of the finitely many characteristic frequencies, the total energy was distributed to each in an integer multiple of a definite physical unit of energy, ϵ, characteristic of the respective characteristic frequency.^{[95][109][110][111]} His new universal constant of nature, h, is now known as the Planck constant.

Planck explained further^[95] that the respective definite unit, ϵ, of energy should be proportional to the respective characteristic oscillation frequency ν of the hypothetical oscillator, and in 1901 he expressed this with the constant of proportionality h:^[112][113]

 

Planck did not propose that light propagating in free space is quantized.^{[114][115][116]} The idea of quantization of the free electromagnetic field was developed later, and eventually incorporated into what we now know as quantum field theory.^[117]

Edited April 6 by martillo

[swansont]

Does your model account for the heat capacity of the material?

What happens to the photons?

[martillo]

1 hour ago, swansont said:

What happens to the photons?

The photons incoming from the hotter body at one side of conductor leave it to the colder body at the other side.

1 hour ago, swansont said:

Does your model account for the heat capacity of the material?

In the model H = Q = VT³ 

Heat capacity at constant volume V is defined as:
C = dQ/dT = dH/dT at constant volume V for for dT = 1º K

C = 3V(dT)² 

Then the capacity per unity of volume

C_V = C/V = 3(dT)² for dT= 1º K

C_V = 3 the same as in the Debye model

 

[swansont]

24 minutes ago, martillo said:

The photons incoming from the hotter body at one side of conductor leave it to the colder body at the other side.

So they are only in the material for a short time. How do you get new photons to take their place?

24 minutes ago, martillo said:

In the model H = Q = VT³ 

Heat capacity at constant volume V is defined as:
C = dQ/dT = dH/dT at constant volume V for for dT = 1º K

C = 3V(dT)² 

Then the capacity per unity of volume

C_V = C/V = 3(dT)² for dT= 1º K

C_V = 3 the same as in the Debye model

 

Your units don’t work. VT^3 doesn’t have the units of Q

[martillo]

43 minutes ago, swansont said:

So they are only in the material for a short time. How do you get new photons to take their place?

The hotter body continuously provides photons.

 

43 minutes ago, swansont said:

Your units don’t work. VT^3 doesn’t have the units of Q

Yes, something is wrong.

Actually in the model:

H = Q = µ(f,T) = [constant].T³ 

where [constant] = 2.π.K³/(h².c³.(e-1) = 1,01660413x10² in SI units: K^-1.kg^-1.s^-5 (precision of 8 digits)

So [constant] is well approximated by 300 

Actually is H = Q = V.[constant].T³ 

Heat capacity at constant volume V is defined as:
C = dQ/dT = dH/dT at constant volume V for for dT = 1º K

C is approximately = 300V(dT)² 

And so strictly:

C_V is approximately = 300 much greater than in the Debye model.

Can this be right?

Something I don't understand is why this capacity is independent of the kind of the material in the conductor but in the Debye model is the same thing.

Edited April 7 by martillo

[martillo]

There is still another error: the units given for the [constant].

The right thing is:

[constant] = 2.π.K³/(h².c³.(e-1) = 1,01660413x10² in SI units: K^-1.kg^-1.s (precision of 8 digits)

My apologies...

I want to mention other thing. The model I was working on the enthalpy of the systems stucked now. The discussion we had about have let me know that the model to propose unavoidably needs two main assumptions that I cannot prove and will not be accepted by you for the model to be taken into consideration:

_ Instantaneous Electric and Magnetic fields and forces.

_ Strong disagreement with Relativity Theory.

I would need to provide some experimental evidence on them, something I don't have and so it is a proposition unfortunately not suitable to discuss here in the forum. Who knows if in some future I could, I don't know.
I have just realized that this model I provided for the conduction process in solid conductors transmitting photons involves some calculation based on the enthalpy so it will also suffer the same limitations.

Edited April 7 by martillo

[Mordred]

46 minutes ago, martillo said:

 

_ Instantaneous Electric and Magnetic fields and forces.

_ Strong disagreement with Relativity Theory.

I would need to provide some experimental evidence on them, something I don't have and so it is a proposition unfortunately not suitable to discuss here in the forum. Who knows if in some future I could, I don't know.

Not sure how you define instantaneous as rates of change in the EM field have been measured to extremely small units of measure.

Also not sure on what your referring to on strong disagreement with relativity with regards to heat.

However you may may not experimental evidence however there may be possible research and experiments already done you can draw upon via arxiv. It's common practice to draw upon other lines of research and experiments done by others as supportive evidence provided those lines of research are applicable.

Edited April 7 by Mordred

[martillo]

40 minutes ago, Mordred said:

Not sure how you define instantaneous as rates of change in the EM field have been measured to extremely small units of measure.

Also not sure on what your referring to on strong disagreement with relativity with regards to heat.

However you may may not experimental evidence however there may be possible research and experiments already done you can draw upon via arxiv. It's common practice to draw upon other lines of research and experiments done by others as supportive evidence provided those lines of research are applicable.

Thanks very much for the comment. +1.  I will try arxiv, something I have never done. May be I could have some luck...

40 minutes ago, Mordred said:

Not sure how you define instantaneous as rates of change in the EM field have been measured to extremely small units of measure.

I mean instantaneous action at a distance of forces. May be this is actually experimentally inobservable at subatomic scales, I'm not sure.

40 minutes ago, Mordred said:

Also not sure on what your referring to on strong disagreement with relativity with regards to heat.

The model unavoidably ends assigning mass to photons for them to have separated internal energy and kinetic energy. I have thought, in the same kind of reasoning for considering virtual particles, to assign an hypothetical mass but this will be easily disproved by some relativistic experiments I think. I think those experiments can actually be interpreted in a different way providing a new way to explain them but is something not so easy. Too much things to be explained at the same time. A huge task...

Edited April 7 by martillo