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

[martillo]

1 hour ago, swansont said:

.03/3 x 10^8 ≠ .03 x 10^-8

.03 x 10^-8 sec is the interval of time t.
.02 m is the thickness d.
v = d/t = .02/(.03 x 10^-8) = (2/3)c in accordance with the refractive index n = c/v = 1.5.

1 hour ago, swansont said:

The frequencies are the same. The wavelength are the same.

Not at all. What is the same is the velocity c. The frequency varies as the wavelength varies: f = c/λ.

1 hour ago, swansont said:

Yes. Do you understand this is what we have been telling you and you keep denying?  The radiative heat transfer depends on T^4

More attention please, if not is useless to discuss. In the reference it is said: "the heat transfer rate varies as the difference in the 4th powers of the absolute temperatures".
I'm tired to say to you that heat transfer of energy rate is different than heat transfer of energy even with different units. I always stated:
transfer of energy rate is proportional to T⁴.
transfer of energy  is proportional to T.
So you are wrong to state "radiative heat transfer depends on T^4".
Radiactive, conductive or whatever heat transfer depends on T (according to the Thermodynamics laws).
And: whatever heat transfer rate is proportional to T⁴ (according to Stefan-Boltzmann law).

 

Edited April 3 by martillo

[swansont]

1 hour ago, martillo said:

.02/(.03 x 10^-8) = (2/3)c 

No, that’s 2/3 x 10^8 m/s

(2/3) c is 2 x 10^8 m/s, since c = 3 x 10^8 m/s

 

1 hour ago, martillo said:

Not at all. What is the same is the velocity c. The frequency varies as the wavelength varies: f = c/λ.

c is a constant. Light that has the same frequency must have the same wavelength 

 

1 hour ago, martillo said:

More attention please, if not is useless to discuss. In the reference it is said: "the heat transfer rate varies as the difference in the 4th powers of the absolute temperatures".
I'm tired to say to you that heat transfer of energy rate is different than heat transfer of energy even with different units. I always stated:
transfer of energy rate is proportional to T⁴.
transfer of energy  is proportional to T.
So you are wrong to state "radiative heat transfer depends on T^4".
Radiactive, conductive or whatever heat transfer depends on T (according to the Thermodynamics laws).
And: whatever heat transfer rate is proportional to T⁴ (according to Stefan-Boltzmann law).

 

Take a heat transfer rate and multiply it by a time interval, and you get the heat transferred. If the rate depends on T^4, the energy transferred will, too.

I have come to the conclusion that you don’t have the math skills to analyze any of this. All of your errors here involve simple algebra. If you can’t get this right, there’s no point to the discussion.

[martillo]

1 hour ago, swansont said:

No, that’s 2/3 x 10^8 m/s

(2/3) c is 2 x 10^8 m/s, since c = 3 x 10^8 m/s

I apologize, you are right on this.

1 hour ago, swansont said:

I have come to the conclusion that you don’t have the math skills to analyze any of this. All of your errors here involve simple algebra. If you can’t get this right, there’s no point to the discussion.

The point is that the problem, as you present it, involves two variables, not only one. Please give me some time to rethink a right answer for you to all this mess. I will focus in the way you calculate it.

Edited April 3 by martillo

[swansont]

50 minutes ago, martillo said:

The point is that the problem, as you present it, involves two variables, not only one. Please give me some time to rethink a right answer for you to all this mess. I will focus in the way you calculate it.

You can look at frequency or wavelength but they are not independent, so it doesn’t matter which one. 

In your estimation, how long does it take for heat to pass through glass?Is it appropriately measured in nanoseconds, milliseconds, seconds?

[martillo]

Let we see if the following is the right answer to the mess.

We are looking at the problem with two views:
1) The relation of energy transfer with the temperature T.
2) How heat energy is transferred through some transparent but good insulator glass in terms of the velocity of propagation, the wavelength and the frequency of the incident heat.

Case 1:
The radiance is P/A where power P = ΔQ/Δt. ΔQ is the total amount of energy transferred between two bodies with different temperatures T₁ and T₂ (T₁ - T₂ =  ΔT). A is some area of a considered surface for the transfer. P is the rate of transfer. Then:
ΔQ = ∫₁²(PA)dT = A∫₁²PdT
Now the problem is that in the Newton cooling formula is assumed a small difference in the temperatures for which P can be considered constant while in the general case it varies non linearly:
If P = constant it is ΔQ = PAΔT
If P not constant P = P(T) which is given by the the Stefan-Boltzmann law and is proportional to T⁴. Nevertheless the Thermodynamics laws can give the amount of ΔQ transfer for this case which result being linearly proportional to T. For a small difference in the temperatures this case falls into the Newton's case above.

Case 2:
The transmission of radiant energy by a glass depend on its frequency :
a) Glass can absorb and emit in the "light" spectrum quite transparently (velocity of propagation inside the glass has a little difference with that in air (refractive index approximately 1).
b) Glass quite cannot absorb in the "heat" spectrum and the small quantity it can transmit propagates at a slow velocity (refractive index is high much greater than 1).
Note: I looked for technical data of this variation in the web but couldn't find a credited source.

 

Edited April 3 by martillo

[martillo]

I have found what I needed,an interesting site (https://www.ist.fraunhofer.de/en/newsletter/answers/can-glass-reflect-heat.html) talking about glass transmission of EM radiation and present a very illustrative typical graphics of the transmission capability of the radiation by glasses varying with the wavelength of the incident radiation in the blue line in a logaritmical scale. It well shows how there is a sharp lose of that capability on the range between 1000 nm (1 microns) and  the 10000 nm ( 10 microns) which is precisely the range @swansont presented in his link. It shows how glass stopped transmitting sharply at about the 3000 nm (3 microns) and so how glasses is transparent to "light" while an becomes an insulator  for "heat". It also talks that the mechanism involved in the transmission is about absorption, emission and reflection of EM radiation as I sustained. So my approach about material particles (atoms/molecules) storing EM radiation in the form of photons absorption keeps up.

 

Edited April 5 by martillo

[swansont]

47 minutes ago, martillo said:

It shows how glass stopped transmitting sharply at about the 3000 nm (3 microns)

That particular glass. The ones I cited transmit out past 10 microns.

http://rmico.com/znse-specifications

 

[Mordred]

13 minutes ago, swansont said:

That particular glass. The ones I cited transmit out past 10 microns.

http://rmico.com/znse-specifications

 

Thanks for the link I had lost track of that site.

[martillo]

1 hour ago, swansont said:

That particular glass. The ones I cited transmit out past 10 microns.

http://rmico.com/znse-specifications

 

I looked in the site that publish that graphics and found the page where it is shown the Transmission curves for several glasses compounds: https://rmico.com/support/tech-notes/material-data. It can be seen the glasses sharply losing transmission capability in a range from 5 to 20 microns. That means that some glasses well transmit "light" while transmitting "heat" with different capabilities. May be there are glasses that well transmit EM radiation in the "heat" spectrum but would be not the case of the common glasses we use at our home windows on which we easily observe they transmitting "light" but not "heat". 

I'm losing the real subject of all of our discussion. Can you please summarize it?

Edited April 5 by martillo

[martillo]

Trying to summarize the real point of our discussion:
The Kinetic Theory explains Temperature as produced by the motion of material particles only. I said that there are photons involved which would be being ignored by the Kinetic Theory.
I have commented I'm working on a model where the energy of those photons strongly contribute to the temperature of the system and you are trying to debunk it. You know, I'm realizing now that the concept of molecular vibrations is totally compatible with the presence of those photons. The atoms just would store the energy of absorbed photons in the form of vibrations in the atomic binding of protons with their electrons. That is also compatible with the discrete levels of the possible energies in the atom considering vibrations in a possible discrete set off possible modes of vibration. 
My model can be compatible with that. What I would disagree is if the photons are ignored and only translations in the center of mass of the particles are considered.

Edited April 5 by martillo

[Mordred]

46 minutes ago, martillo said:

you are trying to debunk it. 

If you understood how robust and accurate models are developed, then you would give thanks to anyone that attempts to debunk your theory under development. That's precisely how one develops a good strong working theory. A good physicist tries to debunk his own theories as well. For any evidence counter to his theory or mathematics etc. That same theorist would seek ways to explain or improve his theory to cover a given experiment that ran counter to it previously.

 I come up with theories all the time. I typically debunk my own and spend far more effort to debunk it than build it more often than not.

Edited April 5 by Mordred

[martillo]

5 hours ago, Mordred said:

If you understood how robust and accurate models are developed, then you would give thanks to anyone that attempts to debunk your theory under development. That's precisely how one develops a good strong working theory. A good physicist tries to debunk his own theories as well. For any evidence counter to his theory or mathematics etc. That same theorist would seek ways to explain or improve his theory to cover a given experiment that ran counter to it previously.

 I come up with theories all the time. I typically debunk my own and spend far more effort to debunk it than build it more often than not.

Yes you are right and I appreciate the time and effort you dedicate discussing with me.

[swansont]

7 hours ago, martillo said:

I'm losing the real subject of all of our discussion. Can you please summarize it?

If your hypothesis is true it must be true for all solids. I only have to find one example where it fails.

A material that transmits radiation in the thermal range does so almost instantaneously. (for 2 cm, this was about 0.1 nanoseconds) It must do so regardless of whether the radiation comes from a thermal or non-thermal source, because photons are photons.

But you agree that heat is transmitted much more slowly through such materials. I contend that this falsifies your model. Zinc Selenide does not behave the way you predict it should.

I await your next tap-dance

6 hours ago, martillo said:

My model can be compatible with that

You don’t quantify things, so you don’t have a model. If you did the math (or comprehend that math I’ve done for you) you would see that your conjecture is not compatible. 

[martillo]

20 minutes ago, swansont said:

If your hypothesis is true it must be true for all solids. I only have to find one example where it fails.

A material that transmits radiation in the thermal range does so almost instantaneously. (for 2 cm, this was about 0.1 nanoseconds) It must do so regardless of whether the radiation comes from a thermal or non-thermal source, because photons are photons.

But you agree that heat is transmitted much more slowly through such materials. I contend that this falsifies your model. Zinc Selenide does not behave the way you predict it should.

I await your next tap-dance

Ok, just give me a little time now.

20 minutes ago, swansont said:

You don’t quantify things, so you don’t have a model. If you did the math (or comprehend that math I’ve done for you) you would see that your conjecture is not compatible. 

I did some quantification. I talked about the enthalpy equation H = U + PV balancing the energies in the system. As evidence for the presence of photons in the system I mention that the Kinetic Theory predicts the total energy for an ideal gas to be H = (3/2)NKT (K is Boltzmann constant) while for solids the total energy is 3NKT. I think the Kinetic Theory is right here including the vibrations energy in the system among the kinetic energy of the particles (due to movement of their center of mass). Is missing then to consider the other half of the total energy in the case of ideal gases. The total energy must be the addition of the kinetic energies of the particles plus the energy of all the photons present. My theory predicts that for the ideal gases exactly the half equal to (3/2)NKT is missing and is related to the photons present in the system. In my model the total energy of the system is H = 3NKT and is equal to the total heat that was absorbed by the system to reach the considered thermal equilibrium, H = Q.

Edited April 5 by martillo

[martillo]

1 hour ago, swansont said:

If your hypothesis is true it must be true for all solids. I only have to find one example where it fails.

A material that transmits radiation in the thermal range does so almost instantaneously. (for 2 cm, this was about 0.1 nanoseconds) It must do so regardless of whether the radiation comes from a thermal or non-thermal source, because photons are photons.

But you agree that heat is transmitted much more slowly through such materials. I contend that this falsifies your model. Zinc Selenide does not behave the way you predict it should.

I await your next tap-dance

There are two phenomena related to heat transmission.
One is the velocity of the photons in a material dense medium. This is described by the refractive index of the material which relates the average velocity of the photons in the material in relation to the vacuum velocity c. At the Wikipedia page of refractive index (https://en.wikipedia.org/wiki/Refractive_index) you can also see how the refractive index varies with the wavelength in the "Typical values" and the "Dispersion" sections of the page.
The other phenomenon is about the time needed to heat a piece of a material, is about hotness and our sensorial perception of hotness. Let us consider the case of a cup of coffee. We fill a cold cup with hot coffee. It takes some perceptible time for the cup to be heated as we can sense it with our fingers. This phenomenon is related to the capacity of the material to absorb heat, to absorb the photons. When we fill the cup it takes some time to its external surface to be heated what is what we can perceive with our fingers. The cup is being heated from its internal surface and it begins to absorb some small percentage of the radiation. While absorbing the heat radiation the heat does not reach the outer surface. The heat reaches the external surface when all the atoms of the thick wall of the cup were already heated. Only after we can sense then that the cup is hot and so we perceive the phenomenon as the heat of the coffee lates some time to propagate through the wall of the cup from the internal surface to the external surface.

I hope to have been clear in my explanation.

[swansont]

34 minutes ago, martillo said:

When we fill the cup it takes some time to its external surface to be heated what is what we can perceive with our fingers. The cup is being heated from its internal surface and it begins to absorb some small percentage of the radiation. While absorbing the heat radiation the heat does not reach the outer surface.

If it only absorbs a small percentage, the rest must be transmitted. We already know the transmission is quite high, so this radiation is not being absorbed.

But we also know the heat in not being transmitted nearly as quickly as this prediction 

[martillo]

4 minutes ago, swansont said:

If it only absorbs a small percentage, the rest must be transmitted. We already know the transmission is quite high, so this radiation is not being absorbed.

You must review this. We have a very hot radiation in one side (the internal surface off the cup) and a not so hot in the other side (the external surface of the cup) so, the difference must be or reflected or absorbed. there is no other way.

[swansont]

59 minutes ago, martillo said:

You must review this. We have a very hot radiation in one side (the internal surface off the cup) and a not so hot in the other side (the external surface of the cup) so, the difference must be or reflected or absorbed. there is no other way.

Of course there’s another way: the radiation level is much lower than you think it is. You don’t quantify anything, so you’re not comparing the numbers that would show this.

The net amount of radiation is actually quite small near room temperature. The heat transfer is via conduction, which is relatively slow, because it depends on the vibration of the atoms.

 

[martillo]

35 minutes ago, swansont said:

Of course there’s another way: the radiation level is much lower than you think it is. You don’t quantify anything, so you’re not comparing the numbers that would show this.

The net amount of radiation is actually quite small near room temperature. The heat transfer is via conduction, which is relatively slow, because it depends on the vibration of the atoms.

 

The hot coffee can "burn" my fingers while the cup does not. There is a not negligible (not so small as you consider) difference in the intensity of the radiation of both.
To quantity the case properly I would need to work with a thermometer here at home. I don't want to do that.

Edited April 5 by martillo

[sethoflagos]

4 hours ago, martillo said:

I did some quantification. I talked about the enthalpy equation H = U + PV balancing the energies in the system.

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?

4 hours ago, martillo said:

Theory predicts the total energy for an ideal gas to be H = (3/2)NKT (K is Boltzmann constant)...

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.

4 hours ago, martillo said:

... while for solids the total energy is 3NKT. I think the Kinetic Theory is right here including the vibrations energy in the system among the kinetic energy of the particles (due to movement of their center of mass). Is missing then to consider the other half of the total energy in the case of ideal gases.

 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. 

5 hours ago, martillo said:

My theory predicts that for the ideal gases exactly the half equal to (3/2)NKT is missing and is related to the photons present in the system. In my model the total energy of the system is H = 3NKT...

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

5 hours ago, martillo said:

and is equal to the total heat that was absorbed by the system to reach the considered thermal equilibrium, H = Q.

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

[swansont]

1 hour ago, martillo said:

The hot coffee can "burn" my fingers while the cup does not. There is a not negligible (not so small as you consider) difference in the intensity of the radiation of both.
To quantity the case properly I would need to work with a thermometer here at home. I don't want to do that.

Heat, yes. Radiation, no. 

You don’t burn yourself by being close to the hot coffee. You have to touch it. That’s not consistent with radiation.

[martillo]

30 minutes ago, swansont said:

Heat, yes. Radiation, no.

Radiation is heat energy transfer rate. The rate of transmission is what makes the difference for my fingers "burn" or not. You can have lot of heat energy but transmitted at small rate then it would not affect too much while if it is transferred at a high rate it will affect ("burn").

30 minutes ago, swansont said:

You don’t burn yourself by being close to the hot coffee. You have to touch it. That’s not consistent with radiation.

I can feel the coffee radiation even with my face at some distance. More the distance less it affects of course.

Edited April 5 by martillo

[sethoflagos]

54 minutes ago, martillo said:

I can feel the coffee radiation even with my face at some distance.

No you can't. That's convection currents.

[martillo]

1 minute ago, sethoflagos said:

No you can't. That's convection currents.

Not convection is present. Convection is transfer of heat by a fluid moving. The fluid absorbs heat in some place and emit it in other place. The case is of radiation, not convection.

By the way I'm thinking in my answers on the several questions in your previous post. Please give me some time.

[swansont]

1 hour ago, martillo said:

Radiation is heat energy transfer rate.

No, that’s the radiation transfer.

You can’t use your idea here, that’s circular reasoning. You need empirical evidence to support it. What we’re doing is some basic testing of the idea.

1 hour ago, martillo said:

The rate of transmission is what makes the difference for my fingers "burn" or not. You can have lot of heat energy but transmitted at small rate then it would not affect too much while if it is transferred at a high rate it will affect ("burn").

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

1 hour ago, martillo said:

I can feel the coffee radiation even with my face at some distance. More the distance less it affects of course.

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.