Heat Flow

[exchemist]

3 minutes ago, Genady said:

Yes, it is.

You're talking about enthalpy, right?

No, enthalpy would include various forms of internal potential energy associated with chemical bonding etc. One can only measure changes in that, rather than absolute values.

I am thinking here of kinetic theory. What I mean is thermal energy, i.e. the famous 3/2RT at constant volume for monoatomic gases or 5/2RT for diatomic gases, due to thermal kinetic energy of the molecules. This does have an absolute value which goes to zero at absolute zero. (Though there is still some residual zero point motion at absolute zero, this is by definition unextractable.)  

[swansont]

20 minutes ago, Willem F Esterhuyse said:

No, because there is no negative free paths for molecules.

Can you explain what you mean by this? This is something new you’ve introduced, with no foundation.

[Willem F Esterhuyse]

30 minutes ago, swansont said:

Can you explain what you mean by this? This is something new you’ve introduced, with no foundation.

The "mean free path" of molecules is not new (I may not have the term 100 percent equivalent). One can picture in mind molecules going around, bumping into each other and the walls of a container and easily see there can be no negative distance between them.

[Genady]

1 hour ago, exchemist said:

No, enthalpy would include various forms of internal potential energy associated with chemical bonding etc. One can only measure changes in that, rather than absolute values.

I am thinking here of kinetic theory. What I mean is thermal energy, i.e. the famous 3/2RT at constant volume for monoatomic gases or 5/2RT for diatomic gases, due to thermal kinetic energy of the molecules. This does have an absolute value which goes to zero at absolute zero. (Though there is still some residual zero point motion at absolute zero, this is by definition unextractable.)  

Ah, I see. Yes, it is a thermal component of internal energy. It can change without any heat flow, though (in a reference to the OP).

[exchemist]

2 hours ago, Genady said:

Ah, I see. Yes, it is a thermal component of internal energy. It can change without any heat flow, though (in a reference to the OP).

Can it in all cases, though? Rotational and vibrational excitation can change by absorption or emission of radiation, but translational motion? Doesn't that have to change by heat flow?  

[swansont]

2 hours ago, Willem F Esterhuyse said:

The "mean free path" of molecules is not new (I may not have the term 100 percent equivalent). One can picture in mind molecules going around, bumping into each other and the walls of a container and easily see there can be no negative distance between them.

And the connection to cold is? This is your proposal. You need to share the details.

You can have a high-temperature sample of gas with a large mean free path or a small one. Same for low temperature.

 

3 minutes ago, exchemist said:

Can it in all cases, though? Rotational and vibrational excitation can change by absorption or emission of radiation, but translational motion? Doesn't that have to change by heat flow?  

Can you clarify this? Radiation can be heat flow, and radiation can cause changes in translational KE. Photons have momentum.

[Genady]

6 minutes ago, exchemist said:

translational motion? Doesn't that have to change by heat flow?  

What about, e.g., an adiabatic process? No heat flow, but the temperature changes.

[exchemist]

3 hours ago, swansont said:

And the connection to cold is? This is your proposal. You need to share the details.

You can have a high-temperature sample of gas with a large mean free path or a small one. Same for low temperature.

 

Can you clarify this? Radiation can be heat flow, and radiation can cause changes in translational KE. Photons have momentum.

Oh you mean transfer of momentum by inelastic (i.e. Raman) scattering? 

[sethoflagos]

On 2/17/2023 at 12:14 PM, Genady said:

dU = δQ + δW

It does not matter what one calls δQ as long as the equation holds. You can also say, "the warmness flaws out and the coldness flows in" or "the coldness replaces the warmness", etc.

 

Remember that engineers and physicists employ a double negative on the work term by a change of both sign and sense. It still works, but we have to remember the correct sign convention in all applications where the work term arises.

Someone with an odd turn of mind could paraphrase the 'unchemist' version as delta U = heat - unwork.

In principle one could apply this double negative process to all three terms and get delta Cold = cool + unwork.

It's not really unphysical, simply perverse.

Just as absolute U cannot be -ve, absolute Cold could never be +ve. It could work in principle but it would only serve to add confusion to the world.

Which I'm guessing is the OPs intent.

 

[studiot]

24 minutes ago, sethoflagos said:

Remember that engineers and physicists employ a double negative on the work term by a change of both sign and sense. It still works, but we have to remember the correct sign convention in all applications where the work term arises.

Someone with an odd turn of mind could paraphrase the 'unchemist' version as delta U = heat - unwork.

In principle one could apply this double negative process to all three terms and get delta Cold = cool + unwork.

It's not really unphysical, simply perverse.

Just as absolute U cannot be -ve, absolute Cold could never be +ve. It could work in principle but it would only serve to add confusion to the world.

Which I'm guessing is the OPs intent.

I wish I could give plus one for each line of this.

Just so funny.

[swansont]

1 hour ago, exchemist said:

Oh you mean transfer of momentum by inelastic (i.e. Raman) scattering? 

Elastic scattering does this, too. There’s a net outward radiation pressure from any source.

[sethoflagos]

55 minutes ago, studiot said:

I wish I could give plus one for each line of this.

Just so funny.

Much appreciated. Thank you.

[studiot]

9 minutes ago, sethoflagos said:

Much appreciated. Thank you.

The idea of unchemists doing unwork fantastic.

Reminds me of that great letter about setaside that was doing the rounds about 10 years ago.

[Willem F Esterhuyse]

14 hours ago, swansont said:

And the connection to cold is?

Mean free path is a minimum for objects at 0 K.

[exchemist]

10 hours ago, swansont said:

Elastic scattering does this, too. There’s a net outward radiation pressure from any source.

Surely elastic scattering, by definition, does not transfer any energy to the medium doing the scattering? If it idid, it would not be elastic, would it? If there were energy transfer there would have to be either absorption of photons or a change in frequency. This is a question of energy rather than momentum. 

Inelastic (Raman) scattering involves excitation of vibrational and rotational modes in the molecules and a concomitant change in frequency of the radiation. My issue is whether there is a process by which EM radiation can directly excite translational motion in molecules. I cannot think of such a process - but then I am a chemist, of course. 

Edited February 19, 2023 by exchemist

[studiot]

2 hours ago, exchemist said:

Surely elastic scattering, by definition, does not transfer any energy to the medium doing the scattering? If it idid, it would not be elastic, would it? If there were energy transfer there would have to be either absorption of photons or a change in frequency. This is a question of energy rather than momentum. 

Inelastic (Raman) scattering involves excitation of vibrational and rotational modes in the molecules and a concomitant change in frequency of the radiation. My issue is whether there is a process by which EM radiation can directly excite translational motion in molecules. I cannot think of such a process - but then I am a chemist, of course. 

Laser vapourisation ?

https://www.mayoclinic.org/tests-procedures/prostate-laser-surgery/about/pac-20384874

 

Laser welding  and other uses of lasers as heat sources to melt materials ?

[exchemist]

3 hours ago, studiot said:

Laser vapourisation ?

https://www.mayoclinic.org/tests-procedures/prostate-laser-surgery/about/pac-20384874

 

Laser welding  and other uses of lasers as heat sources to melt materials ?

In the visible, photons are absorbed by electronic transitions, creating excited states that relax by non-radiative means (i.e. intermolecular collisions) to populate vibrational and rotational excited states  - and thence eventually translational motion.  In the IR, they are absorbed by vibrational states, which then similarly cascade down by collisional relaxation. None of these radiative processes, so far as I know, can directly stimulate excitation of  translational motion. The exceptions would be ionisation or bond breaking, where the absorption causes an electron to jump right out of an atom, or two part-molecules to fly apart because the bond strength has been exceeded.

It seems to me you need a transition dipole moment in order to absorb a photon and I can't see how you get that in translational motion. So I think, subject to correction by one of the physicists here, that translational excitation is always the consequence of collisions, rather than radiation directly. 

Edited February 19, 2023 by exchemist

[swansont]

6 hours ago, exchemist said:

Surely elastic scattering, by definition, does not transfer any energy to the medium doing the scattering? If it idid, it would not be elastic, would it? If there were energy transfer there would have to be either absorption of photons or a change in frequency. This is a question of energy rather than momentum. 

Inelastic (Raman) scattering involves excitation of vibrational and rotational modes in the molecules and a concomitant change in frequency of the radiation. My issue is whether there is a process by which EM radiation can directly excite translational motion in molecules. I cannot think of such a process - but then I am a chemist, of course. 

Elastic scattering just means there’s no change in the energy level of the atom or molecule. If the photon is absorbed and re-emitted in a different direction, there is momentum transferred to the atom. It’s the basis of laser cooling.

7 hours ago, Willem F Esterhuyse said:

Mean free path is a minimum for objects at 0 K.

How much “coldness” is contained in an object at 0K?

[exchemist]

8 minutes ago, swansont said:

Elastic scattering just means there’s no change in the energy level of the atom or molecule. If the photon is absorbed and re-emitted in a different direction, there is momentum transferred to the atom. It’s the basis of laser cooling.

How much “coldness” is contained in an object at 0K?

OK fair enough. So any absorption process will also, as a side effect, alter the translational kinetic energy of the entity that absorbs the photon. This is not elastic scattering, however.   

[swansont]

29 minutes ago, exchemist said:

OK fair enough. So any absorption process will also, as a side effect, alter the translational kinetic energy of the entity that absorbs the photon. This is not elastic scattering, however.   

If a photon is re-emitted when the atom drops back to its original state, what is inelastic about the process?

[exchemist]

1 hour ago, swansont said:

If a photon is re-emitted when the atom drops back to its original state, what is inelastic about the process?

Surely elastic means with no energy loss, does it not? In which case, if there is a net energy transfer from the radiation to the matter, the process is not elastic  - and there has to be a net frequency shift.

But, the issue of whether we call it elastic or inelastic aside, what I take out of this is that if photons are absorbed by matter, there will be a translational energy gain, in addition to the excitation responsible for the absorption.

However radiation can't alter the translational kinetic energy of molecules alone - it has to be a side effect of absorption (or emission).

Is that fair?     

 

Edited February 19, 2023 by exchemist

[swansont]

1 hour ago, exchemist said:

Surely elastic means with no energy loss, does it not? In which case, if there is a net energy transfer from the radiation to the matter, the process is not elastic  - and there has to be a net frequency shift.

No kinetic energy loss, from mechanics. Energy can be transferred from one particle to another.

But there is no energy loss. The atom is in the same energy state. Photon energy shifted to or from the atom’s kinetic energy. 

 

1 hour ago, exchemist said:

But, the issue of whether we call it elastic or inelastic aside, what I take out of this is that if photons are absorbed by matter, there will be a translational energy gain, in addition to the excitation responsible for the absorption.

However radiation can't alter the translational kinetic energy of molecules alone - it has to be a side effect of absorption (or emission).

Is that fair?     

Why does that matter? The net result is translation. Radiation pressure is responsible for clearing out the gas near a star after fusion is initiated. For the tail on a comet.

And you can have e.g. Rayleigh scattering, another elastic process.

[exchemist]

50 minutes ago, swansont said:

No kinetic energy loss, from mechanics. Energy can be transferred from one particle to another.

But there is no energy loss. The atom is in the same energy state. Photon energy shifted to or from the atom’s kinetic energy. 

 

Why does that matter? The net result is translation. Radiation pressure is responsible for clearing out the gas near a star after fusion is initiated. For the tail on a comet.

And you can have e.g. Rayleigh scattering, another elastic process.

OK. The issue I am trying to bottom out is by what means, if any, radiation can transfer energy to the translational motion of molecules, as part of a heating process. To do that, energy has to be lost by the radiation and gained by the matter it is interacting with. That is the energy loss I am talking about. 

I started from the viewpoint that radiation can't directly stimulate translational motion, as there was no dipole for it to couple to. What you have pointed out is that when a photon is absorbed, it not only excites the molecule to a new internal energy state but also imparts its momentum to it, adding to its kinetic energy of translation. That's fine and seems to confirm my original thought that translational excitation directly, on it own, is not a possible process: it has to be a byproduct of an absorption due to excitation of some other degree of freedom rotation, vibration or electronic).

Rayleigh scattering won't add kinetic energy to the scattering particle, presumably, as the wavelength of the scattered photon is the same as that of the incident photon.      

 

[swansont]

48 minutes ago, exchemist said:

OK. The issue I am trying to bottom out is by what means, if any, radiation can transfer energy to the translational motion of molecules, as part of a heating process. To do that, energy has to be lost by the radiation and gained by the matter it is interacting with. That is the energy loss I am talking about. 

I started from the viewpoint that radiation can't directly stimulate translational motion, as there was no dipole for it to couple to.

Photon absorption is often a dipole interaction. There is a transition dipole moment, which gives rise to certain selection rules.

 

48 minutes ago, exchemist said:

What you have pointed out is that when a photon is absorbed, it not only excites the molecule to a new internal energy state but also imparts its momentum to it, adding to its kinetic energy of translation. That's fine and seems to confirm my original thought that translational excitation directly, on it own, is not a possible process: it has to be a byproduct of an absorption due to excitation of some other degree of freedom rotation, vibration or electronic).

Rayleigh scattering won't add kinetic energy to the scattering particle, presumably, as the wavelength of the scattered photon is the same as that of the incident photon.      

If the photon changes direction, momentum has been imparted, and thus, kinetic energy. The amount of energy imparted is small, so the wavelength is only changed by a small amount - to first order it can be ignored.

[exchemist]

1 minute ago, swansont said:

Photon absorption is often a dipole interaction. There is a transition dipole moment, which gives rise to certain selection rules.

 

If the photon changes direction, momentum has been imparted, and thus, kinetic energy. The amount of energy imparted is small, so the wavelength is only changed by a small amount - to first order it can be ignored.

Do you think Rayleigh scattering can produce a significant heating effect, then?