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From: Bill Ward on 4 Sep 2007 22:53
On Tue, 04 Sep 2007 15:37:45 -0700, kdthrge wrote:

> On Sep 3, 10:04 pm, Bill Ward
> ''''If you're talking about thermal conductivity of gases, it approaches
> zero as the pressure drops. That's why a Dewar works.'''
>
> Just to clarify, the question is not the thermal conductivity of the
> gases. This is the rate that heat is transfered through the gases.
>
> The question is the quantity of energy that is transfered through the
> gases to the metal. Therefore the experiment is probably better with the
> air mixed or moved through the duct.
>
> This shows that the temperature of the metal plate and the energy it
> radiates out of the duct, must be from the gases within the duct. It is
> impossible that this energy is being transfered by the collisonal energies
> alone.

I understand what you're saying, but not why you say it. Why can't the
energy be carried through the gas to the plate by molecular collisions per
the kinetic theory?

> This is true of O2 and N2. This proves, and many other examples
> prove, that O2 and N2 absorb infrared radiation, and radiate this
> radiation in continous spectra.

But you haven't shown the requirement for radiative transport in the gas
if molecular collisions can carry the energy.

> That CO2 or other gases have dark regions in this continous spectra at
> specific pressures, does not mean that these gases absorb these
> frequencies while O2 and N2 do not. These bands only mean low re-
> emision in these regions, probably the direct result that the molecule
> radiates these energies better at other frequencies.

The spectrum is very well explained by internal molecular oscillations.
Each line can be attributed to a particular resonance mode. It's easy to
confirm from a number of sources.

> In Planck's theory, radiation energy is in sums of hv. Not integrals of
> vibrations. Present theory makes a grevious error in ignoring Planck and
> working with wavelengths. Only the discreet values of the sums of hv
> exist.

Planck's constant is pretty small. It doesn't take much energy to broaden
the lines.

>
> Therefore there is not a continuum for wavelengths, and integrating
> energy theoretically by wavelength produces very invalid theoretical
> reckoning. A proper Planck curve is done in frequency, has peak
> intensity according to Wien, and the area beneath the curve is
> proporional to temperature by Stefans law and Boltzman's calculations
> for energy.

Frequency and wavelength are just two ways of looking at the same thing,
so I don't think using the "wrong one" would invalidate anything.
Plotting the graphs is independent of the basic calculations.

>
> The distribution is only the theoretical probability for the oscillator
> to be in a particular energy state at any given transmission.This
> probability is for a single oscillator over a sufficient period of time.
> Therefore it is also the flux at each frequency in a sample of the
> radiation from a great number of oscillators of an emission system of a
> gas. Any samle of the radiation field will have this same relative
> distribution of energy at each frequency.

I think the point is that different molecules have different oscillating
modes. Simple molecules like O2 and N2 don't have any modes in the IR,
while more complex ones like H2O and CO2 do.
>
> Since the oscillator exists in this radiation field and it's relative
> distribution of energy for frequency, the oscillator closely mimicks
> this relative distribution in emissions.
>
> Lose of energy transmission in any frequency is made up in other
> frequencies. This is why no detectable effect on temperature can be
> documented according to the theoretical estimations of the spectroscopy.

But there's those NDIR analyzers that depend on exactly that heating
effect.

The kinetic theory has a long, solid history and makes quite valid
predictions. Disproving it would be a long row to hoe.



From: kdthrge on 4 Sep 2007 23:17
On Sep 4, 7:08 pm, Bill Ward .
>
> From the kinetic energy of the molecules of the air.
>
>
> > This proves the gases are also transmitting energy according to the
> > density of the radiation field of Boltzman Stefan for temperature.
>
> There's where you lose me. It seems to me the gas is equilibrating
> with the plate, then the plate radiates the energy. The radiation is
> emitted from the plate, not the gas.

But it must transfer through the gas in order to radiate from the
metal.

It is impossible that the kinetic energy of the gases is being
transmitted into the metal, and that on the other side of the metal
there is radiation of Boltzman Stefan from the transference of energy
in this fashion. This proves the existence of the energy of the
radiation field.

The average kinetic energy of the molecules of a gas is kT. At 300K
this is about .025eV. For 1 mole this is 2490 Joules.
This is the kinetic energy of all the velocities of every molecule in
1 mole. This is the sum total of all of the energy of all of the
velocities of the mole of gas.

At 300K, 459 Joules per second per sq meter is leaving the surface of
the metal. This means in 5.4 seconds, all of the energy of all of the
velocities of the molecules in 1 mole of gas must be delivered through
the metal of 1 sq meter from the gas for the metal to maintain this
temperature.
This mole of gas if 1 meter square would be .38 meter deep.

If one is using a monatomic gas at constant volume, the heat capacity
of this gas is 3/2RT or 3735 Joules. Although not exact, this is the
total heat in one mole of monatomic gas above absolute zero.

At 460 Joules per second per sq meter, this energy would be lost in
8.1 seconds.

What if you cut the source of heat into the gas and evaluated the time
of total heat loss, rate of radiated energy loss with decreasing
temperature and how much time this takes?

You see nothing wrong with these numbers for the kinetic energy of the
velocities and the quantity of energy that is radiated, or that it is
possible for the gas to deliver this energy at this rate to the metal
by collisional energy?

The idea of the metal and the transference of heat to and through the
metal is to prove the idea false that gases do not obey to Boltzman
Stefan for temperature and energy.
The existence of the radiation field is proved. There is no other
possible means for this energy to be transferred to the metal.

It is impossible that the energy of the velocities of the molecules is
transfering the energy to the metal which then leaves the other side
at the rate that it does.

Let me see some numbers in Joules that show this to be possible. Or
show me where this figuring is wrong and where the energy comes from
that is radiating from the metal if it is at 300K. The room could be
cooled, and the temperature in the duct adjusted so that the exterior
of the metal was at 300K. This is to reduce the radiation from the
room on the outside of the metal for the accounting of the energy that
is being radiated.

At this point I damn sure need to see your accounting for the energy
supposedly transfered into the metal by the collisions of the
molecules. Or maybe you could explain your viewpoint and education
that can quantify this energy according to collisional transfer of
energy.

>
> > If the pressure is decreased radically, there becomes too few molecules
> > to adequately absorb radiation and re-emit this radiation.
>
> If you reduce the gas pressure, you reduce the temperature, and must add
> energy to come back to the same temperature. You haven't yet shown that
> there's any radiation interacting with the gas.

Of course you do. But now at the same temperature, the transfer of
energy is very nearly the same, although the number of collisions is
reduced. No viable evidence that the collisions are the sole means
that energy is transfering to the metal.

KD

From: kdthrge on 4 Sep 2007 23:23
On Sep 4, 9:53 pm, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote:
> On Tue, 04 Sep 2007 15:37:45 -0700, kdthrge wrote:
> > On Sep 3, 10:04 pm, Bill Ward
> > ''''If you're talking about thermal conductivity of gases, it approaches
> > zero as the pressure drops. That's why a Dewar works.'''
>
> > Just to clarify, the question is not the thermal conductivity of the
> > gases. This is the rate that heat is transfered through the gases.
>
> > The question is the quantity of energy that is transfered through the
> > gases to the metal. Therefore the experiment is probably better with the
> > air mixed or moved through the duct.
>
> > This shows that the temperature of the metal plate and the energy it
> > radiates out of the duct, must be from the gases within the duct. It is
> > impossible that this energy is being transfered by the collisonal energies
> > alone.
>
> I understand what you're saying, but not why you say it. Why can't the
> energy be carried through the gas to the plate by molecular collisions per
> the kinetic theory?
>
> > This is true of O2 and N2. This proves, and many other examples
> > prove, that O2 and N2 absorb infrared radiation, and radiate this
> > radiation in continous spectra.
>
> But you haven't shown the requirement for radiative transport in the gas
> if molecular collisions can carry the energy.
>
> > That CO2 or other gases have dark regions in this continous spectra at
> > specific pressures, does not mean that these gases absorb these
> > frequencies while O2 and N2 do not. These bands only mean low re-
> > emision in these regions, probably the direct result that the molecule
> > radiates these energies better at other frequencies.
>
> The spectrum is very well explained by internal molecular oscillations.
> Each line can be attributed to a particular resonance mode. It's easy to
> confirm from a number of sources.
>
> > In Planck's theory, radiation energy is in sums of hv. Not integrals of
> > vibrations. Present theory makes a grevious error in ignoring Planck and
> > working with wavelengths. Only the discreet values of the sums of hv
> > exist.
>
> Planck's constant is pretty small. It doesn't take much energy to broaden
> the lines.
>
>
>
> > Therefore there is not a continuum for wavelengths, and integrating
> > energy theoretically by wavelength produces very invalid theoretical
> > reckoning. A proper Planck curve is done in frequency, has peak
> > intensity according to Wien, and the area beneath the curve is
> > proporional to temperature by Stefans law and Boltzman's calculations
> > for energy.
>
;> Frequency and wavelength are just two ways of looking at the same
thing,
;> so I don't think using the "wrong one" would invalidate anything.
;> Plotting the graphs is independent of the basic calculations.

That is not true when integrating. According to Planck, if there were
wavelengths, there would be no wavelength longer than 2.99cm since
this would require a division of h. There would be no intermediate
wavelengths between, 2.99cm divided by 2,,, divided by 3,,,

There are only false graphs of distribution available these days on
the web. AGW utilizes these false graphs to support their lies. AT
some point someone will do a direct analyses of intensity of radiation
at particular frequencies for temperature and hang these charlatans by
the nearest tree for their theoretical curves.

KD
>
>
>
> > The distribution is only the theoretical probability for the oscillator
> > to be in a particular energy state at any given transmission.This
> > probability is for a single oscillator over a sufficient period of time.
> > Therefore it is also the flux at each frequency in a sample of the
> > radiation from a great number of oscillators of an emission system of a
> > gas. Any samle of the radiation field will have this same relative
> > distribution of energy at each frequency.
>
> I think the point is that different molecules have different oscillating
> modes. Simple molecules like O2 and N2 don't have any modes in the IR,
> while more complex ones like H2O and CO2 do.
>
>
>
> > Since the oscillator exists in this radiation field and it's relative
> > distribution of energy for frequency, the oscillator closely mimicks
> > this relative distribution in emissions.
>
> > Lose of energy transmission in any frequency is made up in other
> > frequencies. This is why no detectable effect on temperature can be
> > documented according to the theoretical estimations of the spectroscopy.
>
> But there's those NDIR analyzers that depend on exactly that heating
> effect.
>
> The kinetic theory has a long, solid history and makes quite valid
> predictions. Disproving it would be a long row to hoe.- Hide quoted text -
>
> - Show quoted text -


From: Phil. on 4 Sep 2007 23:48
On Sep 4, 9:57 pm, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote:
> On Tue, 04 Sep 2007 15:18:02 -0700, Phil. wrote:
> > On Sep 4, 4:58 pm, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote:
> >> On Tue, 04 Sep 2007 11:49:24 -0700, Phil. wrote:
> >> > On Sep 4, 1:50 pm, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote:
> >> >> On Tue, 04 Sep 2007 06:47:19 -0700, Phil. wrote:
> >> >> > On Sep 3, 11:04 pm, Bill Ward <bw...(a)REMOVETHISix.netcom.com>
> <snip prior posts>
>
> >> >> >> I think they are way overstating the magnitude of the effect of
> >> >> >> traces of CO2 compared to thousands of times more water vapor.
>
> >> >> > Which planet is that? 380ppmv times one thousand is 38% that's a
> >> >> > very wet (and hot) place!
>
> >> >> You're right. I dropped a decimal point. Mea culpa. Or I guess I
> >> >> could claim a "colloquialism".
>
> >> > Actually two and since when has 'thousands' been a 'colloquialism' for
> >> > tens? Incidentally this is a science newsgroup.
>
> >> >> > More like 20X max to 1/3 in the troposphere, much less in the
> >> >> > stratosphere.
>
> >> >> >> And they are invoking dubious positive feedback properties of WV
> >> >> >> while ignoring its obvious negative feedbacks from cloud
> >> >> >> formation. I agree it looks totally bogus, biased, and
> >> >> >> unscientific.
>
> >> >> > The feedback of temperature on Water vapor is well known and
> >> >> > described by the Clausius-Clapeyron equation.
>
> >> >> Impressive way to talk about humidity. Consider the possibility
> >> >> that you may not be the only person in the NG to have taken P.Chem.
>
> >> > Maybe but I imagine that you can use Google? Again it's a science
> >> > newsgroup, you said that the link between water vapor pressure and
> >> > temperature
>
> >> It makes your position look desperate when you resort to misquoting. I
> >> said "dubious positive feedback properties of WV", not "the link
> >> between water vapor pressure and temperature".
>
> > The positive feedback of WV is that it's vapor pressure increases with
> > temperature, what else were you referring to?
>
> Your hidden assumption that increased WV would increase the temperature.
> in spite of your link to the contrary below.


Ah so you think that it's "totally bogus, biased, and unscientific" to
think that an increase in the most powerful greenhouse gas in our
atmosphere would increase the gh effect. I suppose that it's unbiased
and scientific to assume as you do that instead that increase will
create more clouds and cool the atmosphere despite the absence of any
evidence to suuport you.

>
>
>
> >> "looks totally bogus, biased, and unscientific" whereas I
>
> >> > showed that there is a precise scientific relationship, then you
> >> > complain that I'm being too scientific,
>
> >> No, I think you are attempting to intimidate by being pretentious, and
> >> not being scientific enough. If, as you apparently assume, you are the
> >> only educated person in the group, then you should use phrases that
> >> would be more familiar to us. It makes you look like you are trying to
> >> confuse, rather than convince.

What's intimidating about mentioning the Clausius-Clapeyron equation?
It's no different to referring to Boyle's Law or Stefan-Boltzmann (I
don't recall you calling Deatherage pretentious for his continued
reference to the latter, but then he's on your side)

>
> > OK I assumed that you either had a HS knowledge of chemistry or that you
> > could use Google, apparently I was wrong on both counts. Apparently you
> > have a chip on your shoulder about your science education, what was I
> > supposed to do give the thermodynamic derivation of the C-C equation on
> > here without the ability to adequately present differential equations,
> > which laws is it allowed to refer to without being pretentious, Boyle's
> > Law, Stefan-Boltzmann or is it just any one you haven't heard of?
>
> You assume way too much. You don't know anything about my education, and
> you don't need to. I was referring to your apparent assumptions of the
> educational level in this group.

So you thing it's wrong to assume HS level chemistry as background for
this subject?

> This is a discussion group, and trying to
> impress by throwing around unnecessary technical jargon is
> counterproductive. If you can't explain in terms most readers can
> understand, you don't really know the subject.


Who's trying to impress, to say that vapor pressure is related to
temperature through the Clausius-Clapeyron equation is about the
minimal description which any one not familiar with the term can
easily find on google. I suppose I should have just posted this:

ln(P2/P1) = (DeltaHv/R)(1/T1-1/T2)
where
T1 and P1 are a corresponding temperature (K) and pressure
T2 and P2 are the corresponding temperature and pressure at another
point
DeltaHv is the enthalpy of vaporization
R is the gas constant

Do you like that better?

>
> > You, on the other hand are trying to prosletyse on here and deliberately
> > exaggerate by 100-fold, in other words lie, yet you have mounted a
> > campaign against another poster here for what you have termed lying!
>
> I made a careless decimal point error and immediately acknowledged it.
> What's your problem? Do you really not see the difference?


No you didn't immediately acknowledge it you waited 14 hrs until it
was pointed out to and you flippantly implied it was no big deal.

>
> >> > if I'd just said +7%/K you'd have demanded to know where the number
> >> > came from!
>
> >> You're being a bit presumptuous there. Retake that mindreading class.
>
> > No just your past history on here Bill.
>
> I'll leave that one for readers to decide.
>
> >> > If you want to be taken seriously on sci.environment be prepared to
> >> > talk science
>
> >> With a little more experience you may be able to "talk science"
> >> effectively. From what I've seen, you are far too impressed with your
> >> education to use it wisely. Knowledge is not wisdom. To teach, you
> >> have to communicate, and you apparently don't yet know how.
>
> > From 30+ years experience of teaching I've had no problems, however
> > the students want to learn, are prepared and have a grounding in the
> > necessary prerequisites, and don't bring an attitude and agenda to the
> > class.
>
> I'm surprised. You seem much less experienced than that. I've no doubt
> you were aware of no problems, but I'm not so sure about your students,
> considering your attitude.

Well they were certainly happy enough judging by their course
assessments which were extremely good! Of course as I said before the
students do their homework and don't come into the classroom with
their preconceived ideas and political agendas.

>
> > And yet my reply has told you exactly what the link between vapor
> > pressure and temperature is and an estimate of its magnitude for water,
>
> And apparently the thought never went through your mind that I might
> already know that.
>
> >I can only conclude that you're pissed off because you really didn't want
> >to know that because it doesn't fit with your agenda!
>
> Your logic is as faulty as your assumptions.
>
> >> To communicate, you have to learn what the students know, then expand
> >> on it to get your point across. Trying to impress them by talking over
> >> their heads comes across as just being a jerk, no matter how well
> >> intentioned.

Something I have no problem with.
>
> > I wasn't aware that we have a professor/student relationship on here,
> > it's supposed to be a discussion, but since you've just said that you
> > don't have a HS level knowledge and that I'm supposed to educate you,
> > that's tough to do when the student (you) doesn't want to know the
> > answers and disputes everything he's been told.
>
> When you get rattled, you have a tendency to misquote as above. That's a
> dead giveaway. I haven't (and don't) say anything about my educational
> level, and particularly nothing about needing you to "educate" me.

Ah, but you think I should educate the rest of the group apparently?

>
> >> otherwise take off to alt.fan.rush-limbaugh where you'll
>
> >> > not hear from me again (except by cross postings).
>
> >> That's what I'm talking about (jerk-wise).
>
> > You're one who's making false statements and then getting in a huff when
> > called to account.
>
> Are you talking about me dropping a couple of decimal points? So sue me.

See there you go again, you exaggerated by a factor of 100's and I
don't believe it was accidental, which is borne out by this remark.
It's OK for you to lie in pursuit of your agenda but you go after
anyone you think does that, so let's see your apology to the group for
misrepresentation?

>
> >> >> > Also it's far from obvious that increased water vapor will lead to
> >> >> > more cloud.
>
> >> >> Depends on who's looking. Try getting out more. Preferably in a
> >> >> sailplane.
>
> >> > Done that, also a Cessna, however while it seems intuitive that more
> >> > evaporation due to increased temperature would lead to more cloud as
> >> > far as I'm aware there's no evidence that it's true.
>
> >> And you're a pilot? You haven't seen cumulii popping when the surface
> >> gets hot? Pull the other one, it whistles.
>
> > Yes and as discussed below that doesn't mean that in the steady state a
> > warmer world will have a different cloud cover, it isn't that easy.
>
> News flash: The world is not now, nor has it ever been, in a steady state.


Check out the difference between climate and weather.

>
> >> or as to any change in
>
> >> > type it makes a difference what sorts of cloud we're talking about,
> >> > status, cumulus, cirrus.
>
> >> Do you have some explanation of what happens to the water vapor if it
> >> doesn't condense?
>
> > Well it will condense at a different altitude also it might rain much
> > harder which could lead to shorter lived clouds.
>
> Here's where you apparently think I would demand a cite for that. But
> you'd be wrong. It's just a speculation and needs no cite.
>
>
>
> >> > Here's a couple of sites which discuss the difficulties:
> >> >http://web.mit.edu/cgcs/www/clouds.html
>
> >> <begin excerpt>
> >> Recent observational studies show that these effects almost balance,
> >> but that the cooling effect is somewhat more important. From the point
> >> of view of global change, however, it is crucial to note that this
> >> small difference is about five times larger than the radiative effect
> >> anticipated from a doubling of atmospheric carbon dioxide (CO2), and
> >> that the individual components of the difference are orders of
> >> magnitude larger. In existing climate models about one third of the
> >> predicted warming due to increasing CO2 arises because of the predicted
> >> cloud changes. These predictions, however, are highly speculative
> >> because none of the models include interactive cloud physics. <end
> >> excerpt>
>
> >> Sounds OK to me.
>
> >> >http://www.cahmda.wur.nl/Trenberth_37.pdf
>
> >> Five pages to tell us he hasn't got a clue.
>
> >> > It's actually a very complex problem, you could just as intuitively
> >> > say, 'more rain and shorter cloud lifetime'.
>
> >> But I didn't. That's your intuition, not mine.
>
> > Indeed and both equally valid on that basis, the trouble is nobody knows
> > which is right or even if they're both wrong, the difference is that I
> > recognise that but you dogmatically state that you're right. When not
> > even the Head of the Climate Analysis Section at the National Center for
> > Atmospheric Research knows the answer but you do, that's one smart
> > student!
>
> For a teacher, you haven't learned much, have you?
>
> > So which is it, Bill are we to treat you as some one who doesn't have
> > basic HS knowledge and shouldn't be expected to know anything about
> > thermodynamics and must have everything explained to him step by step or
> > are we to treat you as some omniscient genius who knows the answer to
> > everything but can't explain it in a scientific way, something of an
> > idiot savant?
>
> I only expect to be treated with the same respect as you would anyone else
> on the NG, which is apparently the case. But attempted intimidation is
> not the key to success. Someday you may learn...


You have to have a very thin skin to consider mention of the Clausius-
Clapeyron equation an attempt at intimidation!


From: Bill Ward on 5 Sep 2007 01:56
On Tue, 04 Sep 2007 20:17:04 -0700, kdthrge wrote:

> On Sep 4, 7:08 pm, Bill Ward .
>>
>> From the kinetic energy of the molecules of the air.
>>
>>
>> > This proves the gases are also transmitting energy according to the
>> > density of the radiation field of Boltzman Stefan for temperature.
>>
>> There's where you lose me. It seems to me the gas is equilibrating with
>> the plate, then the plate radiates the energy. The radiation is emitted
>> from the plate, not the gas.
>
> But it must transfer through the gas in order to radiate from the metal.
>
> It is impossible that the kinetic energy of the gases is being transmitted
> into the metal,

Why do you say that? The molecules of gas are pounding on the molecules
of metal, and transferring KE. What about that is "impossible"?

> and that on the other side of the metal there is radiation
> of Boltzman Stefan from the transference of energy in this fashion. This
> proves the existence of the energy of the radiation field.

There's radiation from the metal, but not from the gas. The gas to metal
transfer is by direct molecular impacts.

>
> The average kinetic energy of the molecules of a gas is kT. At 300K this
> is about .025eV. For 1 mole this is 2490 Joules. This is the kinetic
> energy of all the velocities of every molecule in 1 mole. This is the
> sum total of all of the energy of all of the velocities of the mole of
> gas.

I don't think those numbers are exactly correct, (IIRC, KE is 3/2 kT)
but I'd rather not tackle that now and stay on a qualitative level.
>
> At 300K, 459 Joules per second per sq meter is leaving the surface of
> the metal. This means in 5.4 seconds, all of the energy of all of the
> velocities of the molecules in 1 mole of gas must be delivered through
> the metal of 1 sq meter from the gas for the metal to maintain this
> temperature.

Don't forget that the T will drop rapidly unless energy in coming in from
somewhere else. And you're assuming a 0K black body sink.

> This mole of gas if 1 meter square would be .38 meter deep.
>
> If one is using a monatomic gas at constant volume, the heat capacity of
> this gas is 3/2RT or 3735 Joules. Although not exact, this is the total
> heat in one mole of monatomic gas above absolute zero.
>
> At 460 Joules per second per sq meter, this energy would be lost in 8.1
> seconds.

Only if you were radiating into 0K, and no energy was coming in from the
source.

> What if you cut the source of heat into the gas and evaluated the time
> of total heat loss, rate of radiated energy loss with decreasing
> temperature and how much time this takes?
>
> You see nothing wrong with these numbers for the kinetic energy of the
> velocities and the quantity of energy that is radiated, or that it is
> possible for the gas to deliver this energy at this rate to the metal by
> collisional energy?

Yes, actually 460 Wm-2 seems plausible if the gas is hot enough and the
metal thin enough. Remember we're off equilibrium, because heat is flowing
through the system. The amount of heat flowing is proportional to the
temperature drops.

>
> The idea of the metal and the transference of heat to and through the
> metal is to prove the idea false that gases do not obey to Boltzman
> Stefan for temperature and energy.
> The existence of the radiation field is proved. There is no other
> possible means for this energy to be transferred to the metal.

I don't see the proof. Why can't it be molecular collisions?
>
> It is impossible that the energy of the velocities of the molecules is
> transfering the energy to the metal which then leaves the other side at
> the rate that it does.
>
> Let me see some numbers in Joules that show this to be possible. Or show
> me where this figuring is wrong and where the energy comes from that is
> radiating from the metal if it is at 300K. The room could be cooled, and
> the temperature in the duct adjusted so that the exterior of the metal
> was at 300K. This is to reduce the radiation from the room on the
> outside of the metal for the accounting of the energy that is being
> radiated.

If the room (sink) were at 300 K you'd be at equilibrium, with no
radiative transport.
>
> At this point I damn sure need to see your accounting for the energy
> supposedly transfered into the metal by the collisions of the molecules.

It comes from the gas, which presumably gets it from a similar plate on
the other end of your cell, which gets it from ...

> Or maybe you could explain your viewpoint and education that can
> quantify this energy according to collisional transfer of energy.

I'm not really qualified. What I'd recommend is that you take a look at
what hyperphysics has to say about kinetic theory:

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/ktcon.html#c1

I have to warn you that they have unfortunately fallen for the AGW scam,
but most of the traditional physics looks OK to me. It's easy to follow
the hyperlinks to browse around to related areas.

>> > If the pressure is decreased radically, there becomes too few
>> > molecules to adequately absorb radiation and re-emit this radiation.
>>
>> If you reduce the gas pressure, you reduce the temperature, and must
>> add energy to come back to the same temperature. You haven't yet shown
>> that there's any radiation interacting with the gas.
>
> Of course you do. But now at the same temperature, the transfer of
> energy is very nearly the same, although the number of collisions is
> reduced. No viable evidence that the collisions are the sole means that
> energy is transfering to the metal.

But many experiments show the conductivity is actually proportional to the
pressure. Thats how Pirani vacuum gages work.

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