What Traps Heat the Longest?


Jim Steele

It is not just a matter of how much energy increasing CO2 redirects towards the surface but how long it is trapped for that determines greenhouse gases’ warming effect. The contributions from 3 heat trapping mechanisms are examined: Greenhouse gases, heat domes and ocean barrier layers and how they contribute average global temperature and extreme weather.

Transcript is available at

https://perhapsallnatural.blogspot.com/2023/04/the-greenhouse-theorys-greatest-weakness.html

Jim Steele is Director emeritus of San Francisco State University’s Sierra Nevada Field Campus, authored Landscapes and Cycles: An Environmentalist’s Journey to Climate Skepticism, and proud member of the CO2 Coalition.

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Sweet Old Bob
April 9, 2023 6:42 pm

Read the transcript .

It is well worth your time !

And , worth bookmarking .

Stephen Wilde
Reply to  Sweet Old Bob
April 10, 2023 1:31 am

The convective overturning cycle traps heat in the form of potential energy for as long as an atmosphere is retained. That is what traps heat longest. A heat dome is but one element of that and is due to the uneven distribution of convection.
The oceans have little long term effect due to their heat content being dependent on atmospheric pressure bearing down on the surface and that is static for the globe as a whole.
Radiative effects from atmospheric gases are indiscernible compared to mechanical processes such as conduction and convection within both air and water.
Convective overturning on a global scale causes the surface temperature enhancement beneath mobile atmospheres.
All other potential imbalancing processes are neutralised by changes in convection.
There are climate zone shifts as a result of any imbalances but way too small to measure compared to natural variability. We could never isolate and measure the minuscule effect from our emissions.
Ultimately the retention of an atmosphere depends on the balance between the downward force of gravity and the upward pressure gradient force being reliably maintained. Only convective adjustments will achieve that.

Henry Pool
Reply to  Stephen Wilde
April 10, 2023 4:53 am

Amazing. It is just like I always said: there is no warming effect of more CO2 in the air
An evaluation of the greenhouse effect by carbon dioxide | Bread on the water

Of course more CO2 does make the world greener = blacker = more warmth sticks around…
Apparently even the oceans are getting greener. That could become a big problem with all that extra stench of rotting seaweed. There is also a report that a greener environment traps heat, especially the reaction at night where the leaves and the fruit and the wood grow….that reaction is exothermic.

DMacKenzie
Reply to  Henry Pool
April 10, 2023 6:41 am

….no warming effect….”

A little, Henry…. 3 watts roughly from 400 to 800 ppm increase. see under Fig 3 following by Happer and vanW.
The trouble is that 3 watts is about 1% of incoming sunlight, and a 1% increase in Earthly absolute temperature is about 3 degrees, so there can be much room for dissension about emissivities, albedos, emission wavelengths, heat balances, etc. by people skilled in only one branch of knowledge.

https://arxiv.org/pdf/2103.16465.pdf

Henry Pool
Reply to  DMacKenzie
April 10, 2023 2:57 pm

D
If you look during the day, you should get what is pictured in my fig4
If you measure during the night you get what was shown by Lillisand and Kiefer 1994, my fig 5
If you evaluate the absorption of CO2, by looking at the ir spectrum, the nett result is zero.

So when did Happer and Wijngaarden measure? And why did they disregard the SW that is back radiated by the gh gas to the sun / space?

Last edited 1 month ago by Henry Pool
slindsayyulegmailcom
Reply to  DMacKenzie
April 10, 2023 3:47 pm

3 watts is 86km not surface heat.

Tombstone Gabby
Reply to  Henry Pool
April 10, 2023 7:56 pm

G’Day Henry,

“…with all that extra stench of rotting seaweed.”

Well shucks. Some of our retired politians might regret their ocean fronting mansions. Tough.

David Solan
Reply to  Henry Pool
April 16, 2023 11:39 am

  Agree, Mr. Pool, and would like to comment further.

  Time and space are very important concepts that scientists
constantly refer to in trying to prove and elaborate their theories.
Indeed, you can frequently tell a real scientist from a fake one by
the degree to which he constantly does or does not take into account
the element of time — timing, time synchronys, and time durations —
in his explanations of nature. Jim Steele, in “Global Warming
Greenhouse Theory’s Greatest Weakness”, correctly points up this
crucial element of time and time delays in determining the temperature
effects of the Sun on the weather/climate of the Earth, and in that
sense he has proved himself a good scientist, unlike the climatistas
he is opposing.

  The radiative Greenhouse theories being proposed by these
climatistas tell us that infrared absorption blockages of radiation —
from the Earth — are somehow warming it, at least to a significant
degree compared to convective blockages in our atmosphere. That
radiation travels at 186,000 miles per second compared to the average
heat convective rate of our atmosphere (averaged over all the Earth
all the time), which, maybe, is 2 miles per hour. Getting the ratio
of the inversion of these two numbers gives us some idea of the
relative timing of the retention of heat by the Earth from these 2
causes. It comes to 8 orders of magnitude favoring convection.
Methinks the Greenhouse Warming nutcases have some explaining to
do — specifically, about 8 orders of magnitude worth of explaining.
Good luck, guys.

  But it gets worse than this for the global warming crowd. The
Earth’s atmosphere is approximately a closed system in that the
molecules of air aren’t anywhere near their gravitational escape
velocity from the planet and so they stay corralled, convectively
rising, cooling off, and then coming back down again. But the
radiation of the atmosphere does no such thing. The Earth is not a
black hole and the speed of radiation most assuredly exceeds its
escape velocity from Earth. For it, it is a one-way trip (albeit with
detours) straight into space and it’s gone … forever. Thus, any
residual, secondary heating effect it has is minimal, at best — there
is no second bite of the apple for infrared radiation leaving the hot
surface of the Earth. As a previous commentator said: “Radiative
effects from atmospheric gases are indiscernible compared to
mechanical processes such as conduction and convection within both
air and water.”

  Dr. Steele, however, seems to drop the ball in his later remarks,
at least when it comes to that other crucial element I mentioned above
— space. As real estate agents say: it’s location, location, and
location. It is not just a matter of how long heat gets trapped near
the earth’s surface, but where that heat is being “trapped” “near” the
surface. Both determine the surface global temperature — time,
because longer durations mean cumulative heat buildup, and, location,
because if the heat buildup is not where the action is, it’s nowhere
that counts. If any outgoing heat of the Earth is absorbed,
scattered, or transduced above about 8000 feet altitude, it’s a wash.
At 186,000 miles per second into outer space, or at 2 miles per hour,
and hardly a chance of it ever making it back down to the surface, that
emitted heat, after it goes that high, is going bye-bye. And if it
stays at that altitude for a while, even a long while, so what? It
still will never substantially be going to make it back to the surface
of the Earth, which is where all the weather/climate is, and so
anything happening above that altitude is irrelevant.

And the amounts of different frequencies of infrared entirely escaping
Earth into outer space, as measured by satellites, really tell us nothing about
any purported Greenhouse Effect warming of CO2 or H2O or anything
else in our lower atmosphere, since this mix of frequencies is mostly
just a function of the dance of the molecules in our upper atmosphere
way above that crucial 8000 feet.

David Solan

Jim Steele
Reply to  Sweet Old Bob
April 10, 2023 8:09 am

While I appreciate the more technical debate about heat transfer, remember this post is how to best educate the public, and I guarantee that most of the comments I see here would cause most people to be overwhelmed and stop reading. My quest is how to make the physics understandable in a way that people can relate to vs lots of equations or listing all the possible dynamics. So I am requesting if you want to make a technical argument, then also show how an ordinary person my experience the dynamic you allude to.

Thomas
Reply to  Jim Steele
April 10, 2023 9:12 am

It’s an excellent video, Jim. I will share it with friends. I really appreciate all the great videos you do. Very information and easy to follow.

petercampion2724
April 9, 2023 6:47 pm

Love your work, Jim – inspirational.

When I fight the climateers nutters in newspaper opinion columns, I sometimes append some explanatory notes for editors in an ongoing attempt to introduce them to reality.

This is today’s version…

This is the homework people must do before they’ll have any understanding whatsoever of atmospheric physics and atmospheric CO2 levels. They must understand;
– Planck’s Law, which partially explains the spectral-energy distribution of radiation emitted by a blackbody.
– Wien’s Displacement Law, which describes the difference between the temperature of a blackbody and the wavelength at which it emits the most light.
– Kirchhoff’s Radiation Law, which relates emissive power to absorptivity.
– The Second Law of Thermodynamics, which explains that the transfer of heat from a body at a given temperature to a body at a higher temperature is impossible.
– The Stefan-Boltzmann Law, which shows that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature.
– Stokes shift, on why fluorescence emits with a lower energy than the original absorption.
– Henry’s Law, which proves that the weight of a gas dissolved by a liquid is proportional to the pressure of the gas upon the liquid.

– Water vapour absorbs outgoing longwave infrared radiation (heat) across almost all of the 37,000 spectral bands, leaving just three gaps for CO2 to have any effect whatsoever.
– CO2 primarily absorbs outgoing longwave infrared radiation as photons at 14.8 microns wavelength (Planck’s Law), the most energetic of the three gaps.
– The temperature of absorption at 14.8 microns is -80 degrees Celsius (Wien’s Displacement Law), which is found in the lower stratosphere, well above the troposphere where weather exists.
– Minus 80 degrees Celsius is 95 degrees lower than average surface temperature.
– The Second Law of Thermodynamics proves heat doesn’t flow from cold zones to hotter zones.
– CO2 cannot “trap heat.” It can only delay outgoing heat momentarily so high up in the atmosphere it’s irrelevant.
– Atmospheric CO2 is contingent upon seawater temperature (Henry’s Law). 
– Only the sun has the energy to heat 1.335 billion cubic kilometres of water.
– To control Earth’s climate, you must control the sun.
– The “climate debate” is now and always has been political and anti-scientific.

The factors which bring warmth to this planet’s surface are: 1, the plasma inner core, 2, the heat of nuclear decay of uranium in the outer core and mantle, 3, volcanism, 4, the adiabatic pressure gradient of atmospheric gases under the constant pull of gravity, 5, the electromagnetic (EM) spectrum delivered from the sun, 6, orbital variations (Milankovitch cycles) and 7, the EM spectrum delivered from the galaxy.

The sun has the biggest influence on surface temperature as its contribution is largest by far. “Popular” energy balance equations do not correctly consider all the input factors. The sun’s output is not stable and is subject to a multitude of cycles from the 11-year sunspot cycle to the 12,000-year galactic EM sheet reversal cycle, (an artefact of the Parker instability). 

The planet’s surface is 71% water and water vapour absorbs EM across most of the 37,000 spectral bands, ranging from ultraviolet to the 0.37 Kelvin cosmic background radiation. Even water vapour cannot warm the planet’s surface due to the second law of thermodynamics, but it can delay the escape of heat energy which is why cloudy and/or humid nights are warmer than clear and/or low-humidity nights.

The much vilified and politicised carbon dioxide molecule does intercept escaping heat in three spectral bands but it can only do so in a frozen stratospheric zone and has zero effect on weather or climate. The claimed surface heating properties of CO2 are political fiction pushed by politicised segments of academia and most of the corporate media at the behest of the globalists who gain so much wealth and political control from it.

Phil.
Reply to  petercampion2724
April 9, 2023 7:36 pm

There is no such thing as a temperature of absorption at 14.8 microns, any unexcited CO2 molecule will absorb 14.8 micron radiation regardless of its temperature and regardless of the temperature of the source. The temperature of -80ºC has nothing to do with it.

Alexy Scherbakoff
Reply to  Phil.
April 9, 2023 8:38 pm

I’ve pointed this out a number of times. Some people refuse to get it.

DMacKenzie
Reply to  Alexy Scherbakoff
April 10, 2023 7:43 am

Sure Alexy, but on the surface emissions side, without knowing the Wein’s law peak temp (2890/microns wavelength) you can’t tell which Planck curve to use to calculate the W/sq.M being emitted over a certain bandwidth, say the atmospheric window 8-14 microns….
The atmospheric window is important because it allows photons of those wavelengths to mostly get to outer space directly from the emitting surface. IIRC a ground temp of 300 K emits about 40% of its photons in the 8 to 14 micron band, and about 90% of those make it directly to outer space in clear sky conditions. Using Wein’s law 8 to 14 microns correspond to planck’s curve peaks around +90 C to -65 C which covers all Earthly ground temperatures. Then it works out from Planck’s curve, again IIRC, that about 35% of the energy of any surface in that whole temperature range will be emitted in the 8-14 band and again, about 90% of that will get directly to outer space.
Additional CO2 in the atmosphere affects the 15 micron atmospheric absorption band slightly…about 3 watts compared to incoming solar of 340, solar after albedo of 240, surface SW of 160, etc, net Long Wave of 60, etc…. to the effect of 1 or 3 degrees being arguable by different people.

DMacKenzie
Reply to  DMacKenzie
April 10, 2023 8:06 am

Quick quiz:
The latest NASA energy budget shows 398.2 watts emitted by the surface (289.5 K on my calculator). with 40.1 watts going directly to outer space.
Given that a 289.5 K surface will emit 148 W/ M^2 (37%) of its photons in the 8-14 micron band…..What would it seem that the the planet’s cloud cover is ? Assuming clouds are thick enough to absorb all 8-14 micron photons that hit them.

(148-40.1)/148=73%. This is a little high. 50 watts thru the atmos. window would be closer…..comments on these back of napkin calcs always appreciated…

JCM
Reply to  DMacKenzie
April 10, 2023 12:02 pm

50 watts thru the atmos. window would be closer

Yup – the atmospheric emission temperature is fixed to about 273K. At a condensation surface emissivity averaging 0.6, this represents 190 Wm-2 transmitted to OLR.

The bulk remainder from surface, to account for 240 Wm-2 OLR, must be approximately 50 Wm-2.

slindsayyulegmailcom
Reply to  JCM
April 10, 2023 4:27 pm

Nowhere near 214wm² is reality.
960w-m² 8µm TOA 87°C
Surface 460w-m² Equator (-500w-m²)
160w-m² 12.57µm Antarctic above 85°S.
(-300w-m²)
103w-m² 14µm Stratosphere.
-357w-m²
90% (857/960)
40w-m² is based on local day night difference. Not total system heat.

slindsayyulegmailcom
Reply to  DMacKenzie
April 10, 2023 4:16 pm

398w-m² is local (or reduced coverage) not global. Global is 340w-m². (170w-m² NH 170w-m² SH). This is controlled by the tilt of the earth(More energy in July, less in Jan). The atmosphere follows sun energy (more energy in Jan, less in July).
148w-m² (ice age example) needs to double to increase to all (170w-m² reality) 240w-m² photons in the 8-14µm band. Because in reality earth surface isn’t -50°C.
Earth surface is 5.3°C 340w-m².

Phil.
Reply to  Alexy Scherbakoff
April 10, 2023 9:33 am

Yes Wien’s law is completely misunderstood by many in this sort of discussion, I’ve lost count of the number of times I’ve pointed this out!
This post in particular, it starts off:
This is the homework people must do before they’ll have any understanding whatsoever of atmospheric physics and atmospheric CO2 levels. They must understand;
– Planck’s Law, which partially explains the spectral-energy distribution of radiation emitted by a blackbody.
– Wien’s Displacement Law, which describes the difference between the temperature of a blackbody and the wavelength at which it emits the most light.”
And then reveals that he completely misunderstands Wien’s Law.
A surface at 10ºC emits more than 6x the radiation at 15 microns than a surface at -90ºC.

slindsayyulegmailcom
Reply to  Phil.
April 10, 2023 4:43 pm

2898/183=15.83µm (has less radiation). photons at 10.24µm (10°C) travel at speed of light therefore a long distance is achieved before wavelength is reached 15µm. If that was than absorbed and emitted back at the surface 10.24µm IR won’t change and length will be much longer than 15µm.

It doesnot add up
Reply to  slindsayyulegmailcom
April 10, 2023 7:23 pm

Radiance L is given by

L(λ)=(2hc^2/λ^5)⋅1/(exp(hc/λkT)−1)
so for λ=15 microns or 1.5E-5m
=(2×6.6E-34x9E16/(1.5E-5)^5)/(exp(6.6E-34x3E8/(1.5E-5×1.38E-23xT)-1)
=(118.8E-18/7.59E-25)/(exp(19.8E-26/(2.07E-28xT)-1)
=1.565E7/(exp(956/T)-1)
For T=283K that gives 1.565E7/29.3
For T=183K that gives 1.565E7/184.7

There is roughly 6 times (184.7/29.3) as much radiance at 15 microns in a 283K blackbody as in a 183K blackbody.

slindsayyulegmailcom
Reply to  It doesnot add up
April 11, 2023 2:43 am

Your calculations and assumptions are wrong.
283K = 363w-m²
183K=63w-m²
5.7×63=363w-m²
6×10.5=63
283K^4=6414247921/ 1 bar (100,000) / latitude (111) / 1.59m =363w-m²
Divide 6414247921/5.7=1121513121 (183K^4)
If you were in Antarctic and surrounding IR that was 63w-m² you would have to travel 3000km to have IR be at 363w-m² (every 500km would increase IR by 63w-m²).
Stop think a squared meter has multiple energy of absorbed 15.8µm to make up the heat.

bdgwx
Reply to  slindsayyulegmailcom
April 11, 2023 10:35 am

slindsayyulegmailcom: Your calculations and assumptions are wrong.

His calculations are correct. He got 6.3x, Phil got 6.5x, and I get 6.6x. We’re all roughly the same with differences likely the result of rounding errors.

BTW…https://www.spectralcalc.com/blackbody_calculator/blackbody.php will do the calculations for you if you aren’t familiar with how to do it.

Last edited 1 month ago by bdgwx
It doesnot add up
Reply to  slindsayyulegmailcom
April 11, 2023 12:12 pm

You are littering this thread with gobbledegook. because you don’t appear to understand the basic physics. I suggest you go away and do some study.

Phil.
Reply to  It doesnot add up
April 11, 2023 6:57 am

Yes the number I calculated was about 6.5x.

Phil.
Reply to  slindsayyulegmailcom
April 11, 2023 6:56 am

Total Wien law nonsense! Totally irrelevant.

slindsayyulegmailcom
Reply to  Phil.
April 11, 2023 8:09 am

If there was a burning fire it would be relevant but if you were 1 km from the burning fire you would not know there was a fire. Surface emits radiation (I don’t mean 161w-m², I mean 340w-m²) and 86km where CO2 absorbs (only 70w-m²) how is it relevant if CO2 emits. At the surface you wouldn’t notice it.

Kevin Kilty
Reply to  Phil.
April 9, 2023 8:40 pm

I understand what you were trying to explain and I agree with you, but let me suggest that people should realize that a molecule does not have a temperature. It has energy levels that are occupied or not. Temperature is a macro-scale parameter of an ensemble. Without an ensemble, and one molecule is exactly no ensemble at all, there can be no reasonable talk of temperature. This distinction is what makes what you explained true.

It doesnot add up
Reply to  Kevin Kilty
April 10, 2023 4:43 am

Temperature as a bulk property is directly proportional to the root mean square velocity of the molecules via E=3/2kT=1/2mv^2. A single molecule thus does have a temperature defined by its kinetic energy and the Boltzman constant. When you have an ensemble of molecules the distribution of velocities follows the Maxwell-Boltzman function with T as a summary parameter of the integration of all the individual molecular kinetic energies to calculate the RMS average.

Additionally a molecule will have further energy in modes of vibration, rotation and excited electron orbits when there is sufficient energy to raise it from its ground state. These energy levels are governed by quantum mechanics.

Tim Gorman
Reply to  It doesnot add up
April 10, 2023 7:03 am

Are radiation temperature and kinetic energy the same thing?

It doesnot add up
Reply to  Tim Gorman
April 10, 2023 8:56 am

No. The temperature of a gas of molecules relates to its kinetic energy, and the distribution of energies is given by the Maxwell-Boltzmann distribution. The temperature of radiation relates to a “gas” of photons that obey different, quantized statistics, and the distribution of energies is given by the Planck Law. Energy in one form can be translated to energy in a different form.

Phil.
Reply to  It doesnot add up
April 10, 2023 10:57 am

Yes there are 4 components to the energy of a molecule: translational, rotational, vibrational and electronic. For a room temperature gas it will be mostly translational and some rotational. For CO2 IR radiation at ~15microns will excite it to the first vibrational level from which it will try to decay to the ground state. This can occur via emission which is quite slow compared with collisional deactivation.

It doesnot add up
Reply to  Phil.
April 10, 2023 7:42 pm

Indeed so. Billions of collisions are likely in the lower atmosphere before re-emission of an absorbed photon is likely. In the tenuous upper atmosphere, the frequency of collisions falls off. Collision frequency scales with pressure and inverse square root of temperature.

It doesnot add up
Reply to  Phil.
April 10, 2023 5:02 am

Correct. Blackbody radiation spans the entire range of energies/frequencies/wavelengths (interchangeable via the relations E=hv and ν=c/λ) according to Planck’s Law.

https://en.m.wikipedia.org/wiki/Planck's_law

Wein’s law only serves to identify the temperature at which the Planck distribution is a maximum as a parameter of the distribution.

slindsayyulegmailcom
Reply to  It doesnot add up
April 10, 2023 5:17 pm

Equator average 27°C radiation 460w-m² is felt at sea level. 9.65µm
198w-m² compression heating(internal)
174w-m² IR (out to space)
88w-m² stratosphere (solar heating) 14.6µm
372w-m² loss to space.
1360w-m² TOA (900w-m²) loss to space.
Surface is maximum, stratosphere is maximum for 14.6µm unless solar heating increases.

slindsayyulegmailcom
Reply to  Phil.
April 10, 2023 4:05 pm

CO2 accepts a certain length of photon wave. That wave is proportional to a temperature. CO2 also has a ground state. 17µm (Wiens constant 2898/17= 170K -103°C). Absorption at 4.3µm solar heat brings the level up to excited state 14.8µm 196K -77°C). You denial of this and false assumption CO2 absorbs at any temperature is misguided.

bdgwx
Reply to  slindsayyulegmailcom
April 10, 2023 7:20 pm

Phil is correct. CO2 absorbs 15 um radiation regardless of the temperature of the emitter or the CO2.

Last edited 1 month ago by bdgwx
slindsayyulegmailcom
Reply to  bdgwx
April 11, 2023 1:16 am

Phil is incorrect. If an object radiates, it radiates at a certain wavelength. Temperature can be determined
2898/K=wavelength

For CO2 the wavelength is 15µm and the max temperature is 193K -80°C If the object is at room temperature and radiates. no IR will be absorbed by CO2 from that object.

Phil.
Reply to  slindsayyulegmailcom
April 11, 2023 7:13 am

This is faulty Wien’s law garbage. The absorption wavelength of CO2 is determined by the Vibrational/Rotational energy level. Here’s the spectrum showing the rotational fine structure:
CO2-lines.png

CO2 is not a black body, the Wien’s law equation you are using gives the wavelength of the peak of the emission spectrum of a blackbody at a certain temperature and has nothing to do with the CO2 absorption spectrum.

bdgwx
Reply to  slindsayyulegmailcom
April 11, 2023 9:25 am

There is some confusion here, but I’m not sure where it starts. Do you understand that Wein’s Law relates temperature to the peak wavelength? I have boldened peak to reinforce that it is a statement of the wavelength with highest spectral energy density of the emitting body. It is NOT a statement of the temperature the body must be at to absorb/emit that wavelength.

Jim Gorman
Reply to  bdgwx
April 11, 2023 8:47 am

See the following.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6174548/

There are numerous interactions with absorptivity. Temperature, pressure, water vapor, etc.

It doesnot add up
Reply to  slindsayyulegmailcom
April 10, 2023 7:53 pm

There are many frequencies at which CO2 absorbs, although there are particularly dominant frequencies for particular modedois of vibration – scissoring, symmetric and asymmetric stretches, which are spread by rotational energy levels. The details give rise to the characteristic absorption spectrum of the gas.

That has nothing to do with the temperature of the gas, which only determines (via the Maxwell Boltzman distribution) what proportion of the molecules are moving with sufficient kinetic energy to be able potentially to convert that in part into emission of a photon for a particular difference in energy states.

Blackbody radiation produces photons at all energies/frequencies, with their frequency of occurrence determined by Planck’s Law.

slindsayyulegmailcom
Reply to  It doesnot add up
April 11, 2023 1:10 am

Not the gas the IR.

bdgwx
Reply to  slindsayyulegmailcom
April 11, 2023 2:28 pm

slindsayyulegmailcom: Not the gas the IR.

What does this even mean?

Last edited 1 month ago by bdgwx
MarkW
Reply to  petercampion2724
April 9, 2023 10:08 pm

The second law says nothing of the sort.
The 2nd law deals with net energy flows. Energy flows both from hot to cold and from cold to hot. It’s just that a lot more flows from hot to cold.
That’s why warming a cold object will also warm the warmer objects around it.

mkelly
Reply to  MarkW
April 10, 2023 6:41 am

From my thermodynamics book:”It is common knowledge that heat does not, of its own volition, flow from cold medium to a warmer one.”

He spoke of heat you changed what he said to “energy”. His statement is correct.

I can not understand your last sentence. Can you write an heat transfer equation showing where T(hot) increases in temperature?

Phil.
Reply to  mkelly
April 10, 2023 11:25 am

A classic case is using a thermocouple to measure a flame temperature, the ThC gets red-hot and loses radiant energy to its cold surroundings and therefore measures a lower temperature than the flame. One way to deal with this is to get a quartz tube to surround the ThC, the quartz tube also gets hot and radiates both to the cold surroundings and the ThC. Consequently the ThC is hotter and records a temperature nearer the flame temperature. This was first evaluated by NACA in the 40s.

A more recent version is here:
https://journals.sagepub.com/doi/full/10.1177/0734904120963812
It shows that replacing the cold body with a warmer cold body results in a hotter hot body.

10.1177_0734904120963812-fig7.jpeg

Jim Gorman
Reply to  Phil.
April 12, 2023 4:26 am

I am sorry, but this proves nothing. Beyond the fact that experiment changes the system overall, nothing outrageous is happening here. I know it is hard to understand, but surrounding a measuring device with a chamber that itself heats, is changing the surface area of what is the thermocouple is viewing. Of course, an increase in surface area radiating toward the measuring device will raise its temperature.

Think carefully about what the quartz tube becomes as it gets hot. Is the thermocouple making it get hot or the flame. If the flame is warming it, you end up with two sources and of course the temperature of the thermocouple will rise.

This isn’t rocket science. Look at entropy to see what the results of your example leads to.

Phil.
Reply to  Jim Gorman
April 12, 2023 7:22 am

I didn’t claim anything outrageous, just that by replacing the cold body which the ThC is in equilibrium with a warmer one raises the temperature of the ThC. As a result of the ‘cold’ body radiating more heat towards the hot one. The quartz tube is being heating to a temperature closer to the ThC than the background it replaced so the equilibrium temperature is closer to the flame temperature.

h(Tg-Tt)= 𝜀𝜎(Tt^4-T∞^4)

Is the energy balance of the ThC in equilibrium with the surroundings at T∞, let’s say 300K, replace it with Tq at 1500K and Tt is much closer to Tg.

Jim Gorman
Reply to  Phil.
April 12, 2023 8:05 pm

All I am saying it is not that a cold body is warming a warm body which is what it sounds like you are claiming. The issue is that new energy is being entered into the system by modifying the original system. Of course temperatures will change.

Phil.
Reply to  Jim Gorman
April 13, 2023 9:49 am

No. I suggest you look at the equation again.
It’s the heat balance equation for the thermocouple the ThC is heated convectively/conductively by the flame but loses heat to the surroundings by radiation and therefore reads too low a temperature for the flame. The radiation heat transfer is shown by the  𝜀𝜎(Tt^4-T∞^4) term, the heat transferred from the ‘cold body’ is given by T∞^4. If you increase the temperature of the ‘cold’ body the temperature of the ThC(Tt) will increase and be closer to the flame temperature (Tg). It’s elementary radiational heat transfer.

MarkW
Reply to  mkelly
April 10, 2023 11:42 am

A body heated by an electric current and surrounded by a gas.
If the level of current is constant and the temperature of the gas is increased, the temperature of the body will increase, regardless of whether the gas is warmer or colder than the body.

Jim Gorman
Reply to  MarkW
April 12, 2023 4:29 am

Of course it will if the gas is warmed by an external force. Heat is being added into the system from outside the system. That will cause the temperatures in the system to increase. This is easily explained by using equations for entropy.

bdgwx
Reply to  mkelly
April 10, 2023 2:10 pm

Consider 2 flat plate blackbodies A and B. Body A is at 244 K emitting 200 W/m2 (from both sides) with a constant 400 W/m2 of input. It is in steady-state since ΔF = Fin – Fout = (400 W/m2) – (200 W/m2 + 200 W/m2) = 0 W/m2. Then body B at 205 K emitting 100 W/m2 (on both sides) is introduced to the right of body A. Body A’s energy flux is now imbalanced such that ΔF = Fin – Fout = (400 W/m2 + 100 W/m2) – (200 W/m2 + 200 W/m2) = +100 W/m2. Body A begins to warm. The new steady-state forms when body A achieves 262 K emitting 267 W/m2 (from both sides) and body B achieves 221 K emitting at 134 W/m2 (from both sides). ΔFa = (400 W/m2 + 134 W/m2) – (267 W/m2 + 267 W/m2) = 0 W/m2 and ΔFb = (267 W/m2) – (134 W/m2 + 134 W/m2) = 0 W/m2 in this new configuration.

This is solved via the 1LOT and SBLAW in this manner.

(1) 400 W/m2 = σTa^4 + σTb^4

and

(2) σTa^4 = 2σTb^4

We then substitute (2) back into (1) to get the following.

(3) 400 W/m2 = σTa^4 + 0.5σTa^4

Solving for Ta we get the following.

(4) Ta = (2/3 * 400 W/m2 / σ) ^ 0.25 = 262 K.

In this example A is our warm body and B is cool body. Notice that the placement of a cool body (C) causes the warm body (B) to warm further. This is consistent with the 2LOT since the A/B system is not isolated or viewed via the input/A/B/space isolated system the heat flow is still input (hot) ==> A (warm) ==> B (cool) ==> space (cold).

http://rabett.blogspot.com/2017/10/an-evergreen-of-denial-is-that-colder.html

Last edited 1 month ago by bdgwx
Jim Gorman
Reply to  bdgwx
April 10, 2023 4:42 pm

You’ve been reading Eli Rabbit again haven’t you.

Consider 2 flat plate blackbodies A and B. Body A is at 244 K emitting 200 W/m2 (from both sides) with a constant 400 W/m2 of input. 

Let’s examine what you have just done.

“… a constant 400 W/m2 of input.”

A plate with two sides, so the energy flux gets divided by two.

400 / 2 = 200 out each side.

What will happen with a volume like a cube or eight sides?

Do you divide 400 by eight to get the energy flux flowing from each surface of an 8 sided volume? How about a sphere?

Some text from Max Planck’s Theory of Heat radiation.

It is true that for the sake of brevity we frequently speak of the surface of a body as radiating heat to the surroundings, but this form of expression does not imply that the surface actually emits heat rays. Strictly speaking, the surface of a body never emits rays, but rather it allows part of the rays coming from the interior to pass through.

We shall now consider the interior of an emitting substance assumed to be physically homogeneous, and in it we shall select any volume element dτ of not too small size.

Nevertheless, we shall as a rule be able to treat the phenomenon of emission as if all points of the volume-element dτ took part in the emission in a uniform manner, thereby greatly simplifying our calculation. Every point of dτ will then be the vertex of a pencil of rays diverging in all directions.

We shall next assume our substance to be isotropic. Hence the radiation of the volume-element dτ is emitted uniformly in all directions of space.

The total emission of the volume-element dτ may be obtained from this by integrating over all directions and all frequencies.

Now the condition of thermodynamic equilibrium requires that the temperature shall be everywhere the same and shall not vary in time. Therefore in any given arbitrary time just as much radiant heat must be absorbed as is emitted in each volume-element of the medium. For the heat of the body depends only on the heat radiation, since, on account of the uniformity in temperature, no conduction of heat takes place.

Max Planck. The Theory of Heat Radiation by Max Planck (pp. 28-29). Prabhat Prakashan. Kindle Edition. 

Read these carefully and see if your assumption of dividing the flux by two is correct. Remember, that is what Planck dealt with, e.g,, radiation. Consider why Planck made it plain that the SAME heat is radiatied in all direction.

Here is where you go off the track.

400W/m² => 290K

You state they are at steady-state which much be equilibrium as you show “0” from the S-B equation.

What is the conclusion from your analysis? Two objects can never obtain thermodynamic equilibrium as even defined by Planck. “… the condition of thermodynamic equilibrium requires that the temperature shall be everywhere the same and shall not vary in time”

Think hard about what you are proposing. If the earth’s surface receives 160 W/m² it can only heat to a temperature that radiates half that to the atmosphere. The other half is radiated down into the soil and is scattered and absorbed. Is that your position?

Finally, take a look at the image of an iron bar heated to a uniform temperature. Does it look like it only radiates 1/4th the flux from each side?

white hot steel.jpg
Phil.
Reply to  Jim Gorman
April 11, 2023 8:37 am

““Consider 2 flat plate blackbodies A and B. Body A is at 244 K emitting 200 W/m2 (from both sides) with a constant 400 W/m2 of input. ”
Let’s examine what you have just done. 
“… a constant 400 W/m2 of input.”
A plate with two sides, so the energy flux gets divided by two.
400 / 2 = 200 out each side.
What will happen with a volume like a cube or eight sides?
Do you divide 400 by eight to get the energy flux flowing from each surface of an 8 sided volume?”

To be in steady state it has to radiate the same amount of energy as is being input, therefore each side will radiate half of the input energy if the plate is at a uniform temperature. Similarly a cube would radiate 1/8th of the input energy assuming uniform temperature.

Jim Gorman
Reply to  Phil.
April 11, 2023 9:20 am

When you point an infrared thermometer at an object do you tell how many surfaces are on that object so it can calculate the total temperature?

Look at that bar. If I put a thermocouple on each surface would they all show 1/4th of the temperature or would they show a uniform temperature.

The usage of adding fluxes from all surfaces to balance the S-B equation is faulty. There is a viewing area component in using the S-B equation to determine the flux exchange between two bodies. Radiation not viewed by both bodies is not used in calculating the exchanged flux.

The viewing area is usually assumed to be 1:1, so it is ignored. This example ignores this principle.

From wikipedia:

In radiative heat transfer, a view factor, F(A -> B) (sic), is the proportion of the radiation which leaves surface A that strikes surface B.

I can also give an internet textbook if you need it.

By ignoring this principle you end up in the situation where thermal equilibrium can never be reached thru radiation. You also have the situation where calculating the Earth’s radiation budget would result result in either the wrong temperature or the wrong amount of radiation.

From Planck:

“””””We shall now consider the interior of an emitting substance assumed to be physically homogeneous, and in it we shall select any volume element dτ of not too small size.”””””

“””””We shall next assume our substance to be isotropic. Hence the radiation of the volume-element dτ is emitted uniformly in all directions of space.”””””

Planck didn’t make these on a whim! He had good reason to do so.

Last edited 1 month ago by Jim Gorman
Phil.
Reply to  Jim Gorman
April 11, 2023 10:34 am

Look at that bar. If I put a thermocouple on each surface would they all show 1/4th of the temperature or would they show a uniform temperature”.

They would all show the same temperature and each face would radiate one quarter of the total energy radiated.

Jim Gorman
Reply to  Phil.
April 11, 2023 4:06 pm

So S-B doesn’t work any longer? 1/4th the flux will give a much lower temperature than the bar is at.

So Planck was wrong about a volume radiating the same power in all directions?

You didn’t answer about how an infrared thermometer knows that only 1/4 of the power is being radiated and automatically multiplies by 4 to get the actual temperature.

The last i knew infrared thermometers and other direct reading thermometers were fairly close. You are trying to say that is no longer the case!

Phil.
Reply to  Jim Gorman
April 11, 2023 5:02 pm

My understanding is that the rod you were describing had a square cross-section and therefore four sides of equal area at equal temperature, consequently each of these faces would radiate an equal energy, one quarter of the total.

Jim Gorman
Reply to  Phil.
April 12, 2023 11:26 am

You didn’t answer my question about how an infrared thermometer can give a correct reading for temperature if it only sees 1/4 the flux! Do you enter the number of sides into the thermometer.

Jim Gorman
Reply to  Phil.
April 11, 2023 9:23 am

BTW, a cube has six sides.

Phil.
Reply to  Jim Gorman
April 11, 2023 10:31 am

Yes I was misled by you referring to a cube as an eight sided body.

slindsayyulegmailcom
Reply to  bdgwx
April 10, 2023 5:39 pm

There is no plate B blackbodies in the earth system.
You have 1360w-m² (input) – 900w-m² out to space w/ surface at 460w-m².
460w-m² – (230w-m² + 230w-m²) = 0w-m².
Out of the 900w-m² out to space 90w-m² heats stratosphere leaving
1270w-m² to heat surface. 1270w-m² – 810w-m²(less than 900w-m²) out to space gives 460w-m² at surface.
When the warm body radiations cools and is equilibrium with the cool body warmed by the sun. The cool body does not make the warmer body warmer. You don’t add them together.

bdgwx
Reply to  slindsayyulegmailcom
April 10, 2023 6:18 pm

The two plate system above is a thought experiment to help people understand the concepts of the 0LOT, 1LOT, 2LOT, SBLAW, radiant heat transfer, and why the introduction of another body acting as a thermal barrier in the presence of an energy input into the system necessarily causes both bodies to warm. mkelley asked how a warm body could warm further with equations. That’s what I gave the commenter.

Jim Gorman
Reply to  bdgwx
April 11, 2023 3:44 am

But it can’t. You forgot that the plate experiment also has a (source + back flux) segment. It originates at the second plate and sends flux back toward the source.

The question is, where does the “back flux”, which is brand new energy, originate? You have a system that creates its own energy. It works such that the source ends up putting out more energy, which results in more “back flux”, which results in more energy radiated by the source, and on and on and on. Where does all this new energy come from?

Jim Gorman
Reply to  Jim Gorman
April 11, 2023 9:33 am

By the down vote I see someone didn’t believe me.

The person doing that needs to show how the sun raises the temperature of the earth to the S-B temperature instead of just to a temperature where half of the flux is radiated.

bdgwx
Reply to  Jim Gorman
April 11, 2023 10:20 am

JG said: “What will happen with a volume like a cube or eight sides?”

The thought experiment is of flat plates. There is no mention of a cube or any other shape.

JG said: “Here is where you go off the track.

400W/m² => 290K”

The thought experiment is of a source delivering a constant 400 W/m2 on the left side of the A/B system. The temperature of the source is irrelevant.

JG said: “You state they are at steady-state which much be equilibrium as you show “0” from the S-B equation.”

I chose the phrasing “steady-state” very carefully. It was done intentionally to distinguish it from the concept of equilibrium described by the 0LOT. In this context steady-state means there are no changes occurring within the system such that ΔFa = 0, ΔFb = 0, ΔTa = 0, and ΔTb = 0. The word “equilibrium” is already reserved for the case Ta = Tb by the 0LOT.

JG said: “What is the conclusion from your analysis?”

To satisfy the 1LOT, 2LOT, and SBLAW body A must change such that the temperature change is ΔTa = (262 K – 244 K) = 18 K and the radiant exitance change is ΔI = (267 W/m2 – 200 W/m2) = 67 W/m2.

JG said: “ If the earth’s surface…”

Irrelevant. The thought experiment is of two plates.

JG said: ” take a look at the image of an iron bar “

Irrelevant. The thought experiment is of two plates.

JG said: “The question is, where does the “back flux”, which is brand new energy, originate? “

The “back flux” from plate B onto A is the result of B having a non-zero temperature.

JG said: “You have a system that creates its own energy.”

The A/B system does not create energy. It does allow 400 W/m2 to enter on the left side though.

JG said: “It works such that the source ends up putting out more energy, which results in more “back flux”, which results in more energy radiated by the source, and on and on and on.”

In the thought experiment the source delivers a constant 400 W/m2 into the left side of the A/B system.

JG said: “Where does all this new energy come from?”

In the thought experiment the only source of energy is the 400 W/m2 entering the left side of the A/B system. No other sources of energy exist and the 400 W/m2 remains constant.

Jim Gorman
Reply to  bdgwx
April 11, 2023 5:55 pm

The thought experiment is of flat plates. There is no mention of a cube or any other shape.”

Flat plates is not a valid reason for anything. In fact, the less material between the surfaces, the whole plate will then be homogeneous and at a constant temperature throughout.

The thought experiment is of a source delivering a constant 400 W/m2 on the left side of the A/B system. The temperature of the source is irrelevant.”

So S-B is has no meaning? The last I knew, a volume radiating 400 W/m² with the same surface area as P1 did so following the S-B equation. Basically,

⁴√(400/5.67×10⁻⁸) = 290K

The temperature IS RELEVANT as can be. S-B equation for 2 bodies, one at 290K and one at 244K is:

I = σ(T1⁴ – T2⁴)
I = 5.67X10⁻⁸ (290⁴ – 244⁴) = 200 W/m²

The 2nd Law of Thermodynamics says that HEAT will spontaneously flow from the source to P1. You might investigate the subject of entropy to find out why.

I chose the phrasing “steady-state” very carefully. It was done intentionally to distinguish it from the concept of equilibrium described by the 0LOT.”

In this context steady-state means there are no changes occurring within the system such that ΔFa = 0, ΔFb = 0, ΔTa = 0, and ΔTb = 0. The word “equilibrium” is already reserved for the case Ta = Tb by the 0LOT.

As much as you want your definition of steady state to be true, it is not. Ta = Tb simply can not occur with different fluxes being radiated between two bodies. You obviously don’t want to call the source a body, but that is an error. There must be a body radiating the flux.

Here are good explanations of the 0th and 2nd LOT. Steady state will not exist with different temperatures in two adjacent bodies exchanging radiation. The 2nd Law insures that spontaneous heat flow will occur until entropy has reached its limit.

12.7: The Zeroth Law of Thermodynamics – Physics LibreTexts

What the Zeroth Law of Thermodynamics means is that temperature is something worth measuring, because it indicates whether heat will move between objects. This will be true regardless of how the objects interact. Even if two objects don’t touch, heat may still flow between them, such as by radiation (as from a heat lamp). However, according to the Zeroth Law of Thermodynamics, if the systems are in thermal equilibrium, no heat flow will take place.

lec1.dvi (mit.edu)

The zeroth law of thermodynamics describes the transitive nature of thermal equilibrium. It states:

• If two systems, A and B, are separately in equilibrium with a third system C, then they are also in equilibrium with one another.

Despite its apparent simplicity, the zeroth law has the consequence of implying the existence of an important state function, the empirical temperature Θ, such that systems in equilibrium are at the same temperature.

12.3 Second Law of Thermodynamics: Entropy – Physics | OpenStax

The second law of thermodynamics states that the total entropy of a system either increases or remains constant in any spontaneous process; it never decreases. An important implication of this law is that heat transfers energy spontaneously from higher- to lower-temperature objects, but never spontaneously in the reverse direction. This is because entropy increases for heat transfer of energy from hot to cold (Figure 12.9). Because the change in entropy is Q/T, there is a larger change in ΔS at lower temperatures (smaller T). The decrease in entropy of the hot (larger T) object is therefore less than the increase in entropy of the cold (smaller T) object, producing an overall increase in entropy for the system.(bold by me.)

JG said: “ If the earth’s surface…”

Irrelevant. The thought experiment is of two plates.

It may be irrelevant to you but not in thermodynamics. You are attempting to use laws and equations that don’t distinguish between objects in that fashion. You obviously have not read Planck’s Theory of Heat Radiation or you would know these things.

Do you use the S-B equation to determine the amount of radiation the earth emits? If so, why is it not divided by 2 as you have done in your plate example.

The A/B system does not create energy. It does allow 400 W/m2 to enter on the left side though.

Look at the image I have attached.

Here is the math. From the first image.

a = 2b

From the image i have attached.

b’ > b so, a > 2b
b’ = 2c
so a = b’ + c
and a = 2c + c and, a = 3c
therefore
2c > b and 3c > b

Where does the extra energy come from? Why is the source not radiating 400 + 3c worth of energy? Your assumption is that the “back radiation” is absorbed by the source.

If you look closely at Rabett’s explanation he shows b’ going right by the source. That violates lots of laws of energy and thermodynamics. A body can’t just decide not to absorb energy if it “sees” the body from which it originates.

green plate 2.png
bdgwx
Reply to  Jim Gorman
April 11, 2023 6:58 pm

JG said: “Flat plates is not a valid reason for anything.”

That is the thought experiment. And it is as simple as it gets. If you can’t understand an idealized thought experiment with 2 bodies you won’t be able to understand vastly more complex scenarios. You won’t even be able to understand a 3 body problem.

JG said: “So S-B is has no meaning? The last I knew, a volume radiating 400 W/m² with the same surface area as P1 did so following the S-B equation. Basically,”

No where in the thought experiment is it mentioned that the source is radiating at 400 W/m2. What is mentioned is that the A/B system is receiving 400 W/m2.

And the SBLAW relates a body’s temperature to its radiant exitance. It does not relate a body’s temperature to the flux it receives from another body. You cannot use the SBLAW to determine the temperature of the emitting body by using the ingress flux from the receiving body (at least not without knowing the view factor anyway).

JG said: “The temperature IS RELEVANT”

First…the thought experiment does not provide enough information to determine the temperature of the source.

Second…its not even relevant. The thought experiment is not focused on the energy source. If you think the energy source is of concern then you’ve completely missed the point and/or are wanting to change the thought experiment to a different scenario.

JG said: “The 2nd Law of Thermodynamics says that HEAT will spontaneously flow from the source to P1. You might investigate the subject of entropy to find out why.”

This has to be a joke. You’re lecturing me about the 2LOT when you don’t even agree with it? Seriously?

JG said: “Here are good explanations of the 0th and 2nd LOT. Steady state will not exist with different temperatures in two adjacent bodies exchanging radiation.”

This has to be a joke. Neither the 0LOT nor the 2LOT say that bodies have to be in equilibrium for them to be in a state in which their temperatures are not changing. Your everyday experiences should be enough to convince you of this.

JG said: “Why is the source not radiating 400 + 3c worth of energy?”

Because the thought experiment is for a scenario in which there is no change (or at least insignificant) to the energy source.

JG said: “Your assumption is that the “back radiation” is absorbed by the source.”

Not necessarily. The thought experiment does not say what the properties of the source are. Remember, the source is not the focus of the thought experiment.

But if you have to assume the back radiation from plate A is absorbed by the source then assume the source has a radiant exitance of 18.6e6 W/m2 and a temperature of 4256 K. At a distance of 150e9 m plate A will receive 400 W/m2. And when plate A is at 262 K it will have a radiant exitance of 267 W/m2. The source will be absorbing 0.006 W/m2 from plate A (ya know…because it is really far away). That will increase the temperature of the source by an imperceptible 0.0000003 K. Now, we can solve this 3-body problem exactly using the same procedure as the 2-body problem, but it’s only going to amount to an increase in plate A’s temperature by less than a millionth of a degree K.

My recommendation, however, is that you understand the simpler 2-body problem first. If you can convince me that you understand it I’d be more than happy to discuss the 3-body problem. The 3 body problem is interesting. I’ve discussed it in other forums and have solved the equations so I am prepared to discuss it if necessary.

Jim Gorman
Reply to  bdgwx
April 12, 2023 5:42 pm

No where in the thought experiment is it mentioned that the source is radiating at 400 W/m2. What is mentioned is that the A/B system is receiving 400 W/m2. 

You have an odd interpretation of what a source and sink are in a thermodynamic system. The following site has a good description.

\documentstyle[12pt]{article} (saylor.org)

It says:

Heat (thermal) reservoir

A heat reservoir is a sufficiently large system in stable equilibrium to which and from which finite amounts of heat can be transferred without any change in temperature.

A high temperature heat reservoir from which heat is transferred is sometimes called a heat source. A low temperature heat reservoir to which heat is transferred is sometimes called a heat sink.

If the system is receiving a constant 400 W/m² it is doing so from a “source”. In order to have a constant flux, S-B tells us that the temperature must constant also. That fits the definition above exactly for a heat reservoir which is also called a heat source!

But if you have to assume the back radiation from plate A is absorbed by the source then assume the source has a radiant exitance of 18.6e6 W/m2 and a temperature of 4256 K.

So the example was ill posed. As stated one could assume the source was 1 meter away and was 1 m² and viewed a 1 m² plate exactly. You are claiming an ill posed example.

The source will be absorbing 0.006 W/m2 from plate A (ya know…because it is really far away).

You cannot use the SBLAW to determine the temperature of the emitting body by using the ingress flux from the receiving body (at least not without knowing the view factor anyway).

You are again saying the example was ill posed to begin with. No kidding!

My recommendation, however, is that you understand the simpler 2-body problem first. 

Your recommendation carries no weight because all you are doing is dancing around, changing the example to fit what you want.

You have no idea about thermodynamics. You have not given one reference from a reputable source and just expect people to accept your assertions as genuine fact. Guess again.

You want to redeem your reputation? Give a source that shows radiation from a volume is divided among its sides rather than radiating the S-B flux at a given temperature in all directions.

bdgwx
Reply to  Jim Gorman
April 12, 2023 7:41 pm

If you don’t think plate A warms to 262 K then post your own solution and let us review it.

Jim Gorman
Reply to  bdgwx
April 13, 2023 6:42 am

Fine. If the “source” enters 400 W/m^2 constantly to the system and Pa has reached thermal equilibrium with the source, then the assumptions become:

• The source is at a distance where no reduction in intensity are required
• Pa and Pb are close enough to not require reduction in intensity
• All bodies are homogeneous.
• All bodies at uniform temperature throughout (conductivity is not an issue)
• All bodies are isotropic, i.e., radiate equally in all directions.
• Areas of bodies are equal, assume 1 m^2
• View factors are not an issue between the source and Pa.
• Pb has no view of the source, only Pa
• Pa absorbs/emits all 400 W/m^2 (black body)
• Pb absorbs/emits all radiation (black body)
• Sufficient time has elapsed to allow equilibrium (no gradients)

S-B equations becomes:

Ts = (400 / 5.67E-8)^1/4 = 290K

I = σ(Ts – Ta) = 0 => Ts = Ta = 290K (no net flux, equilibrium)

Pa radiates 400 W/m^2 IN ALL DIRECTIONS.

Pb receives 400 W/m^2 from Pa, therefore Tb = Ta = 290K

You will notice these are basically ideal conditions. Real world conditions complicate things exponentially. But, if you want to use simple algebra, you must use ideal conditions. Things like absorptivity, emissivity, conductivity, view factors, steradians, gradients based on time, heterogeneous material, etc. make for complicated analysis.

Radiation in an isotropic fashion, i.e., spherical is necessary for this analysis. If you find a reference describing a homogeneous body that doesn’t radiate isotropically please post it along with the derivations.

Jim Gorman
Reply to  Jim Gorman
April 13, 2023 6:38 am

Messed up

Last edited 1 month ago by Jim Gorman
bdgwx
Reply to  Jim Gorman
April 13, 2023 8:05 am

Yeah, unfortunately that doesn’t work. To help facilitate the visualization I often adopt this notation for use in forums.

(Fout, Fin) <==> [body: temperature] <==> (Fin, Fout)

So using that notation we see that when Ta = Tb = 290 K we have the following configuration.

(400, 400) <==> [A: 290] <==> (400, 400) <==> [B: 290] <==> (0, 400)

A is balanced since Pin = (400 * 0.5A) + (400 * 0.5A) and Pout = (400 * 0.5A) + (400 * 0.5A). Therefore Pin – Pout = 0 W.

B is not balanced since Pin = (400 * 0.5A) + (0 * 0.5A) and Pout = (400 * 0.5A) + (400 * 0.5A). Therefore Pin – Pout = -200 W.

The A/B system is in equilibrium at least initially in this configuration since Ta = Tb, but it is not in a steady-state since the energy flows in/out of B are unbalanced by -200 W. So the equilibrium is going to break. B is going to cool which means A must cool as well. The question is…how much will both cool?

And again, so there is no confusion I’m using the 0LOT as the definition for “equilibrium” and 1LOT as the definition for “steady-state” and “balanced”.

Last edited 1 month ago by bdgwx
bdgwx
Reply to  petercampion2724
April 10, 2023 6:26 am

petercampion2724 said: “The Second Law of Thermodynamics proves heat doesn’t flow from cold zones to hotter zones.”

Next to the -80 C misunderstanding this is the next most misunderstood item in your item. The 2LOT does NOT say heat cannot flow from cold to hot. What it says is that heat cannot flow from cold to hot when the system is isolated. Phrases you’ll often see for the isolation clause include “when the system is evolving by its own means” or “without some other change” or “when the system is unaided by external agency” or “spontaneously”, etc. They all mean the same thing. Heat can flow from hot to cold. It happens all of the time. It just doesn’t happen spontaneously.

Furthermore, the 2LOT does not mean that energy cannot flow from cold to hot isolated or otherwise. Remember, energy is the ability to do work. Heat is the net flow of energy. The keyword here is net. Energy is moving from cold bodies to hot bodies all of the time. It’s just that more than energy is moving from the hot body to the cold body so the heat flow is from hot to cold.

But that is moot because the flow of heat in this context is Sun (hot) ==> Surface (warm) ==> Atmosphere (cool) ==> Space (cold). The introduction of a thermal barrier (like CO2) does not change the heat flow configuration. It only changes the magnitude of the heat flow.

petercampion2724 said: ” CO2 cannot “trap heat.”

In this context “trap” refers to the 1LOT ΔE = Ein – Eout where Ein > Eout and ΔE > 0. It is the familiar law of conservation of energy. When you allow something into a system, but don’t allow it to leave it is considered “trapped”. It is no different than say when someone is allowed into a mine, but due to a cave-in or whatever the exits are blocked preventing the miners from leaving. At that they are “trapped”. CO2 allows energy from the Sun to enter the climate system. But partially blocks it from leaving the climate system. Therefore CO2 “traps” energy.

petercampion 2724 said: “To control Earth’s climate, you must control the sun.”

Remember the 1LOT ΔE = Ein – Eout. To cause a change in E you can either change Ein, change Eout, or both. Sure, Earth’s climate can be controlled via changes in the Sun. But that’s not the only way. Changes in aerosols, clouds by day, surface albedo, etc. can also change the Ein side. Changes in GHGs, clouds by night, etc. can change the Eout side. There are a lot of ways to the ΔE side of the equation. The Sun is but only one among many of the ways.

Last edited 1 month ago by bdgwx
Mr.
Reply to  bdgwx
April 10, 2023 8:38 am

End efffects are ultimately governed by the dominant influences (as you mention – clouds, albedo, aerosols, etc).

Why even consider a puny bit-player like CO2 in this naturally-occurring, eternally continuous climate extravaganza?

bdgwx
Reply to  Mr.
April 10, 2023 9:56 am

End effects are ultimately governed by the net effect of all influences.

Mr.
Reply to  bdgwx
April 10, 2023 1:22 pm

And as I said, CO2 is a puny bit-player.

Whose capacity for controlling how climates work is totally overwhelmed by the dominant influences.

What? You think CO2 is a dominant influence on how climates work?

bdgwx
Reply to  Mr.
April 10, 2023 1:34 pm

Mr said: “ You think CO2 is a dominant influence on how climates work?”

Sometimes. The present era and the PETM would be examples.

slindsayyulegmailcom
Reply to  bdgwx
April 10, 2023 6:17 pm

CO2 has no effect on surface only Mesosphere.

bdgwx
Reply to  slindsayyulegmailcom
April 10, 2023 7:13 pm

CO2 impedes the transmission of energy via IR radiation regardless of whether it is emitted by the surface, troposphere, stratosphere, or mesosphere.

slindsayyulegmailcom
Reply to  bdgwx
April 10, 2023 6:15 pm

“CO2 allows energy from the Sun to enter the climate system. But partially blocks it from leaving the climate system”. This is incorrect as CO2 intercepts energy from the sun (4.3µm) coming in and not partially blocks it from leaving (as sun heated earth is nowhere near 15µm). The reduction of heat at other layers means less heat leaving.

Remember surface isn’t -41°C 164w-m² to 175w-m² -37°C (it is usually 328w-m² 2.8°C (Jan)) to 350w-m² 7.5°C).

Not 12°C to 15.8°C (partial earth coverage).
198
+175(absorbed surface heat)
+87 (latent heat)
equator (460)
-87
-175
198
= 198 (compression heating w/ sunlight)  

bdgwx
Reply to  slindsayyulegmailcom
April 10, 2023 7:10 pm

I get a radiance of 0.3 W/m2 in the incoming 4.0-4.6 um band from 5780 K source at 1 AU away.

I get a radiance of 9.4 W/m2 in the outgoing 14.7-15.3 um band from a 276 K (2.8 C) source.

I get a radiance of 10.1 W/m2 in the outgoing 14.7-15.3 um band from a 281 K (7.5 C) source.

I get a radiance of 11.0 W/m2 in the outgoing 14.7-15.3 um band from a 288 K (15 C) source.

Even at the lower 275 K temperature CO2 blocks 31x more outgoing 15 um radiation than incoming 4.3 um radiation.

Jim Gorman
Reply to  bdgwx
April 11, 2023 10:48 am

“””””The 2LOT does NOT say heat cannot flow from cold to
hot. “””””

The Second Law of Thermodynamics DOES say that HEAT can only flow from hot to cold.

From: https://www.thermal-engineering.org/what-is-second-law-of-thermodynamics-definition/

“””””Direction of Thermodynamic Processes
Many thermodynamic processes proceed naturally in one direction but not the opposite. For example, when a temperature difference does exist heat flows spontaneously from the warmer system to the colder system, never the reverse. In fact, such heat flow (from a colder body to a warmer system) would not violate the first law of thermodynamics, i.e. energy would be conserved. But it doesn’t happen in nature.”””””

From:

https://www.sfu.ca/~mbahrami/ENSC%20388/Notes/Second%20Law%20of%20Thermodynamics.pdf

“””””Heat transfer from a hot container to the cold surroundings is possible; however, the reveres process (although satisfying the first law) is impossible.”””””

You seem to not know the laws very well. The 1st law discusses energy. Energy can fow both ways between two bodies. The first law is only concerned with the preservation of energy. Not what that energy does.

The second law has more to do with entropy and how that is handled in a system. Remember, HEAT IS NOT ENERGY. Heat only flows in one direction in nature, hot to cold. Do not equate energy and heat. The first law cares not which way energy flows, there are no restrictions. The second law DOES care which way heat flows.

bdgwx
Reply to  Jim Gorman
April 11, 2023 11:17 am

JG said: “You seem to not know the laws very well. The 1st law discusses energy.”

The irony is palpable because it was only a few months ago that Tim challenged the 1LOT by claiming that it only holds after a certain amount of time had elasped. He also challenged the SBLAW by saying it only works on bodies that are in equilibrium with their surroundings.

JG said: “The Second Law of Thermodynamics DOES say that HEAT can only flow from hot to cold.”

That is patently false. The 2LOT does NOT say that heat can only flow from hot to cold. What the 2LOT says is that heat flows from hot to cold in an isolated system. Your own references make this abundantly clear. And if it wasn’t clear from your own references then surely your everyday experience with refrigerators would be convincing that heat CAN flow from hot to cold.

This whole discussion is beyond absurd. First it’s the SBLAW that’s wrong. Then it’s the 1LOT that’s wrong. Now it’s the 2LOT that’s wrong. Your and Tim’s challenges of the thermodynamic laws almost defies credulity.

Last edited 1 month ago by bdgwx
bdgwx
Reply to  bdgwx
April 11, 2023 2:15 pm

bdgwx said: “And if it wasn’t clear from your own references then surely your everyday experience with refrigerators would be convincing that heat CAN flow from hot to cold.”

Typo…that should have read heat CAN flow from cold to hot.

Jim Gorman
Reply to  bdgwx
April 12, 2023 12:56 pm

“””””Your own references make this abundantly clear. “””””

From: https://www.sfu.ca/~mbahrami/ENSC%20388/Notes/Second%20Law%20of%20Thermodynamics.pdf

“””””The Second Law of Thermodynamics”””””

“””””Heat transfer from a hot container to the cold surroundings is possible; however, the reverse process (although satisfying the first law) is impossible..”””””

I have attached an image that has this statement!

You didn’t bother studying either document did you? Just stopped when you found something that might fly!

PSX_20230412_145346.jpg
Jim Gorman
Reply to  bdgwx
April 12, 2023 1:20 pm

Heat can flow from cold to hot ONLY IF WORK IS PERFORMED. Do you understand what that compressor in your fridge or air conditioner does? You can only reverse an irreversible process by doing work. In effect, adding energy to the system.

Tim Gorman
Reply to  Jim Gorman
April 12, 2023 1:37 pm

The statisticians on here just don’t seem to have a good grasp of how things work in the real world. A fridge just transfers cold to hot and you can unplug it from the wall and it will keep right on doing it!

Jim Gorman
Reply to  bdgwx
April 12, 2023 12:33 pm

“”””The 2LOT does NOT say that heat can only flow from hot to cold. What the 2LOT says is that heat flows from hot to cold in an isolated system. “””””

You didn’t read beyond the first paragraph did you.

“””””The entropy of any isolated system never decreases. In a natural thermodynamic process, the sum of the entropies of the interacting thermodynamic systems increases.”””””

I’ll paste it again!

“””””In a natural thermodynamic process, the sum of the entropies of the interacting thermodynamic systems increases.”””””

Entropy INCREASES. What does this mean? It means HEAT IS TRANSFERED until maximum entropy is reached. Y ou can introduce new energy or subtract energy but entropy (2nd law) dictates that heat will be rearranged to achieve maximum entropy with what is left.

You have never studied thermodynamics have you? I have 9 hours of upper level dude. As an EE I was trained to help design and to supervise the operation of a supercritical steam generating plant. I am not someone who has just read a book.

bdgwx
Reply to  Jim Gorman
April 13, 2023 5:20 am

So are you amending your original statement: “The Second Law of Thermodynamics DOES say that HEAT can only flow from hot to cold.” or not? The gist I’m getting from your posts suggests you still think your original statement is correct.

JCM
Reply to  Jim Gorman
April 11, 2023 12:26 pm

Entropy is a most interesting and useful way to think about the climate system, as the sun’s beam is converted through the Earth system on its way to being transmitted as relatively longer waves back to space.

The two lines of reasoning are as follows: such that (1) the rate of entropy production must be maximized, suggesting a very low climate sensitivity measured in terms of temperature. i.e. the rate of entropy production must increase with radiative forcing.

Conversely, the one in use by climatology such that (2) the rate of entropy production remains unchanged, and so the temperature rise is relatively large with an initial radiative forcing by trace gases.

The primary means of entropy production is by turbulent dissipation in the fluid system.

It is widely agreed among both streams of thought that the entropy production directly associated with the absorption of solar radiation into the climate system is irrelevant to the properties associated with turbulence.

So it is largely the optical properties to longer waves which either (1) increase the rate of entropy production by turbulent dissipation or (2) that the rate of entropy production remains unchanged.

These are important questions because in the world of GCM these are parameterized processes. However, there are some streams of theoretical discussion leaning towards the case (1) maximum rate of entropy production.

The entropy system (S) is that turbulent boundary layer between the “hot” radiating surface and the colder zone.

It is not a coincidence Clausius wrote on water phases changes quite a lot, in his efforts towards more important discussions of entropy budgets.

Conveniently, one could study the entire climate sensitivity paradigm purely by using entropy budgets, which use the same units for which we are all familiar W K-1 m-2.

System.png
Jim Gorman
Reply to  JCM
April 12, 2023 1:13 pm

You could do this but boy, it gets complicated in a hurry.

I have always wondered how a heat reservoir in the atmosphere can ever heat the surface because of the mass difference. Certainly the atmosphere doesn’t heat the surface (land/ocean) by conduction. Water does absorb a lot of near IR and I never see that treated anywhere.

I certainly don’t have the resources to begin to isolate systems and the accompanying heat flows. I do think any true study should address this.

JCM
Reply to  Jim Gorman
April 13, 2023 8:39 am

The most important bit to keep in mind is that the atmosphere is basically a fluid heated at the bottom and cooled at the “top”.

The top boundary condition is debatable but not especially relevant here. Strictly, it really must be the literal top of atmosphere at 0hpa.

When the temperature difference between the bottom and the top becomes larger than a certain critical value, the fluid is no longer stable, and turbulent motions take hold.

This has been known since Lord Rayleigh wrote on the subject in 1916 “On convection current in a horizontal layer of fluid, when the higher temperature is on the under side”.
https://www.tandfonline.com/doi/abs/10.1080/14786441608635602

It is this fluid motion itself that is transporting the heat energy.

So, radiation enthusiasts are absolutely correct, so long as the system is below the fluid system critical value of instability.

Rayleigh labelled this dimensionless threshold parameter the “Rayleigh number” (naturally).

Where radiation enthusiasts go wrong, is ignoring that the Earth system atmosphere has already reached the critical stage of instability.

Entropy budgeters would note immediately that with any supposed increase of temperature difference between the bottom and top, such as by an initial virtual LW radiative forcing, this can only increase instability.

And so, the fluid motion must increase, along with heat flux (and so-too the entropy budget).

The unnatural radiative forcing by emission of trace gas can be real in terms of virtual physical principle, but in reality it may be difficult to observe or practically non-existent in real atmosphere.

Last edited 1 month ago by JCM
JCM
Reply to  JCM
April 13, 2023 8:57 am

The moral is that greenhouse radiative forcing only “appears” simple because it ignores the real forced dynamic response. It could be argued the atmosphere has practically limitless supply of ‘freedom’ in dynamic response to maintain its so-called greenhouse effect practically unperturbed. At the very least, additional “positive feedbacks” are almost inconceivable.

Last edited 1 month ago by JCM
Tim Gorman
Reply to  JCM
April 15, 2023 4:24 am

Far too many scientists today are “blackboard scientists”. They lack any experience with the real world. It used to be that research scientists used math to explain real world observations. Today they use math to describe model outputs whose outputs are based on incomplete inputs.

It’s a sad state of affairs.

Jim Gorman
Reply to  bdgwx
April 12, 2023 10:33 am

“”””””The 2LOT does NOT say heat cannot flow from cold to hot. What it says is that heat cannot flow from cold to hot when the system is isolated.”””””

From: http://physics.bu.edu/~duffy/py105/Secondlaw.html

“””””The second law – The level of disorder in the universe is steadily increasing. Systems tend to move from ordered behavior to more random behavior.”””””

“”””””One implication of the second law is that heat flows spontaneously from a hotter region to a cooler region, but will not flow spontaneously the other way. This applies to anything that flows: it will naturally flow downhill rather than uphill.”””””

Energy may flow from cold to hot and not violate the 1st law. But HEAT will not flow from cold to hot without WORK being done. That is, new energy added to the system. ENTROPY RULES.

You would know this if you took the time to study Planck’s book, The Theory of Heat Radiation.

Planck calls it compensation. When energy is absorbed from a cold body it is simultaneously radiated away in the flux already being radiated. Thus, the hot body does not warm from an addition of heat.

“””””On the other hand for T i = 0, the increase in entropy is only equal to , i.e., the emission of a black body of temperature T without simultaneous absorption of heat radiation is irreversible without compensation, but can be reversed by a compensation of at least the stated finite amount. For example, if we let the rays emitted by the body fall back on it, say by suitable reflection, the body, while again absorbing these rays, will necessarily be at the same time emitting new rays, and this is the compensation required by the second principle.”””””

“””””Generally we may say: Emission without simultaneous absorption is irreversible, while the opposite process, absorption without emission, is impossible in nature.”””””

Read this closely, “absorption without emission, is impossible in nature”. This is entropy at work.

Steve Case
April 9, 2023 7:00 pm

I don’t sleep well knowing that climate propaganda is going to continue for a long time. Regulations claiming to be designed to reduce CO2, when the destruction of the world’s economy seems to be the goal.

What about latent heat? Doesn’t that also contribute to the transport of heat out to higher altitudes where it can radiate on out of the atmosphere?

There was this:

     CO2 emits any observed heat in less than a second it emits those
     wavelengths in all directions with half redirected back to the Earth’s
     surface. It is this redirected heat that is believed to warm the surface.

Paint me skeptical, CO2 absorbs at 15µ microns. A black body that radiates predominantly at 15µ microns would be a brick of dry ice. That’s not going to warm anything. Will it prevent cooling by canceling out upwelling 15µ micron radiation? Yes. Maybe that’s a matter of semantics but it’s the sun that does any warming not the 15µ microns of back radiation.

I liked the explanation that heat domes can be caused by an omega loop in the jet stream. I learn something new every day.

It doesnot add up
Reply to  Steve Case
April 10, 2023 5:24 am

Remove your scepticism. Take a look at plots of the Planck distribution at different temperature parameters and you will see that there is always some energy at 667cm-1, 15μ. Or simply plug in the numbers into a Planck formula. The amounts will be smaller the further the temperature parameter diverges, but they are always positive.

https://en.m.wikipedia.org/wiki/Planck's_law

mkelly
Reply to  It doesnot add up
April 10, 2023 6:45 am

But they are not “heat”. Heat does not flow from lower temperature to higher.

It doesnot add up
Reply to  mkelly
April 10, 2023 7:21 am

True. However from E=hv we can calculate the energy of absorption and emission. From the Maxwell Boltzman distribution we can calculate the proportion of molecules thermally capable of emission. We know that the direction of emission is random, but the probability of absorption declines with reducing density and altitude. In turn that means there will be a net surplus of photons heading to space. The atmosphere acts rather like a pinball machine delaying the passage of the ball by increasing the length of the route it must take, and occasionally trapping it for a while.

slindsayyulegmailcom
Reply to  It doesnot add up
April 10, 2023 6:55 pm

Proportion of molecules is 0.01% of the mass for absorption and emission at high temperatures. And 99.9% is mass x mean square root / heat capacity.

1360w-m² – (460w-m² (900w-m²) – 370w-m²)) = (900 + 370 out to space)=90w-m² greenhouse gases in stratosphere and mesosphere(slowest molecules in the atmosphere).
460=198+90+172. (172×2=344w-m² surface average).

bdgwx
Reply to  mkelly
April 10, 2023 7:56 am

Careful…heat does not flow from lower temperature to higher when the system is evolving by its own means. I’ve noticed a lot of people forget or ignore the isolation clause. We need to make sure the isolation clause is front and center when discussing the 2LOT.

Jim Gorman
Reply to  bdgwx
April 12, 2023 10:58 am

The isolation assertion is to be able to simplify the equations derived for entropy. Systems do not have to be isolated for entropy to apply, but it sure complicates the math involved to derive what happens for each body. Unisolated systems also allow energy to enter/leave the system further complicating derivations.

You should notice that the 2nd law allows the transitive property for thermodynamic equilibrium to be derived when three bodies are involved.

RickWill
April 9, 2023 8:19 pm

Radiative imbalance has no bearing on surface temperature. That attached group of charts proves that.

In the past 16 years of the availability of the CERES net radiation balance at the top of the atmosphere, there is no correlation between the net radiation at any location and the surface temperature at that location. Temperature change is independent of net radiation

The tropical oceans and tropical rainforests and their atmosphere have absorbed an enormous amount of heat, which has been predominantly transferred within Earth’s climate system poleward to be released to space.

The average global temperature change over those 16 years has been small but the high latitudes in the NH have warmed the most and the high latitudes in the SH cooled the most. Despite all the energy uptake in the tropical regions, there is very little to no temperature change in the tropics.

What is clearly evident, is that there is no correlation between radiation imbalance and surface temperature at any location on Earth.

The primary focus of any theory that aims to predict how Earth’s surface temperature will change has to be based on internal heat transfer; certainly not any local or global radiation balance.

The idea that Earth is in some near term radiative balance is clearly nonsense. There is little understanding of energy conversion processes; such as what energy is involved in terraforming and biomass semi-permanent accumulation and near permanent accumulation like the expansion of coral atolls.. We do have good indication that the living biomass is expanding due to the increase in CO2.

Net_Images.png
JCM
Reply to  RickWill
April 9, 2023 10:33 pm

energy conversion processes

everywhere, all the time.

such as in atmosphere – where increasing optical depth at the 15 micron band results in enhanced emission at the 20 micron band.

How.

This is accomplished via energy conversion.

Inhibited radiative flux is converted to turbulent flux;

Via the phase changes of water. 

Surplus 15 micron wavelines are taken up in evaporation, and converted back to radiative emission in condensation.

Now at 20 micron, at some altitude. 

Further afield into the infrared. Slightly cooler, but more transmittable.

The suspended liquid and solid particles acting as broadband radiators. A miracle.

slindsayyulegmailcom
Reply to  JCM
April 10, 2023 7:04 pm

No. Complete nonsense. Water doesn’t exist below -48°C 12.9µm so why would something non-existent evaporate. 15µm is dry ice as solid carbon dioxide from a changed state as gas (14µm).

JCM
Reply to  slindsayyulegmailcom
April 10, 2023 8:12 pm

Any hypothetical hindrance of surface radiative transmittance to space, however trivial, must result in an increase to a virtual instantaneous surface radiation budget.

The surface radiation budget has no awareness of the temperatures for which you are speaking. Under such a scenario, the surface budget only responds to the perturbation of atmospheric transmissivity to terrestrial radiation temperatures.

Any hypothetical hindrance creates an instantaneous virtual increase to the surface radiation budget.

Phil.
Reply to  slindsayyulegmailcom
April 11, 2023 7:19 am

More of your faulty Wien’s law nonsense.

Phil.
Reply to  slindsayyulegmailcom
April 11, 2023 8:21 am

Here’s the phase diagram of water, as you can see at -48ºC it’s predominantly in the solid phase with a vapor pressure of about 10 Pascals so it sublimates rather than evaporates.
water-phase-diagram.png

Bob
April 9, 2023 8:31 pm

Very nice Jim.

Kevin Kilty
April 9, 2023 8:40 pm

I always like your presentations, Jim. I wish you had platforms for their presentation that could reach a wider audience.

I could quibble about one thing in this one — when ascending air reaches the stratosphere, it hasn’t cooled because pressure is lower but because to reach this lower pressure the ascending air had to do work against the surrounding atmosphere — it expanded. The ascent being rapid (adiabatic or pseudo-adiabatic) without significant heat exchange the only source of energy for this work is the internal energy (temperature) of the ascending parcel. Hence it cools.

To say that low pressure makes for lower temperature only encourages people who insist the ideal gas law demands that lower pressure equals a lower temperature and vice versa.

Last edited 1 month ago by Kevin Kilty
AndyHce
Reply to  Kevin Kilty
April 9, 2023 10:47 pm

Air, any gas, has constant molecular motion if above absolute zero, where it is unlikely to be a gas anymore. The average kinetic energy of its gas molecules is its temperature. The kinetic energy of individual molecules can change through collisions but the total can only change if there is gain or loss energy to something outside itself (i.e. to something that is not any of the molecules of the gas at time zero). Is it not the case that each molecule will keep moving in the same direction, at the same speed, unless acted upon from without that molecule? That is generally referred to as inertia.

The molecules of some parcel of gas suddenly surrounded by vacuum will very rapidly, in human terms, move apart because there is nothing constraining them to the original volume. No energy is expended in doing so. As the parcel becomes more tenuous, its temperature drops, does it not, even though the total energy kinetic energy of its molecules remains constant?

If this is so, I fail to see any difference in a parcel of atmosphere expanding as it moves into a larger space, at higher altitude, where there is now more room for each molecule to continue in its original velocity before interacting with anything that can change that velocity. That motion will result in the molecules moving further apart, lowering any temperature measurement, without any change of total energy.

Alexy Scherbakoff
Reply to  AndyHce
April 9, 2023 11:57 pm

It’s potential energy till it hits something. Then it becomes kinetic.
Defenestration is potential energy till you hit the ground, it becomes messy kinetic after the stop.

AndyHce
Reply to  Alexy Scherbakoff
April 10, 2023 2:52 am

I don’t see this has any relevance. Do you think I’m missing something?

Alexy Scherbakoff
Reply to  AndyHce
April 10, 2023 3:11 am

You are right. I just added to it. Travelling without collision doesn’t impart energy to anything. A particle would have to hit a thermometer and transfer energy to impact (and register a temperature). Heat is generated through impact and not simply travelling at high velocity. First thing I learned in high school when I was 12.

It doesnot add up
Reply to  AndyHce
April 10, 2023 6:26 am

As a parcel of warm air rises it does work against gravity, converting kinetic energy to gravitational potential energy. That is without any interaction with its surroundings. If we use the approximation E=mgh at 10,000m and taking g=10m/sec^2 we get a reduction in kinetic energy of about 3kJ per mole, dividing by R=8.3J/mol/K or about 0.35K

It doesnot add up
Reply to  It doesnot add up
April 10, 2023 6:56 am

Strike the last calculation with R. It’s nonsense, and wrong by a factor of 1,000. But 288K air has kinetic energy of about 2.4kJ/mol. It must draw on 3kJ/mol to ascend to 10,000m. Some of that will come from the energy released by other parcels of (cool) air or precipitation moving in the other direction. Some will come from cooling as it ascends.

Tim Gorman
Reply to  It doesnot add up
April 10, 2023 7:10 am

Ummmm, Buoyancy force also figures in here. For a parcel of air the pressure it exerts against the air surrounding it and the pressure exerted by the air below it works to counteract the force of gravity. Thus less dense air (e.g. wet air) rises.

You can’t just look at one molecule at a time.

JCM
Reply to  Tim Gorman
April 10, 2023 8:12 am

Buoyancy force

and yet “wet air” is cooler, offsetting such buoyancy.

There must be another process. It is the condensation – for which the condensed matter occupies less space than in vapor phase, creating the necessary pressure vacuum. A force in its own right.

It doesnot add up
Reply to  JCM
April 10, 2023 9:16 am

Wet air is less dense. Condensation only occurs when the conditions for it are met, and certainly when it does happen it results in other changes, such as cloud formation and precipitation which change the whole dynamic.

Of course around the edges of the warm parcel there will also be continuos diffusion of molecules to and from the surrounding air which will tend to equalise the conditions, shrinking the warm parcel and cooling it as it rises. It’s just that the rate of convection is more rapid that the rate of diffusion.

JCM
Reply to  It doesnot add up
April 10, 2023 9:28 am

you seem pretty certain. But this is a climate forum. Condensation always occurs. it is the ‘force’ majeure.

It doesnot add up
Reply to  JCM
April 10, 2023 10:41 am

Condensation only occurs when the conditions are right – for the formation of fog or cloud. In this discussion about convection much of the elevation of warm parcels of air occurs before condensation takes effect. Particularly in thunder clouds it can be extremely important, I’d agree.

JCM
Reply to  It doesnot add up
April 10, 2023 10:53 am

Condensation only occurs when the conditions are right

The condensing atmosphere occupies 2/3rds of the planar earth surface.

It is not the exception. Rather, it is the prevailing condition. It is all other circumstances which represent the exception.

That condensing atmosphere represents 2/3rds of the all sky emission process.

Tim Gorman
Reply to  JCM
April 12, 2023 6:25 am

You are talking about how the heat is lost. The buoyancy force only causes the parcel of air to rise to where the heat can be lost. They aren’t the same thing. No offsetting.

JCM
Reply to  Tim Gorman
April 12, 2023 12:39 pm

your buoyancy “force” is coincident with condensation and emission. Too much “cause and effect” chit chat in these forums. it is a system.

It doesnot add up
Reply to  Tim Gorman
April 10, 2023 9:05 am

I made no specification of where the energy to raise the parcel comes from, simply pointing to some examples. If you think carefully about buoyancy you will realise that air ahead of the rising parcel ends up below it – it is displaced downwards occupying the space previously taken by the warm parcel, giving up gravitational potential energy.

Tim Gorman
Reply to  It doesnot add up
April 12, 2023 6:27 am

I think you have a pretty good handle on this. Buoyancy doesn’t cause the heat loss. It only puts the parcel of air in a position to lose heat. Not the same thing.

Tim Gorman
Reply to  AndyHce
April 10, 2023 4:31 am

Andy,

I suspect the difference of view here has to do with volume. A thermometer has a fixed volume. As you go up that thermometer is impacted by fewer and fewer molecules thus it indicates a lower and lower temperature. If the volume of the measuring device expanded along with the initial parcel of air you wouldn’t see any temperature change other than that lost from radiation and collisions with matter outside the expanding volume.

It doesnot add up
Reply to  Tim Gorman
April 10, 2023 6:32 am

A thermometer should measure the average root mean square velocity of the molecules, assuming it is shielded from heating by radiation. That does not depend on their density.

Tim Gorman
Reply to  It doesnot add up
April 12, 2023 6:33 am

Of course density matters! What do you think the “n” factor in pv = nRT is?

The thermometer measures the *sum* of the kinetic energy imparted to it from colliding molecules. It’s why a thermometer in a total vacuum would read 0K – nothing is impinging on it.

bigoilbob
Reply to  Tim Gorman
April 16, 2023 6:49 am

“Of course density matters! What do you think the “n” factor in pv = nRT is?”

Gosh Tim Gorman, hate to nitpick, per the standard whine here. But “n” is not density related, that’s for sure. It’s the number of moles in the sample. Are you mistaking it for the Z factor in the non-ideal gas law? That makes more sense…

Last edited 1 month ago by bigoilbob
Tim Gorman
Reply to  bigoilbob
April 18, 2023 5:59 am

The number of moles in a given volume IS the density!

0 moles = 0 density And it goes up from there!

bigoilbob
Reply to  Tim Gorman
April 18, 2023 6:09 am

I agree that n/V has units of molecular density. I’m missing that term in PV=nrT. Can you help me find it?

Tim Gorman
Reply to  bigoilbob
April 18, 2023 7:23 am

troll

bigoilbob
Reply to  Tim Gorman
April 18, 2023 7:34 am

Piqued? Then you might try thinking before writing, and limiting your responses to areas of expertise.

Last edited 1 month ago by bigoilbob
Kevin Kilty
Reply to  AndyHce
April 10, 2023 4:49 pm

 As the parcel becomes more tenuous, its temperature drops, does it not, even though the total energy kinetic energy of its molecules remains constant?

No. If an ideal gas expands into vacuum it does no work against anything. The average K.E. of its various molecules does not change. Thus, it does not drop in temperature though it becomes infinitely tenuous. Real gasses are possessed of a Joule-Thompson coefficient and may have a temperature that rises or falls upon a free expansion depending on temperature vis a vis the inversion temperature — this is the basis of a “throttle” for refrigeration.

Jim Gorman
Reply to  Kevin Kilty
April 10, 2023 6:02 pm

I’ve got to think about that one. Assume throwing a ball up on the moon. Gravity will exert a force against the ball, and the ball basically does work as it travels in the up direction. At some point the kinetic energy dissipates to zero in the up direction such that the ball returns.

AndyHce
Reply to  Kevin Kilty
April 9, 2023 10:50 pm

If expansion expends energy, what form does that energy take, and where does it end up?

Alexy Scherbakoff
Reply to  AndyHce
April 10, 2023 12:01 am

Dreamer. You expect physics with climate science.

mkelly
Reply to  AndyHce
April 10, 2023 6:53 am

Work. W = P dV.

Frank from NoVA
April 9, 2023 9:46 pm

Concise and effective, as usual. Thank you.

Alexy Scherbakoff
April 9, 2023 10:05 pm

I hate the online calculators for the planck curve. They give different peaks for wavelength and wavenumber.

It doesnot add up
Reply to  Alexy Scherbakoff
April 10, 2023 7:36 am

Of course they will. One is essentially the reciprocal of the other.

The Dark Lord
April 9, 2023 10:32 pm

Wow … redirects towards the surface … I.e. from cold to hot … not on this planet … use better phrases … CO2 is insulation … talk about it like that …

Last edited 1 month ago by The Dark Lord
JCM
Reply to  The Dark Lord
April 9, 2023 11:19 pm

CO2 can only narrow the window of surface transmitted flux to space around 15 microns.

This increases the virtual instantaneous surface radiation budget.

Coincidently, atmospheric instability is enhanced, along with turbulent flux.

Atmospheric emission to space increases at 20 microns.

The surface budget is restored.

Alexy Scherbakoff
Reply to  The Dark Lord
April 10, 2023 12:52 am

At what temperature do you think things stop radiating?

Peta of Newark
April 10, 2023 12:08 am

The following is a snippet from today’s (10th March 23) weather forecast for the UK:

Quote:Cold air aloft (between -20C to -30C at 500 hPa) associated with upper trough extending from the west, will overspread aloft atop of moist maritime flow which will be heated in sunny spells to yield around 300 – 600 j/kg CAPE across Ireland, Wales and large parts of England. This will support heavy showers and scattered thunderstorms to develop here from surface heating creating steep lapse rates.

(My highlight)

Do we see what it says:= The sun heats the air directly and when that air is moist, it creates (stormy) weather

No mention of trapped heat, downwellings (the exact opposite in fact) or CO2

“CAPE” Convective Available Potential Energy

The very last thing that is going on in there is trapped heat.
Yes, energy from the sun is captured/trapped (=CAPE) by moist air but it is immediately ## dumped into the stratosphere by the thunderstorms – which block out the sun while releasing avalanches of cold water and ice onto the ground below

Takeaway: The atmosphere cools the surface

## Inside a few short hours – not years, decades and centuries

JCM
Reply to  Peta of Newark
April 10, 2023 12:23 am

The emission spectra shifted slightly to the left in Wien wavenumber space. This, by turbulent conversions (weather). Budged ‘just’ that amount needed.

mkelly
Reply to  Peta of Newark
April 10, 2023 7:40 am

April?

atticman
April 10, 2023 4:02 am

I’m puzzled by the sentence in paragraph 7 which says: “This illustration graphicly represents the collection of data describing how the energy absorbed [sic] by the sun each day escapes back to space via waves of long wave radiation.”

Should that not read “…absorbed from the sun…” or, perhaps “…absorbed by the Earth…”? It doesn’;t make sense to me otherwise.

Jim Steele
Reply to  atticman
April 10, 2023 8:17 am

Agreed atticman, I was grammatically incorrect. I should have said “from”

David Dibbell
April 10, 2023 4:22 am

“What Traps Heat the Longest?”
I appreciate these presentations by Jim Steele.

For perspective on the headline question, consider the planet as a whole, as indicated by the 2 meter air temperature. It gains and loses enough heat energy at and near the surface during the annual trip around the sun to produce a cycle of ~3.8C warming and cooling.

Think about the implications, and lose the fear that non-condensing GHGs in the atmosphere are capable of “trapping” heat energy down here. They don’t end up doing that to any harmful extent, or even to any extent we can reliably detect or compute as a climate trend response. The static radiative effect is not in dispute, but it does not control the end result in a highly mobile atmosphere in which dynamics dominates the outcome.

Yes, this is from a model – a reanalysis. But it makes no claim of prediction. It simply represents an advanced attempt at computing mass and energy flows to explain what is observed directly.

https://climatereanalyzer.org/clim/t2_daily/

Last edited 1 month ago by David Dibbell
Tim Gorman
April 10, 2023 4:42 am

Jim,

Very nice.

You’ve brought into the discussion the factor that the Earth is not a fixed radiation model based on “averages”. It is a time dependent system. One must look at the entire integral, not just some kind of an “average”.

I understand your need for brevity but I would have also added a couple of sentences concerning the fact that the amount of radiation from a body is based on the fourth power of its temperature. As the Earth is warmed by the sun the Earth also radiates at a higher level. Again, a time function. If the maximum temperature of the Earth goes up by 270K to 271K it’s radiation changes from 270K^4 to 271K^4. At least a partial offset occurs.

Jim Gorman
April 10, 2023 5:15 am

Excellent! I am most appreciative of the work you have expended to gather this information into an understandable transcript. There are several points inside the transcript that will provide exciting rebuttals to folks on twitter.

E. Schaffer
April 10, 2023 7:19 am

Lots of of misunderstanding here presented by Steele.

  1. A common misunderstanding is to say something would emit 1/2 of the radiation up/down. The SB law applies to surfaces naturally emitting into a hemisphere. The atmosphere (and GHGs within it) emits a 100% both up and down.
  2. There is no time component and no issue over “how long the heat is trapped”. Once CO2 reduces emissions, the “damage” is already done. While there will be more emissions through the atmospheric window, that is because and after(!) surface temperature increased. It is called the “planck feedback” and will NOT lower temperatures.
  3. The Happer/Wijngaarden chart is about how CO2 reduces emissions TOA. This figure is about 3W/m2 in their paper.
  4. The 2W/m2 is a different story. And although Wijngaarden names it in the Tom Nelson podcast, its origin is rather here..

https://greenhousedefect.com/the-holy-grail-of-ecs/why-2w/m2-co2-forcing-is-actually-quite-certain

Jim Steele
Reply to  E. Schaffer
April 10, 2023 8:13 am

Schaffer there is no misunderstandings presented. You are quibbling with words. 100% both up and down is a non-sensical statement and to say there is no time component is utter BS.

E. Schaffer
Reply to  Jim Steele
April 10, 2023 10:58 am

To be fair, I do not know what exactly you mean with half the radiation going up and down respectively. But I know this idea of halving the SB law is circulating, just because the atmosphere so to say has two sides, and that is confusing people. Nothing has just one side, the SB law needs no modification.

Apart from that thing, I would recommend to read carefully if you get advice from someone who knows a lot more than you do.

mkelly
Reply to  Jim Steele
April 12, 2023 6:25 am

Jim the 100% in both direction is exactly how Willis sells his steel greenhouse idea.

http://wattsupwiththat.files.wordpress.com/2009/11/steel_greenhouse_diagram.jpg?resize=345%2C454

Jim Gorman
Reply to  Jim Steele
April 12, 2023 11:16 am

It is not non- sensical. Radiation flux is emitted in a spherical fashion on an isotropic basis. That is 100% in all directions. Many people believe that you take the absorbed power flux divide by two and that is the power that goes up and down.

I don’t know how to simplify it without simply saying that radiation is emitted as a sphere around a point kind of like a balloon or basketball expanding. It is also why the power at a point reduces by r^2. It is also the need for steradians.

JCM
Reply to  E. Schaffer
April 10, 2023 8:31 am

There is only a virtual instantaneous reduction of surface transmitted flux, by slightly shuttering the windows.

Once the window is closed, the radiative flux never occurs. No need to worry about the ups and downs and back radiations.

E. Schaffer
Reply to  JCM
April 10, 2023 11:01 am

Please do not say “radiative flux”, it is a horrible term. Radiation does not flow, it just radiates. The “flux” then suggests a flow of energy, that usually does not occur. Two objects radiating onto each other, which is the norm btw., are NOT two energy fluxes. Rather there is no flux at all.

JCM
Reply to  E. Schaffer
April 10, 2023 11:03 am

The flux is real – it is that portion which is transmitted to space. Else, you are correct, there is no flux at all.

PS, “I would recommend to read carefully”

Last edited 1 month ago by JCM
JCM
April 10, 2023 8:45 am

Notice here how it all goes wrong.

The top plot that emission spectra of desert (a), vs the bottom that of a moist regime (b).

Does the desert exhibit a stronger greenhouse effect judged by that difference between the emission spectra and the 320K curve?

Does the moist regime exhibit a 3x lower greenhouse effect judged by that difference between the emission spectra and the 280K curve?

of course not.

Desert_v_Wet.png
JCM
Reply to  JCM
April 10, 2023 9:00 am

look at all that excess “greenhouse” heat trapping occurring in the desert H20 band… LOL. WRONG.

desert.png
It doesnot add up
Reply to  JCM
April 10, 2023 9:40 am

The desert will of course exhibit substantial diurnal temperature fluctuations. During the day the surface will endure strong solar heating, and it warms up. It is during the night when it gets much cooler, even down to freezing at times. That rather complicates the analysis.

JCM
Reply to  It doesnot add up
April 10, 2023 10:19 am

Greenhouse theorists would have us integrate the Watts per square metre per cm difference between the 320K desert planck curve and the observed spectrum. A massive greenhouse effect by CO2 and H20.

E. Schaffer
Reply to  JCM
April 10, 2023 11:15 am

There are a couple of details to consider here.

First of all the surface temperature there will likely be significantly higher than just 320K. We are talking about 330, 340K. The sand turns very hot in direct sun light, and turns cold during the night. Daily temperature range (DTR) for sand (not for air!) is around 40K. On the other side sand is a poor emitter within the atmospheric window, especially in the right half.

The atmosphere on the other side has a very small DTR and is not super-heated like the sand at noon. What the dessert scenario suggests as GHE, is largely just the relative heat of sand vs. atmosphere at noon. I have explained this here..

https://greenhousedefect.com/2/deception-with-emission-spectra-part-2

JCM
Reply to  E. Schaffer
April 10, 2023 11:40 am

yup. it is the effective radiating surface properties – those surfaces transmitting radiation to space – where one must focus.

Here is your chart annotated with the primary radiating surfaces.

The terrestrial surface (red), the boundary layer surface (blue), and the tropopause surface (grey).

Here the blue surface is fixed to the phase point temperature of water, always.

emission.png
LJ
April 10, 2023 9:53 am

There’s one thing I just can’t fathom — if we have a body that has half of it at a temperature of 50C and the other half at -50C, it’s average temperature will be 0 degrees as well as its stored heat energy. However, its radiation losses would be some 18% higher than if its temperature was the average all over that body.

So how come one would average the insolation at 1/4 of 1362 W/m^2 on the left side and then take the average temperature of the Earth on the right side and then believe these two would be equal? Yet that’s what most of the CO2-blaming graphs try to do…

In other words, if we take Sahara for example, it receives some 1000 W/m^2 at noon and its temperature gets up to some 40C, while at night it receives 0W/m^2 of insolation and the temperature falls to -3.9C. If we calculate average It would seem there is an imbalance, while if we integrate it will be a different story.

Last edited 1 month ago by LJ
E. Schaffer
Reply to  LJ
April 10, 2023 11:23 am

But Earth is not half 223K and 323K respectively. Instead about 70% is covered by water, with a temperature range between 0 and 30°C. That is not making much difference. Only the remaining 30% of land and ice have a significant spread. If you account for all of it, that is not a 18% deviation, neither 6% (as Trenberth 08 argued), but only 2.5%. And there is no problem accounting for it.

LJ
Reply to  E. Schaffer
April 10, 2023 11:07 pm

Take the Sahara example. During daytime it receives some 1000 W/m^2. If we assume CO2 adds some 3 W/m^2, that’s less than 0.3% of the total irradiance. So if the daytime temp of Sahara is some 38C, this 3W/m^2 more would be negligible. And at night, the temperatures fall to under 0C, and the explanation for that phenomenon is the dry air…

The same goes with the areas near (and not quite near) the poles, which either receive a lot of insolation, or quite little.

As a result, the so-called “warming” is mainly observed wherever there are low temperatures.

bdgwx
Reply to  LJ
April 10, 2023 11:51 am

LJ said: “So how come one would average the insolation at 1/4 of 1362 W/m^2 on the left side and then take the average temperature of the Earth on the right side and then believe these two would be equal? Yet that’s what most of the CO2-blaming graphs try to do…”

That’s not how it is done. Energy budget diagrams use full integration. See Trenberth et al. 2009. Pay particular attention the section discussing spatial and temporal sampling and the “rectification effect”.

LJ
Reply to  bdgwx
April 10, 2023 11:12 pm

The article you refer to mentions “an estimated imbalance from the enhanced greenhouse effect of 0.9 W m^2”. So how much temperature increase would that amount to, using Stefan-Boltzmann? At emissivity of 0.6 it’s some 0.2-0.3C depending on the temperature.

bdgwx
Reply to  LJ
April 11, 2023 5:45 am

It depends on the climate sensitivity (in C per W/m2). Right now it is about 0.6 C per W/m2. So 0.9 W/m2 * 0.6 C/W.m2 = 0.5 C. However, it is important to note that the climate sensitivity is not believed to be static due to feedback and tipping points.

LT3
April 10, 2023 12:29 pm

Great article, however, this statement does not compute.

“Satellites observe that the stratosphere is cooling twice as fast as the lower atmosphere is warming.”

There has been almost no trend in the Lower Stratosphere temperature since the mid 90’s after the effects of Pinatubo subsided.

TheLowerStratosphere4-10-23.png
Phil.
Reply to  LT3
April 11, 2023 8:09 am

The upper stratosphere was cooling at about -0.7K/decade (at 40km, same source), however that has flattened out over the last decade.

LT3
Reply to  Phil.
April 12, 2023 8:15 am

The Last 2.5 decades have been flat. the 80’s and 90’s are meaningless to utilize in linear regression in the context of AGW, because of two large volcanic SO2 events.

There is no trend, just two large step functions.

slindsayyulegmailcom
April 10, 2023 12:42 pm

Energy with increasing CO2 redirects towards the surface is less than 1w-m²(0.13w-m²).
Incoming energy 1360w-m² at tropics
Surface temperature 299.7K 457.4w-m²
902.6w-m² out to space.
457.4w-m² 299.7K 9.66µm
Out of 902.6w-m² 88w-m² water removes heat from surface.
Else surface would be 545w-m² 313.2K 40°C.
And less at 814.6w-m² goes out to space.
From tropics to pole 247.4w-m² is released to space.
As 210w-m² is compression heating, -26.3°C
11.74µm.
No solar radiation needed.
CO2 cools the mesosphere because 210w-m² isn’t there.
Ozone heats stratosphere (Dec 11th over Antarctic) and CO2 cools stratosphere everywhere else.

CO2 15µm(-80°C) is peak (can go up to 14µm -66°C, like human body is 9.36µm (36.5°C) is peak (can go up to 9.26µm 40°C).

210w-m² is 314.83 m/s (1.13 million km/hour). Add solar heat speed goes up to 334.5 m/s(1.2 million km/hour)

Speed can’t get down to zero in 8 or 16 hrs.

Lowest in Antarctic -78°C 280 m/s (1 million km/hour).

So scientist ignore this and add it to greenhouse gases heat instead.

BurlHenry
April 11, 2023 8:57 am

Jim Steele:

I have to be blunt.

Just about everything that you say about our climate is wrong, because you base your comments on the premise that CO2 actually has a warming effect, which it does not, and ignore the actual cause of our changing climate, which is changing levels of SO2 aerosols in our atmosphere, due to volcanic eruptions and Industrial activity (since about 1850).

For example, you say that Heat Domes are caused by Omega loops that trap air and prevent it from moving along, causing warming

More correctly, they are stalled High Pressure weather systems, where the atmospheric SO2 aerosols within the trapped area settle out, in a week, or less, and temperatures always soar to disastrous levels because of the cleansed air.

This is an extremely important fact, since it offers a method of quickly ending them, saving thousands to hundreds of thousands of lives around the world each year.. The method would simply be to seed the atmosphere high above the stalled areas with SO2 aerosols, or some other dimming substance.

See: “Stalled High Pressure Weather Systems”

https://doi.org/10.30574/wjarr.2022.13.3.0264

And on a previous thread, I had pointed out that changing levels of SO2 aerosols were responsible for all La Nina and El Nino events.

“The Definitive Cause of La Nina and El Nino Events”

https:doi.org/10.30574/wjarr.2023.17.1.0124

Every aspect of Earth’s climate that I can think of can be explained by changing levels of SO2 aerosols in our atmosphere, with no hint of any additional warming due to CO2…

Jim Steele
Reply to  BurlHenry
April 11, 2023 10:01 am

Burl, I investigated your sulfate claims a long time ago, but I didnt find any support for your obsession. It seems whenever I post, you see it as an opportunity to push your pet theory as the only possible explanation. So I’ve relegated your “All things are Possible with SO2″ to the dust bin of climate religions along with All things are Possible with CO2. I find your claim that SO2 causes all ENSo events to be totally ridiculous. For example Holocene records show no correlation

HOLOCENE SULFATE VS ENSO .png
BurlHenry
Reply to  Jim Steele
April 11, 2023 1:44 pm

Jim Steele:

The bottom graph which you show as proving there is no correlation with SO2 and ENSO events shows temperature peaks for the Minoan Warm Period, The Roman Warm Period, and the Medieval Warm Period.

Those were eras where there were very few VEI4, or larger, volcanic eruptions, with the result that most of the time there were few to zero volcanic SO2 aerosols in the atmosphere, and temperatures rose (i.e. many “ENSO” events caused by the lack of SO2 aerosols in the atmosphere).

Rather than disproving what I have claimed, it actually provides COMPLETE correlation.

You said that you investigated my sulfate claims long ago, but you didn’t find any support for my “obsession”. .

I have 11 articles on Research Gate that support my SO2 claims. You should read them.

I see that the link to my La Nina/El Nino article is not working. Here it is again:

https://doi.org/10.30574/wjarr.2023.17.1.0124

A good place for you to start

ATheoK
April 13, 2023 9:42 pm

“Wavelengths between 20 and 15 microns should act like interstate highways carrying the greatest flow of heat,
While wavelengths of 5 or 40 microns act like dirt paths allowing very little heat transport.”

20 and 15 microns are between 5 or 40 microns…
So, they all should be like dirt paths.

From Joe Bastardi’s twitter:

The fact is that 99% of the radiation absorbed by atmospheric CO2 is in the 15 micron wavelength band, a micron being 1 millionth of a meter in length. This equates to the lower end of the cold Far Infrared spectrum with a frequency of 20.02 Tera Hz and a photon energy of 1.33X10^-20 Joules.
It coincides with the peak radiation from a source at -80 deg. C°, the minimum temperature occasionally reached in Antarctica.”

Doesn’t change your conclusion at all, “Global Warming Greenhouse Theory’s Greatest Weakness

Phil.
Reply to  ATheoK
April 15, 2023 7:11 pm

Yes the same Wien’s law nonsense I referred to above. It also is the radiation that is radiated at a spectral radiance of 6.6841 W/m2/sr/µm at 300K 6 times more than the oft mentioned radiance from the -80ºC source (1.10107 W/m2/sr/µm).
You appear to not have seen the difference between “Wavelengths between 20 and 15 microns” and the discrete “wavelengths of 5 or 40 microns”.

Last edited 1 month ago by Phil.
Bob W
April 20, 2023 3:32 pm

The video is interesting as far as it goes, but it has a some serious problems.

It focuses its attention on localized “delays” in heat transport (in the form of convection) near the surface. Such localized reductions in heat transport can indeed produce localized, temporary, warming.

But, these phenomena have little, if any, effect on the big picture of Earth’s overall energy balance.

If one focuses on localized heat transport within the atmosphere or ocean, one will almost inevitably get lost in details that obscure, rather than illuminate, the big picture.

The video starts out by looking at a chart of energy radiated to space vs. energy emitted by the surface, as a function of frequency, agreeing that this chart offers correct science.

The video starts to go wrong as it describes the significance of that chart, using the language of CO2 “delaying” heat loss, and talking about this as if it’s about the amount of time it will take until the heat escapes.

It’s really regrettable that anyone ever used the word “delay” with respect to how the Greenhouse effect works, because that way of talking about things is intensely misleading, and virtually always leads to false reasoning. (I’ve never seen any valid mathematical analyses related to the topic that involve time delays.)

A much more accurate way of describing what Greenhouse gasses do would be to say that they reduce the rate of radiative cooling to space, for any given surface temperature. “Delaying” or “trapping” heat isn’t about any time delay; it’s about a reduction in the RATE of heat loss to space.

In the curve of emissions to space, the upper curve represents the rate of heat loss to space that would exist if there were no Greenhouse materials (GHGs and clouds) in the atmosphere. The lower curve represents the rate of heat loss to space when GHGs are present.

The gap between the two curves represents the amount by which GHGs reduce Earth’s rate of cooling to space, for a given surface temperature. As a result of that reduction, Earth’s mean surface temperate naturally rises until the planet can shed energy to space as fast as it is absorbing heart from the Sun.

In other words, that curve (or a version of it that incorporated the entire Earth’s surface rather than just one location) ultimately determines the planet’s average surface temperature.

The sort of phenomena the video talks about can temporarily affect the surface temperature in a localized region. However, they will only affect the planet’s overall energy balance, and hence the global mean surface temperature, only to the extent that they alter the chart of emissions to space.

The video hasn’t shown any such effect, so it hasn’t shown that the subject being discussed has any effect on the big picture of overall planetary temperature.

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