The Effect of a Colder Solid’s Thermal Radiation on a Warmer Solid Exposed to Sunlight
Abstract
It is experimentally shown that the thermal radiation from a transparent, colder solid has the capacity to influence a solid warmer than it to become even warmer, under the right circumstances.
This dispels the critique of the greenhouse effect that, as heat only flows from hot to cold, the effect is thermodynamically impossible. Even so, significant portions of the theory of the greenhouse effect remain experimentally unproven, signaling caution rather than uncritical acceptance of the theory.
Introduction
A long-running debate about the physical reality of the greenhouse effect centers around whether the thermal radiation from the colder atmosphere can possibly have a warming effect on the warmer Earth’s surface.
The argument goes as follows: as heat only flows from hot to cold, the radiation emitted by a colder object cannot possibly cause a warmer object to become warmer. It might have a reduced cooling effect, but under no circumstances can it result in a warming effect.
The most salient features of such debates are that neither side provides experimental evidence in their defense, and that the debates frequently devolve into heated arguments, which is generally an indicator that solid arguments are lacking.
As genuine scientific knowledge is acquired via experimentation performed in physical reality, and not theory or “experiments” (sic!) consisting of computer simulations, I sought to settle the debate once and for all with a properly-performed experiment.
I constructed a de Saussure-style hotbox, above which I placed a clear glass plate separated by a good amount of space. I pointed it towards the Sun, and, left to its own devices, the bottom of the box got to around 100ºC, with the glass plate getting to around 30ºC.
I then swapped the glass plate with an identical one that was pre-heated to 60-70ºC, and the result was unmistakable: the already much hotter bottom of the box got even hotter as a result.
I ruled out any possible conductive or convective effects, concluding it was due to increased thermal radiation coming from the +30-40ºC warmer glass plate. Although this neatly dispels this particular critique of the greenhouse effect, significant obstacles remain before being able to accept that the greenhouse effect behaves as-described.
The Demonstration
The apparatus (Figure 1, Figure 2) consists of a hotbox made of styrofoam, with the inside floor being cardboard and the inside walls and floors spray-painted black.
Two layers of thin plastic polyethylene film suppress convection with the outside air. Surrounding these is another piece of styrofoam to further prevent heat loss through the sides. Cotton padding between the styrofoam pieces provides further insulation.
On top of the hotbox are two styrofoam walls that support an extra-clear glass plate on top. This is the glass that is swapped with an identical, hotter version during the experiment. A middle mount point allows the optional mounting of another plate in the middle.
Temperatures are measured with Type K thermocouples, labeled and color-coded for ease of reference. Styrofoam pieces immediately above yet not touching them serve as radiation shields.
The solar insolation is measured with an Apogee SP-510-SS pyranometer, and the net infrared gain or loss is measured with an Apogee SL-510-SS pyrgeometer.
I pointed the apparatus at the sun and let it heat up on its own. In the meantime, I set aside an identical extra-clear glass plate, on top of which I placed an aluminum plate with a pot of water on top.
An immersion heater kept the water temperature at a constant 80-90ºC, which heat diffused downwards to heat the glass plate.
Once the bottom plate exceeded 100ºC, I swapped the cool glass with the hot glass, first by hovering the hot glass above it, then removing the cool glass and placing the hot glass in the same position.
This swapping technique provides a dip in measured solar insolation due to both glasses absorbing the sunlight rather than just one. This clearly delineates the swaps, and ensures there is no extra heating effect due to slightly higher insolation that would happen if we first removed one glass and then replaced it with the other glass.
I performed five swaps as above, with Tcoolglass measuring 30-38ºC and Thotglass measuring between 67-75ºC at the start of the swaps. The result was clear and unambiguous in every case: Tbottom shot up rapidly in response. The following two runs are representative (Figure 3):
As is evident, before the first swap on the graph, Tbottom was increasing at a certain steady pace. After the swap to the hot glass (delineated by the purple vertical line), the rate of increase rapidly shot up.
Once swapped back to the cool glass (delineated by the gray vertical line), Tbottom stabilized around 108ºC. It remained there for ~5 minutes, after which another swap caused it to increase rapidly again.
It is notable that at the exact moment of the swaps, the net infrared radiation Net IR picked up by the pyrgeometer remains negative yet increases. That is, the bottom’s radiative cooling is measurably lessened as a result of this swap.
To rule out differences between the two glass plates, such as the possibility of slightly more solar insolation which could cause a temperature change, I did two swaps with both glass plates at around the same temperature. With these swaps, no difference was observed in the evolution of Tbottom or in the Net IR.
Confirming the Radiative Nature of the Effect
It is important to rule out any convective effects as having caused the increase. The hotter glass will, of course, heat the air around it as well. How do we know that Tbottom didn’t get hotter because the hotter glass heated the air in-between, which then heated the thin plastic films, which then reduced the convective loss of the bottom and caused its temperature to increase?
The evidence is three-fold. First, we can observe that Tlowair did not noticeably increase as a result of the swap. Indeed, it was warmer before the swap than after (Figure 4):
Second, by including Tthin1 and Tthin2 in the graph, we can see that Tbottom increased first, and only nearly half a minute later did Tthin1 start to increase, soon after which Tthin2 increased as well (Figure 5).
Thus the temperature increase started from bottom to top, not from top to bottom, meaning the observed increase could not be due to the air being heated from the top.
Third, I repeated the experiments, but with a thin borosilicate plate mounted in the middle of the apparatus (Figure 6). Borosilicate is highly absorbent of infrared radiation.
If the effect is radiative, we would expect the borosilicate to absorb any extra thermal radiation from the hotter glass, preventing Tbottom from increasing.
This is precisely what happened. During these runs, the Net IR did not change when the hotter glass was swapped in, nor did Tbottom respond to the swaps.
Thus we have to conclude that the increased thermal radiation from the hotter glass is what caused the bottom to increase in temperature, even though the hotter glass itself was much cooler than the bottom.
Analysis
How does this not violate the laws of thermodynamics, wherein heat only flows from hot to cold? The answer is that one must consider all the heat flows in the system.
When Tbottom is at a steady temperature around 108ºC, it is because the amount of heat it is gaining from all sources is equal to the amount of heat it is losing to all sinks.
At that temperature, the only heat it is gaining is from the sunlight, which is actually the thermal radiation emanating from our nearby star with its surface temperature of 5900K.
It is at the same time losing heat conductively with the bottom of the hotbox and the walls, convectively with the air in the box, and radiatively with the plastic wrap layers, the glass, and the sky above. The hotbox itself is also losing heat convectively with the outside air.
When the hotter glass is swapped in, effectively the only change is that now the bottom is losing less heat radiatively to the glass.
With all else being equal, it is thus now gaining more heat from the sunlight than it is losing to all other sources, and the result is an increase in temperature. Notably, without the much hotter sun as a heat source, the increase in temperature would not be observed; only a reduced cooling effect would occur.
Conclusion
There are several key differences between the experiment performed here and the theorized radiative greenhouse effect, such that this experiment does not serve as verification of the latter.
- The hotbox is heavily insulated and enclosed to suppress convective heat loss. By contrast, the majority of the heat loss by the Earth’s surface is due to convective loss to the air.
- The temperature ranges are different, with the Earth’s surface being around 15ºC and the air ranging from 15ºC to -55ºC with increasing altitude, contrasted with 100ºC for the black bottom and 30-70ºC for the glass.
- The colder object is a solid plate of glass as opposed to a column of atmospheric air, i.e. a large volume of gas.
- The warmer object is a uniform pitch-black plate, as distinct from the Earth’s surface with its varied terrain, soil, plant, foliage, ice, snow, water, etc.
- The greenhouse effect is due to the air becoming more absorbent of and emissive of thermal radiation at the same temperature, while the experimental result was due to the colder object retaining the same emissive properties yet becoming hotter.
- The observed time duration was minimal and no conclusions can be drawn about the total magnitude of the effect or what effects it may have in the long run.
Indeed, the greenhouse effect theory has a large gap when it comes to the empirical demonstration of the greenhouse effect’s total effect on surface temperatures. It largely relies on a simplified calculation that shows the Earth’s average temperature would be -18ºC without any atmosphere. This calculation even includes the albedo effect of clouds, which would not be present without an atmosphere.Further it ignores all the distinctions listed above, and doesn’t account for the adiabatic lapse rate, which is the main reason why the bottom of the grand canyon is +50ºC warmer on average than the peak of Mount Everest. Such a lapse rate occurs entirely due to non-radiative effects, and therefore would have some effect even if the air were fully transparent to infrared radiation.
- The observed temperature increase here was due to swapping in a hotter glass plate that was externally heated by another heat source (i.e. neither the sun nor the black bottom). This is in contrast with the greenhouse effect, where the atmosphere’s thermal radiation that is theorized to result in a much warmer surface temperature, is initially warmed by the surface itself. This gives the appearance that it presents a situation where an object is able to heat itself up with its own heat – first the atmosphere at the same temperature emits thermal radiation according to its own temperature, which then results in a warmer surface, that then in turn heats the atmosphere further, etc., in a (diminishing) feedback loop.
Although the experiment performed here indicates that the presence of the sun ought to make this possible, and Infrared Halogens appear to exploit an analogous mechanism to reduce energy consumption, an experiment should be done to confirm this is the case. It is notable that past experiments have failed to definitively show a powerful effect: R W Wood’s 1909 experiment showed “scarcely a difference of one degree” between two hotboxes, one with a radiatively-absorbent glass lid and one with a radiatively-transparent rock salt lid, where we would expect the radiatively-absorbent lid to result in a much hotter temperature due its higher radiative emissions as well. More recently, a re-do of Wood’s experiment “performed more carefully” by Pratt actually replicated the result, despite being heavily critical of Wood: a difference of 1.1ºC between the floor of a hotbox with a glass lid vs one with a rock salt lid. Pratt further showed that the glass lid was actually 6.2ºC hotter than the rock salt lid. This further complicates the analysis as convective heat transfer is proportional to temperature difference and indicates the +1.1ºC result is at least in part due to reduction of convective heat loss.
- Finally, even if the basic greenhouse effect were to be conclusively demonstrated, it is well-known that the current levels of CO2 are already saturated with regards to higher concentrations causing higher thermal emissions. The enhanced greenhouse effect is said to occur due to the much colder higher-altitude layers of the atmosphere absorbing and emitting more thermal radiation, thus having a cascading effect on the air immediately below it, which in turn has an effect on the air below, and so on up to the surface. In other words, the hotter absorbing element in the enhanced greenhouse effect is also atmospheric air, and not a solid object. This provides further complications as atmospheric air can freely travel vertically up and down, and hot air tends to expand and rise.
In conclusion, although the thermodynamic possibility of colder objects causing warmer objects to increase in temperature has finally been conclusively demonstrated, a lack of experimental verification of other facets of the greenhouse effect theory warrants caution before accepting its conclusions.
Please Donate Below To Support Our Ongoing Work To Defend The Scientific Method
PRINCIPIA SCIENTIFIC INTERNATIONAL, legally registered in the UK as a company 
incorporated for charitable purposes. Head Office: 27 Old Gloucester Street, London WC1N 3AX.
Trackback from your site.
Herb Rose
| #
Mr Cloudman retains all the erroneous assumptions of the proponents of the GHGT.
1: A thermometer is not measuring the kinetic energy of molecules but how much energy is being transferred to the measuring device. If it was measuring the KE adding more fuel to a combustion would not raise the reading of the thermometer (causing it to get warmer).
2: The loss of energy by radiation is a minor factor for the transfer of energy from the surface. (Use the search function at PSI to lookup “An interview with Tom Shulah” to see how the Pirani gauge shows how much energy is lost by radiation versus convection.) In order to stop convection you must conduct your experiment in a vacuum. Plastic film will not work, just adding a layer with different conductivity.
3: With convection energy is transferred by collisions from the object with the greater energy per unit mass to the object with less energy per unit mass regardless of the amount of kinetic energy of the two objects. It obeys the aw of conservation of momentum, not radiation. In elastic collision no mass is transferred, only energy. The second law of thermodynamics is wrong bother radiated energy and convection.
4: The atmosphere is not being heated by the surface of the Earth but by the O2 and N2 molecules absorbing 95+% of the UV radiation coming from the sun and converting that energy to kinetic energy. How can anybody believe that the Earth is providing the 492,000 joules/mole to split O2 molecules in the stratosphere creating ozone yet avoiding all the O2 molecules in the troposphere. All matter absorbs radiated energy!
5: The measured temperature of the atmosphere has nothing to do with the kinetic energy of the molecules in the atmosphere. When a gas absorbs energy it expands and fewer collisions occur. Look at the graph of the temperature of the atmosphere. That zigzag line with pauses is not how energy flows. Try making a graph of the temperature at an altitude divide by the density at that altitude to ge an image for the energy of a constant number of molecules instead of a constant volume of molecules. If a loss of energy causes a gas to contract (at 0 K it would be a layer on the surface of the Earth) why does the density of the atmosphere continually decease despite increases and decreases in temperature?
Garbage in – Garbage out applies to experiments as well as computers.
Reply
Dale Cloudman
| #
Hi Herb,
Thanks for the feedback.
This is true, but doesn’t change the result or the conclusion. More energy was transferred to the thermometer as a result of swapping the cooler glass plate for the warmer one, yet the warmer one was still cooler than the thermometer’s temperature. So the analysis and conclusion holds.
a. The relative effect of radiative vs convective loss of the surface isn’t relevant for the experiment, which involved a small styrofoam box with insides spray-painted black.
b. Convection of course was not completely stopped, yet it was certainly impeded by the plastic film, which allowed the box to get to the ~100C it did (it gets to only ~70C with an uncovered top)
3-5. None of these points appear to be relevant to the purpose of the experiment, which was to show that when a colder object becomes less cold, the extra thermal radiation it emits can lead to warming an even-warmer surface, when there’s an external heat source such as the Sun involved.
Best,
Dale
Reply
Dale Cloudman
| #
The formatting seems to not have been preserved — the “This is true” should be prefixed by a “1”, and the “a.” should be prefixed by a “2”
Reply
Herb Rose
| #
Hi Dale,
I have stated that the 2nd LoT is invalid because objects transfer energy not mass. With convection the law of conservation of momentum applies. With the transfer by radiation it is geometry that prevents equalization. An object radiates energy in all directions but can only transfer energy going from higher energy level to lower. Since energy declines with distance any object seperated from another object will only receive a fraction of the decreased energy coming from the other object and cannot equalize with it, only achieve stasis where the energy it is radiating is equal to the energy level coming to it (less than the energy being radiated from the other object).
Your glass plate is achieving stasis with the energy coming from the blackened interior. Replacing it with a heated piece of glass will change the equilibrium point to a distance below the glass. It will no longer absorb heat coming from the interior as the equilibrium point where the flow of energy halts, has moved into the box.
If you have a large and a small source of heat as you move the thermometer from the larger source towards the smaller source the temperature will decline, until it enters the heat being radiated from the smaller source where the temperature will begin to rise.
By adding the heated glass you have in essence added insulation stopping the flow of energy from the interior. You are not making the interior hotter by adding energy, you are preventing it from cooling by blocking it from losing energy.
Herb
Reply
Dale Cloudman
| #
Hi Herb,
“By adding the heated glass you have in essence added insulation stopping the flow of energy from the interior. You are not making the interior hotter by adding energy, you are preventing it from cooling by blocking it from losing energy.”
As long as this statement applies to radiative heat exchange — we agree in principle.
The simplified equation for radiative heat loss is basically “σ(Thot^4 – Tcold^4)”. What I have shown is that if Tcold increases, this lessened radiative heat loss can lead to Thot increasing, when there are other heat flows involved.
This has been denied by many (see Chapter 2 of Slaying the Sky Dragon) and I used to think it was possibly not true — at the very least, I hadn’t come across any experiments that convincingly showed it was true or not true.
“An object radiates energy in all directions but can only transfer energy going from higher energy level to lower.”
I think it’s more correct to say every object radiates and absorbs energy from every object it sees. But heat can only transfer from a higher-temperature object to a lower-temperature object. If the net energy transfer is positive (more being absorbed), then heat is flowing into the object from what it sees, the object is warming — if the net energy transfer is negative (more being radiated), then heat is flowing away from the object to what it sees, it is cooling.
The relevance to the atmosphere is that the atmosphere is actually radiatively insulating the surface, analogous to the hotter glass in the experiment. It can be measured with a handheld IR thermometer or a pyrgeometer — objects on the surface lose less heat than they would if facing outer space. As the experiment shows, when the sun is warming the surface, it seems like this must be causing the surface to be warmer than it would be without the radiative insulation (although how much warmer is in question, and it would be good to validate this with an experiment with gases).
At night of course it can’t lead to a higher temperature as in causing the surface to warm up. But it could prevent cooling, e.g. instead of cooling from 20C to 5C it might cool from 20C to 15C instead, thus being warmer than it otherwise would be.
The impact of added CO2 in particular to the atmosphere is an entirely different story, which we can get into later.
Minor last note: “I have stated that the 2nd LoT is invalid because objects transfer energy not mass. ” I believe the 2LoT applies to both energy and mass transfer. Whatever mode of transfer or operation or whatnot happens, heat can only flow from hot to cold.
Best,
Dale
Reply
Herb Rose
| #
Hi Dale,
Heat is kinetic energy. The 2nd LoT says that an object with more kinetic energy will transfer energy to an object with less kinetic energy.
If you are in a car and collide with the rear end of a large slower truck moving in the same direction your twill slow down (lose kinetic energy) and the truck swill increase speed (gain kinetic energy) even though the truck has more kinetic energy.
Energy equalizes with all mass so the total ke does not change but the flow of energy will be from the object with the greater energy per unit mass to the object with lower energy per unit mass. Use the PSI Search function to look up my article “How Cold Heats Hot”.
Water in the atmosphere has the same temperature as the gases, which is less than the temperature radiated from the surface. They obviously have more ke than the water molecules on the surface.
The amount of energy being radiated depends on both the number of molecules radiating energy and the level of energy of those molecules. In a gas there are fewer molecules radiating (transferring) energy so even if they have more energy they will appear to have less.
The thermometer was designed to measure the flow of energy from one medium (water) to another medium (air). As the portion of measuring liquid in the one medium gains energy the amount of measuring medium in the other medium expands causing it to transfer more energy into that medium.
In the atmosphere the entire thermometer ids exposed to one medium which means it is measuring the total energy being absorbed by the measuring liquid. The total energy comes from the momentum of the molecules striking it which is a function of both the velocity of the molecules and their mass (number). It is not measuring the mean kinetic energy of the molecules. If you want to determine the energy (v^2) of the molecules you must divide the measured temperature by the density.
As I stated there are 2 ways to transfer energy radiation (slow) and convection (fast). When we speak of an object radiating energy we mean it is losing energy not the method that energy is being lost. This causes confusion. In the troposphere the method of energy transfer is almost all a result of convection (collisions) not radiation. (Again read the article “An interview with Tom Shula).
Do you think the molecules in a 100 C oven have the same ke as the molecules in 100 C pan of boiling water?
Herb
Jerry Krause
| #
Hi PSI Readers,
I used a modified de Saussure hot box for more than a year to study the influence of the solar radiation upon its interior temperature and wrote several PSI articles about my results and thoughts. Here (https://principia-scientific.com/solving-global-warming-de-saussure-device-paradox/) is a link to one of these articles which contains the links to two more. So I ask readers to compare my experimental results and thoughts with Dale Cloudman’s results and thoughts.
Have a good day
Reply
Dale Cloudman
| #
Hi Jerry,
I read all three articles. I found the experimental detail a bit sparse and buried in the writing. It seems to amount to having made a styrofoam box with two layers of thin plastic film, that reached 212F (ie 100C, not 212C as written in the article). The 3rd article just contains a suggestion for someone else to perform another experiment.
Mostly I didn’t find any additional insight from the reports. It is a lot of qualitative evaluation with supposition of how things work, but little experimentation actually done to test the hypothesis. Also I did not see anything directly related to whether the thermal radiation from the colder walls and thin plastic film had an effect on the bottom temperature. I didn’t see any satisfying resolution to the paradox you proposed.
The full resolution of it will be found in a paper that I will finalize and will be published shortly, detailing an experiment of a hotbox I constructed that reached 150C (302F). As this far exceeds the moon temperature and the computed 121C max from the solar insolation, it resolves the matter for good that the thermal radiation of the colder objects has a clear effect on impeding cooling and thus leading to higher temperatures.
A few other notes:
You wrote that the thin film “does not significantly absorb any solar radiation or that being emitted by a surface at a high temperature near the maximum temperatures observed in the cases of the lunar surface and the de Saussure device”. I can contradict both cases, as measured by the pyranometer it does reduce the insolation by maybe 100 W/m2, and from the pyrgeometer the IR readings do go up quite a bit with the film than without.
You wrote “But, given apparently cloudless sky, the observed temperature of the a-e surface was always significant lower than that of the ambient atmosphere.” — I have confirmed this to be the case myself as well at night. The inside was 18C or so while the outside air was 22C or so. I explain that it comes down to the black insides losing a lot of heat radiatively to the cloudless sky, and the mixing of the air is prevented, so the inside gets colder. A pyrgeometer works in similar ways as i understand it.
Best,
Dale
Reply
Matthijs
| #
The “surface” in this example is doing work. Its taking visible light and slowing it down into infra red and turning it in to energy.
That energy is then used to heat the surface. The surface has to shed this energy in the form of radiation. If it cant get rid of this radiation, the temperature goes up. Thats why with a pre-heated plate the temperature goes up.
Now, what work does CO2 do? It is claimed that back radiation is warming the surface. So, what is driving this energy generation?
Reply
Dale Cloudman
| #
“Now, what work does CO2 do?”
Hmm you appear to have switched contexts? You start by talking about the “surface” (ie the black bottom), and the work it does, but then end by asking what work the CO2 does, whereas CO2 is analogous to the pre-heated plate, not the surface.
My current position is: it is unclear why the thermal radiation from the pre-heated plate would be of a different nature from the thermal radiation from CO2, such that one would cause the effect outlined in the experiment and the other wouldn’t.
I think the thermal radiation emitted by the atmosphere as a whole must have some effect on the surface. However, the specific contribution of CO2 to this in the first place, and additional CO2 afterwards, is highly, extremely questionable and unsupported by anything other than computer models.
Speaking of which, Clauser has a great presentation showing how the models don’t properly handle albedo, leading to an error of +/- 10 W/m2 in the modeling which is almost 20x greater than the supposed heating effect of CO2 that they find. Link here: https://eike-klima-energie.eu/2024/08/01/presentation-files-of-the-eike-conference-in-vienna-now-available/ , the one titled “A Cloud Thermostat Controls …”
Cheers,
Dale
Reply
Herb Rose
| #
Hi Dale
The current concentration of CO2 in the atmosphere is .00442. For every molecule in the atmosphere there are 1000 molecules on the surface. This means 442 CO2 molecules in the air must heat 1 billion molecules on the surface with the energy they (and the other 999, 558 gas molecules) supposedly recieve from those billion surface molecules. How much effect does CO2 have on the surface? Zilch
Herb
Reply
Dale Cloudman
| #
Hi Herb,
Well, we can focus on the entire atmosphere first, the 1 million molecules of the atmosphere vs the 1 billion of the surface, as you put it. How much effect does the entire atmosphere’s radiative emission have on the surface temps?
We can answer this with the pyrgeometer readings. Based on the data here (https://www.e-education.psu.edu/meteo3/l2_p6.html), during the day the pyrgeometer measured a loss of -150 W/m2. If the surface were facing bare outer space the loss would be -400 W/m2 instead of -150 W/m2. So the atmosphere is preventing -250 W/m2 of radiative loss for the surface.
This certainly seems significant considering the peak solar measured during that same day was 800 W/m2.
So we have determined that the supposed million to billion ratio doesn’t prevent the atmosphere from affecting the surface temps. The question now is what the contribution of CO2 is. It is more about the net effect they would have on the thermal radiation rather than the number of molecules, although the small amount does as a first approximation indicate it would not be significant (and certainly not a primal ‘control knob’ for the climate).
Best, Dale
Reply
Herb Rose
| #
Hi Dale,
You’ve again made an assumption that the temperature is indicating the energy in the atmosphere.
Most of the energy in the atmosphere is contained in the water in the atmosphere, not the gases and it does not show up as “temperature”.
When we sweat the water on our skin removes heat cooling the skin. This cooling occurs with all the water on the Earth’s surface. Even when frozen, water will absorb heat and remove it by sublimation. As long as the humidity is not 100%, water will cool the surface. The higher the temperature the more evaporation and cooling.
This water carries the undetected heat to the top of the troposphere where it is released into space and the water falls back to Earth to repeat the process.
It takes 600 calories to evaporate 1 gram of water so every gallon of rain that falls is the result 2,241,000 calories of hat removed from the surface and sent into space.
The amount of water in the air varies from 0 to .5% depending on the temperature. The amount of CO2 in the atmosphere remains at 442 pm regardless of the temperature.
When the water level is .22% it is 50 times greater than the CO2 level. In order to compensate for the cooling by water each gram of CO2 must add 30,000 calories to the surface to keep the temperature constant. If the humidity is not 100% water will continue to evaporate and when it reaches .33% the gram of CO2 must now add 45,000 calories to the surface to keep the temperature constant. If the surface is heating the CO2 and the temperature is constant where does this additional 5,000 calories come from?
Herb
Jerry Krause
| #
Hi PSI Readers,
“It is experimentally shown that the thermal radiation from a transparent, colder solid has the capacity to influence a solid warmer than it to become even warmer, under the right circumstances.”
Why not write: An experiment shows that the thermal radiation from a transparent, colder solid causes a solid warmer than it to become even warmer. Then immediately describe the right circumstances so another experimenter can do the same experiment and confirm the same result+++.
Is not a reproducible result the fundamental basis of the scientific method?
Have a good day
Reply
Dale Cloudman
| #
Hi Jerry,
The circumstances are described in the paper such that anybody can reproduce it. Someone who lacks the patience to read the full paper will certainly not be someone who would put in the much greater effort required to reproduce the experiment. They are more likely to spend their time doing something much less time consuming, like critiquing sentence construction.
Cheers,
Dale
Reply
Jerry Krause
| #
Hi Dale,`
I quoted from the abstract. And what I quoted implies one doesn’t always get the same result. Thank you for responding but even you response still doesn’t describe these special circumstances. It seems to me that the purpose of your response was to be critical of me. John O,Sullivan, the founding editor of PSI taught me to be brief because the readers of PSI didn’t, have time read long articles. Which advise I unfortunately didn’t immediately take.
Later I discovered that Einstein had stated: “If you can’t explain it simply, you don’t understand it well enough.” So, I now first scan the length of an article and don’t waste my time for reading for understanding.
Have a good day
Reply
Dale Cloudman
| #
Hi Jerry,
I am surprised to see you writing this considering the nature of your own article which you linked to in another comment. You mention comparing its experimental results, yet the only results in it are remarkably sparse, about midway through an article linked from that article. It is certainly not clear and concise.
The “right circumstances” are explained in the title, namely in this experiment being exposed to sunlight. More generally it would be when the hottest surface is being heated by an external heat source. Of course running the experiment in the same circumstances will produce the same result. I don’t see how the abstract implies it wouldn’t.
Best,
Dale
Reply
Jerry Krause
| #
Hi Dale CLOUDMAN,
I totally agree (understand) that a colder cloud surface at the bottom of a cloud scatters a portion of the radiation being constantly emitted upward by the earth’s surface regardless of the temperatures of either surface. Is this your point? See how simply this can be stated.
Have a good day
Reply
Dale Cloudman
| #
Hi Jerry,
That is not the point at all. I suggest you read the paper before commenting further.
Best,
Dale
Jerry Krause
| #
Hi James McGinn,
Are you familiar with “Strokes of lightning are preceded by. “feelers” which guide the main bolt to its objective, according to Karl B. McEachron, high voltage electrical engineer …?
Have a good day
Reply
Jerry Krause
| #
Hi Herb,
Just read “Eighty percent of our atmosphere is nitrogen–an essential food for plants. … Before plants can take life from it, it must undergo what our food undergoes in our digestive machinery: a series of chemical reactions. Lightning touches off the series. … Air particles are made white-hot by lightning. … Under this intense heat, the nitrogen combines with the oxygen in the air to form nitrogen. oxides that are soluble in water. The rain dissolves the oxides and carries them down to earth as dilute nitric acid.” (The Miracle of Lighting) by Ira Wolfers in Our Amazing world of Nature published by Readers Digest, 1969)
Have a good day
Reply
Herb Rose
| #
Jerry,
There is a layer of the atmosphere, above the stratosphere, that is composed of nitrogen-oxygen molecules. The molecules form when uv energy from the sun (not the surface of the Earth) provides the 492,000 joules/mole to split oxygen molecules and partially split the triple bond of N2 molecules, allowing the oxygen atoms to combine with the nitrogen molecule. Above 90 km the ratio of oxygen to nitrogen increases as the energy from the sun is so great that the nitrogen-oxygen molecules are not stable and the oxygen becomes lighter oxygen atoms. I’ve pointed this out to you on several occasions but you steadfastly believe that it is extremely “cold” at higher altitudes because the thermometer is measuring the mean kinetic energy of the molecules.
Reply
Jerry Krause
| #
Hi Herb,
And you have never seen the lightning which I have just drawn to your attention that you seem to be ignoring?
Have a good day
Reply
Geraint
| #
oh dear. You didnt eliminate convection at all. The experiment was totally open to the atomsphere. Therefore, a hotter plate, would heat up the air and then redirect the heated air back to the plate below causing it to warm as lower rates of convective cooling are possible with warmer air compared to cooler air.
Are you so dumb as to disagree with that statement?????????????
Are you really such a dullard, you need me to do this in the correct manner for you??????????????????
If you performed this test in a vacuum, you would have found no difference what so ever, as back radiance does not occur in reality, as back radiance is a non valid concept and one which is easily disproven. Silly person.
Good evening.
Reply
Dale Cloudman
| #
Hi Geraint,
Clearly you did not read the “Confirming the Radiative Nature of the Effect” section of the paper before posting. This is unfortunate as the strength and vigor and insulting manner in which you directed your comment, leads me to already not take you as a serious person.
Be that as it may, if you have a particular critique of that section — which showed beyond a reasonable doubt that the effect was due to radiation and not convection (that I obviously considered) — then I am open to hearing it.
I also direct you to this Green Plate experiment performed in a vacuum, that also shows the validity of the effect of colder objects’ thermal radiation: https://drive.google.com/file/d/1DkLzG9EafmXk_-ALbq_gpC1xJuphaszG/view?usp=sharing .
On top of this, consider a soon-to-be-published experiment where I attained a temperature of 150C (302F) for a surface heated only by the Sun. This far exceeds the max limit possible of 121C if the colder thermal radiation has no effect.
Best,
Dale
Reply
Herb Rose
| #
Hi Dale<
You are confused by the different meanings of “radiation” and “convection”.
Radiating heat can refer to an object losing heat or a manner in which heat is lost. An object loses heat by convection or conduction when there is contact with another object and a heat transfer. With this method there is an equalization of energy. between the masses according to the law of conservation of momentum M1V1 +M2V2 = M1V3 =M2V4. “Convection” also means the rising of warmer air in the atmosphere.
Your plastic films will prevent the rising of air but when molecules strike them they will absorb and transmit energy just as like other object.
The rate at which the energy transfers is known as conductance and it depends on the nature of the matter. If you were to substitute aluminum foil for the plastic and put a heat source under the box equal to simulate the heat coming from the sun the temperature in the box would be much lower than the results you got with the plastic film.
With the transfer of heat by radiation there is never contact between the objects and because the energy emitted from an object decreases with distance, there will never be equalization as occurs in convection.
The rate of energy depends on the difference between the temperature of the objects and declines as the difference between their temperatures declines. Theoretically, just like by going half the distance to an object, means never reaching the object, there will never be a time where the radiating of energy from the object absorbing energy will equal the energy being absorbed.
The reason why objects not in contact with each other do equalize (even in a vacuum where there is no convection), even though one object does not absorb the wavelength of energy being emitted by the other object, is because both objects are equalizing with the energy field they are in, not each other.
HerB
Reply
Dale Cloudman
| #
Hi Herb,
“You are confused by the different meanings of “radiation” and “convection”.
Radiating heat can refer to an object losing heat or a manner in which heat is lost.”
There’s no confusion here – by “radiating heat” I am always referring to thermal radiation, ie hot objects emitting light, and not convection, which is a different heat transfer mode.
“The reason why objects not in contact with each other do equalize (even in a vacuum where there is no convection), even though one object does not absorb the wavelength of energy being emitted by the other object, is because both objects are equalizing with the energy field they are in, not each other.”
It’s unclear the point you’re making. The experimental result is thoroughly explainable by the warmer object’s radiative loss being lessened due to its absorption of the colder object’s radiation. If you are proposing an alternate explanation that has exactly the same outcome, then there’s no practical difference between the two as both theories would make identical predictions and have identical consequences
Best,
Dale
Reply
Herb Rose
| #
Hi Dale,
Molecules in the troposphere are colliding hundred of thousands of times a second. When a collision occurs energy equalizes and any radiated transfer of energy between the objects would cease, both now having the same energy.
I will again repeat the reason for your results. When the interior heats the glass that glass will radiated energy until the energy being lost equals the energy being gained. The black bottom of the box is losing energy through the glass. When you put a hotter piece of glass on the box, the glass is now losing more energy than it is absorbing from the bottom. Since the box is now losing less energy (as the loss through the styrofoam remains the same) the temperature in the box will rise due to it losing less energy but still gaining the same amount from the sun.
It is not gaining energy from the warmer glass.
Again, if you have two sources of heat, one greater than the other when you move the thermometer from one source towards the other source the temperature will decline as the energy from the sources decreases. with distance. You will reach a point where the temperature begins to rise because the energy is now coming from the other source (which also declines with distance) and it is now greater than the energy coming from the other source.
If you were to block or turn off one of the sources while the thermometer was recording the temperature of the other source, the temperature reading would not change.
Energy only flows from greater to lesser and when equilibrium is reached the flow of energy stops.
Herb
Dale Cloudman
| #
Hi Herb,
(Since the box is now losing less energy (as the loss through the styrofoam remains the same) the temperature in the box will rise due to it losing less energy but still gaining the same amount from the sun.
It is not gaining energy from the warmer glass.)
Yes, we’re in agreement, this was my conclusion as well, as written in the paper.
By the same principle, the fact that the surface of the earth is losing less energy to the sky radiatively, than it would if it were faced with outer space — this means the surface is warmer than it otherwise would be as heated by the Sun, taking the entire thermal radiation spectrum into account
The question then is just what additional effect CO2 has, this has never been sufficiently proven and this is where the problem with climate alarmism lies.
Best Dale
Herb Rose
| #
Hi Dale,
You state that the increase in temperature in the box, when the warmer glass is added, is due to the thermal radiation from the warmer glass. The thermal radiation from the glass has nothing to do with it Addition is not the absence of subtraction.
The surface of the Earth is not losing significant energy into space nor are the gases storing heat. It is the upper atmosphere that is gaining energy from the sun and losing energy into space.
The surface of the moon is +250 in sunlight -230 when in the dark. The GHGT starts with the premise that the atmosphere is not gaining energy from the sun and is being heated by the Earth. If the atmosphere is not absorbing energy from the sun why isn’t the surface temperature +250? Are you claiming that the heat contained it the atmosphere is somehow preventing the sun from heating the surface?
It is water that s cooling the Earth not any gases. Both Venus and Mars have atmospheres with 90+% CO2. One is hot, one is cold the CO2 has nothing to do with it.
If you divide the temperature at an altitude by the density at that altitude you will see that the ke of the molecules increase in a straight line in the troposphere, where water is present, and in an exponential curve at higher altitudes.
Herb
Dale Cloudman
| #
Hi Herb,
“You state that the increase in temperature in the box, when the warmer glass is added, is due to the thermal radiation from the warmer glass. The thermal radiation from the glass has nothing to do with it”
Earlier you wrote: “Since the box is now losing less energy (as the loss through the styrofoam remains the same) the temperature in the box will rise due to it losing less energy but still gaining the same amount from the sun.”
By what manner is the box losing less energy due to the hotter glass being swapped in? Remember that the hotter glass is spatially separated from the box and is not covering the box and does not enclose the box, and further that the effect cannot be due to conduction or convection through the air, as detailed in the paper in the “Confirming the Radiative Nature of the Effect” section.
“The surface of the Earth is not losing significant energy into space […]”
Yes… so if it were faced with outer space directly, it would be losing more energy, no? And be cooler? Therefore the fact that it isn’t, means it is warmer than it would be due to this.
“The GHGT starts with the premise that the atmosphere is not gaining energy from the sun and is being heated by the Earth.”
Of course the atmosphere (particularly very high up) absorbs some sunlight, but most of it hits the surface directly. Of ~1300 W/m2 at the TOA, about ~1000 W/m2 hits the surface. 300 W/m2 is absorbed by atmosphere, 1000 W/m2 hits the surface.
“If the atmosphere is not absorbing energy from the sun why isn’t the surface temperature +250?”
Because the surface loses heat convectively to the air, thus warming it up. There is also some radiative exchange, but the major mode of transfer at the surface is convection.
“Are you claiming that the heat contained it the atmosphere is somehow preventing the sun from heating the surface?”
Not sure how anything I wrote could be construed as claiming this
Best,
Dale
Herb Rose
| #
Hi Dale,
Energy equalizes with mass. At 80 km the density is 00005 kg/m^3. At sea level the density is 1.2 kg/m^3. If the same energy were at both altitudes the ke of a molecules at 80 km would be 60,000 time the ke energy of a molecule at sea level. It is energy (v^2) that is being radiated not mass. 1,200 gas molecules will radiate more energy than 1 molecule so even if that molecule has greater energy it may not radiate more energy.
The bottom theGrand Canyon is always 10 degrees hotter than the top of the Grand Canyon. If the molecules had more ke they would rise to the top so the greater temperature is due to more molecules transferring energy.
The surface of the Earth is not heating the air. The air is heating the surface (greater energy). While the gas molecules in their do not absorb visible or IR (except CO2 which absorbs IR at -80) they do absorb shorter wavelengths (gamma and X-ray (which create the ions in the ionosphere) and uv (90+ %) and convert this radiation ke of the O2 and N2 molecules. These molecules, with there greater energy (v^2) add energy to the surface (law conservation of momentum). It is water (even when it is ice) that reeves energy from the surface.
Even though the warmer glass is not on the box the transfer of heat is still convection not radiation. Again I urge you to read the PSI article (Interview with Tom Shula) and learn about the Pirani gauge. The only way you can eliminate convection is with a vacuum. If it weren’t for the Pirani gauge you would not have a computer.
During a solar minimum the Earth gets colder. The amount of visible and longer waves doesn’t change as they are produced by the surface of the sun but the shorter X-ray and uv radiation are produced in solar flares. During a solar minimum the energy heating the surface does not change it is the lose of energy from the atmosphere that causes the cooling.
The atmosphere is not absorbing or blocking energy from the sun so what is preventing the surface from reaching +250 like the moon?
Herb
Dale Cloudman
| #
Hi Herb,
“Even though the warmer glass is not on the box the transfer of heat is still convection not radiation”
Again I ask you to read “Confirming the Radiative Nature of the Effect” which completely rules out convection. If you disagree, then for the discussion to proceed further you will have to write a specific critique of this section. Otherwise the conversation will not be able to continue productively.
To summarize it here: if it were convection then the temperatures near the glass should warm up first, which then traverses down to the black bottom and causing it to warm up. But what was observed was the black bottom warming first, followed by the bottom-most layers warming first then the next layer up etc. This is impossible to happen via convection, how can the bottom warm before the air above it warms?
“The surface of the Earth is not heating the air.”
Ehm… this is patently false. The sun heats the surface, and the surface heats the air. You can notice this by seeing that the air above a warm patch of ground in the sunlight is warmer than the air above a cool patch of ground or in the shade.
Best,
Dale
Dale Cloudman
| #
“””The surface of the Earth is not heating the air.”
Ehm… this is patently false. The sun heats the surface, and the surface heats the air. You can notice this by seeing that the air above a warm patch of ground in the sunlight is warmer than the air above a cool patch of ground or in the shade.””
A better point is: the air is always cooler than the surface, cooling with altitude according to the lapse rate. Heat flows from hot to cold, not cold to hot, so the air can’t be heating the ground.
Herb Rose
| #
Hi Dale,
I thought the sun was the source of energy heating the bottom of the box. That, is then heating the bottom of the glass.
You have a problem with what a thermometer measures.
In a gas you are not measuring the flow of energy from one medium to another medium but how mush heat the thermometer is absorbing.
The temperature of objects not a gas is measuring the equilibrium point where the energy absorbed from one medium is equal to the energy being radiated into another medium.
When gas gains heat it expands and there are fewer collisions and less transfer.of energy. You cannot compare the temperature of a gas to the temperature of other objects .Again the comparison of a 100 C oven to 100c water. There are few molecules in the oven and they gain energy by colliding with a red hot heating element or a flame.
The water is gaining ke from a container (never red hot) so there the ke is far less than the air molecules in an oven. (You can boil water in paper cup in a campfire because the water prevents the paper from reaching its ignition temperature.) You can cook food faster in the boiling water faster than in an oven because there are more of them transferring energy (even they have less energy).
Do you think that even when water on the surface and the air above it have the same temperature, that the water in the air has the same ke as the liquid water below? If the water in the air has greater ke than the water below when it collides with that water will it add energy to the water below? Do you believe the gas molecules in the atmosphere have greater velocity than molecules on the surface? If so those air molecules will add energy to the surface. Do you believe the law of conservation of momentum that governs the transfer of energy by convection?
Thin film2 is hotter and gains heat for the same reason that the air near a ceiling is hotter than the air below. When the air hits a barrier, preventing it from rising the energy collects as the change of conductance of the different medium slows the flow of energy. The addition of the heated glass above thin film2 also slows the flow of heat through the film acting as additional insulation.
Herb
Dale Cloudman
| #
Hi Herb,
“Thin film2 is hotter and gains heat for the same reason that the air near a ceiling is hotter than the air below. When the air hits a barrier, preventing it from rising the energy collects as the change of conductance of the different medium slows the flow of energy.”
Yes
“The addition of the heated glass above thin film2 also slows the flow of heat through the film acting as additional insulation.”
If the heated glass slowed the flow of heat convectively, it would have to be by warming the air. But the air in between the top film and the glass did not warm, it actually was cooler.
Further, if this convective effect caused the bottom to warm up, that would only be because the air inside the bottom chamber would heat up first, which would have to be by the thin film heating up first. But it didn’t happen, the bottom got warmer first. This is simply impossible to explain except via thermal radiation being the culprit.
How can the effect be convective if neither the air or the films in between the heated glass and the bottom got warmer before the bottom did? It just does not make physical sense.
Let’s focus on just this part before discussing any others
Best
Dale
Herb Rose
| #
Hi Dale,
I’ve the discussion over so we don’t have to scroll up so far.
Look at the means that energy is being transferred.
Radiated energy from the sun is being converted to ke by the black interior. Air molecules are striking the surface and gaining energy, which causes them to rise. They meet the plastic film striking it and transferring the energy. to the bottom of the film This energy is then transferred by conduction through the plastic.
The rate of conduction is determined by the material. Plastic, with its covalent bonds restricts the amount of motion/energy being transferred. resulting in a build up of air with ke on the one surface. keeping energy there and slowing the rate of energy transfer. (If you had used glass there would be a faster transfer and lower build up.. If you put your hand against a piece of glass it feels cool as it transfer energy away from your hand. If you put your hand against a piece of plastic it feels warmer because the plastic is transferring energy at a slower rate.)
Air molecules between the films are striking the surface of the lower film and absorbing energy from again transferring energy by convection. When these air molecules strike the upper film they transfer energy to it to be transferred again by conduction. Because the conduction rate of both films are the same there is not the build up of heat that occurs at the bottom film. The air above the higher film converts the energy from the top of the upper film into convection, which then transfers it to the glass.
When you substitute a hotter piece of glass the molecules striking it retain more of the ke energy they received from the bottom film, transferring less heat causing the flow of energy to slow.
Herb
Reply
Dale Cloudman
| #
“When you substitute a hotter piece of glass the molecules striking it retain more of the ke energy they received from the bottom film, transferring less heat causing the flow of energy to slow.”
Ok, and what will increase first in temperature if this happens?
The air between the glass and the top film
The black bottom, even before any of the temperatures of the films or air in between are affected
Herb Rose
| #
Hi Dale,
I don’t understand you question.
When any matter, be it the box, an air molecule, the plastic films, or the glass absorb energy they become a source of energy. The energy are radiating will decrease with distance from the source until it reaches equilibrium with another source of energy. Energy will flow until stasis is achieved, where the energy field being radiated by an object can no longer expand..
Radiating energy decreases the strength of the energy field from an object by spreading it over a larger area. (Basically making the object larger. If the hydrogen atoms that fill deeps space are 2 cm apart the size of the hydrogen atoms is 1 cm.). An object only loses energy when that energy becomes part of another objects energy field.
Herb
Dale Cloudman
| #
Hi Herb,
Look at Figure 4 and Figure 5 and read the words around them. Explain to me how those results are possible if the hotter glass increased the temperatures of everything below it because of convection and not radiation.
Best,
Dale
Herb Rose
| #
Hi Dale,
The bottom is where the energy is being added by the sun. From there it decreases with distance. When the flow of energy encounters a barrier it cause the flow to slow, meanwhile energy is still continuing to be added at the bottom. Why do you think that the temperature would not be higher closer to the source?
As a gas expands fewer molecules transfer energy to a thermometer. A rise in temperature could be from an increase in energy (t) or an increase in number (n). (PV=nrt) The air molecules at the bottom of the Grand Canyon are not receiving more energy from the sun than those at the top and yet the temperature is 10 degrees higher.
Herb
Dale cloudman
| #
Hi herb,
This seems to not be going anywhere. I don’t know how it’s possible so many messages have been exchanged and you still are not understanding my question.
Before we continue can you confirm that you read 100% of the words in the section “Confirming the Radiative Nature of the Effect”? Only once you tell me you have read the entire section will I proceed
Best
Dale
Herb Rose
| #
Hi Dale,
I have seen no evidence that you have watched the “interview with Tom Shula” and made any attempt to dispute his evidence. I have seen no evidence that you have converted the temperature at altitude from temperature per unit volume to temperature er constant number of molecules. I have tried to explain why your experiment is wrong but you seem to be once of those who think that your belief makes something true.
Herb.
Dale cloudman
| #
Hi Herb,
I can’t believe we spent so long discussing when you didn’t even read the “Confirming the Radiative Nature of the Effect” section, which precisely addresses your criticism and which has been the entire point this whole time
I’m not sure how you can think it’s sensible to discuss the experiment without reading it, especially when the experimenter is drawing specific attention to a point.
Again, the section “Confirming the Radiative Nature of the Effect” shows that the effect is not due to convection like you have said, but rather due to radiation. From this point I will only be responding to specific critiques of that section as there’s just no point in continuing when you won’t read it.
Best,
Dale
Herb Rose
| #
Hi Dale,
I’ve read that section several times an have been discussing it.
The 30 C glass is being heated by the bottom which is heated by the sun. That 30 C of the glass is the result of energy lost from the box. When you replace that glass with the 70 – 100 C glass the box is no longer losing energy to the glass. The flow of energy has stopped at an equilibrium point below the glass. The sun is still heating the bottom causing it to get hotter. The air in the box expands resulting fewer collisions between air molecules and the thermometer causing the reading of the thermometer to decrease.
I have said multiple times that energy only flows from higher to lower. The energy coming from the heated glass is not flowing to the bottom of the box and making it hotter. The sun adding energy to the bottom and because it is no longer losing energy to the glass it is getting hotter.
You are using your belief that energy from the hotter glass flows to the bottom to try and make the evidence provide support to your belief. when it doesn’t.
The fact that you are unwilling to look at any evidence that conclusively shows your beliefs to be wrong shows that you are ego driven rather than looking for reality.
End discussion.
Herb
Dale Cloudman
| #
Hi Herb,
“The sun adding energy to the bottom and because it is no longer losing energy to the glass it is getting hotter.”
Yeah we already talked about and agreed on this point. This is not what is under contention. I’m at a loss as to how you can think this was not something we already agreed on.
Are you not able to see that what we are disagreeing about is whether the “no longer losing energy to the glass” is due to convective heat transfer or radiative heat transfer (as in exchange of infrared light)?
Are you able to see that this is what we’ve actually been discussing?
Once you are able to see it — will you be capable of comprehending how for the effect to be due to lesssened convective heat transfer, the effect of higher temperatures would be first seen at the glass, then the upper air, then the lower air, then the top film, then the bottom film, and finally the black bottom? Because the air would heat up at the top first. I fail to see how the bottom would react with a higher temperature before even any of the films have reacted with a higher temperature.
I’ve asked this question many times and your answers have led me to believe that you don’t even understand the question. The bottom being hotter than the top in absolute terms is not what is under discussion. It’s how the bottom reacted FIRST to the hotter glass, before anything in between reacted. Talking about the temperature at different altitudes is totally irrelevant as this is all within the span of 50cm.
In short I have seen no reply of yours even indicating that you’re understanding the point of the bottom being FIRST to react to the hot glass being swapped . This led me to believe you haven’t even read the section which details this point. Maybe you can clarify now you can see where I’m not understanding you.
Best,
Dale
Geraint
| #
Read this if you want to see how proper comparison tests are performed.
https://www.thepostil.com/global-warming-is-not-science/
Reply
Jerry Krause
| #
Hi Geraint,
I went to your link but it quickly became obvious that the author was trying to study NATURAL PHENOMENA in a laboratory setting. Yes, artificial instruments must be invented to study the natural phenomena cannot be brought into a laboraotory.
Have a good day
Reply
Max
| #
(Regretfully) I found a similar result with the classic two-pane experiment in a vacuum bell, one irradiated with a lamp, and the other getting the infrared radiation from the first pane. I used thin black aluminum for the two panes under 7×10-4 bar vacuum and compared the result with the “back-radiation theory”‘s prediction finding good agreement. The goal of my experiment was to prove the opposite…but I got this surprise. I’m going to share some material soon. The only doubt I have about my set-up is whether there was some reflection from the surrounding wall that could affect the temperature of the second pane. I plan on repeating the experiment outside to avoid this.
Reply
Dale Cloudman
| #
That’s good confirmation of this experiment I came across recently, that was done in 2018: https://drive.google.com/file/d/1DkLzG9EafmXk_-ALbq_gpC1xJuphaszG/view?usp=sharing .
I also was seeking to disprove back radiation, but alas. Alan Siddons wrote very well-constructed rhetorical arguments but the theory he proposed does not stand up to experimental verification. Such is the progress of science!
Reply
Max
| #
…forgot to mention that to have “visible” results it is important to keep the panes close enough so that radiation does not “escape” out of the gap, otherwise the temperature boost is hardly detected. I used two Aluminum panes approximately 50x70x1 mm, 5 mm distant from each other, the temperature increase (backradiation effect ?) is about 5 degrees after the heating transient comparing with the “single” pane
Reply
Jerry Krause
| #
Hi Max,
Suggest you go to this link (https://principia-scientific.com/the-horace-de-saussure-hot-box/) of a June 23, 2016 article about the Horace de Saussure Hot Box.
Have a good day
Reply
James Bernard McGinn
| #
DC:
Although the thermodynamic possibility of colder objects causing warmer objects to increase in temperature has finally been conclusively demonstrated,
JMcG:
This is the right conclusion.
James McGinn / Genius
Reply
Herb Rose
| #
Hi James,
Right, but only through convection where the law of conservation of momentum is in effect.
With radiation a colder object cannot increase the temperature of another object because the flow of energy stops at the equilibrium point between the two objects snd never reaches the warmer object.
Herb
Reply
Herb Rose
| #
Dale’s correct conclusion contradicts the premise of his argument.
When you tell him that the higher velocity of the molecules in the atmosphere add energy (heat) to the surface of the Earth he denies that this could be true because the surface has a higher temperature. He refuses to consider the possibility that the thermometer is an inaccurate indicator of the ke of molecules so he believes the molecules at higher altitudes have less energy because of the lower temperature measured by the thermometer.
Herb
Reply
Dale Cloudman
| #
Hi Herb,
The atmosphere is irrelevant for the purposes of discussing what the experiment showed. The box is not the atmosphere.
I will elaborate on my point. Say you have a 10 meter long rectangular block of steel. At the left end is applied a constant energy input. The right end has a plate on it that is held fixed at 30C. Say the left end is at 100C in this steady state.
Now we swap the right plate out immediately for one held at a fixed 60C. What will happen?
1- the left 100C end of the block of steel will start heating up FIRST, then this increased temperature propagates to the right, heating up the in-betweens.
2- the right 60C end of the block of still will start heating up FIRST, and the increased temperatures propagate to the left.
The answer should be obvious: the second situation will happen, because the left side doesn’t magically “know” what the temperature of the right side is. The system has to update in a sane, sensible, physical manner. As the heat increased on the right side, the conductive equilibrium will propagate from the right to the left.
Now for convection it’s similar. Say you have a 10 meter tall hollow tube, the top fixed at 30C, with constant energy input on the bottom. In this state the bottom is at a steady 100C. This 100C is reached because the bottom is heating the air, which then rises, which then hits the 30C ceiling, cools, and then falls. The 100C is the steady state at which the flow into the air matches the cooling from hitting the 30C.
Now you swap out the top for one fixed at 60C. What will happen convectively?
1- The bottom will heat FIRST, without ANY OF THE AIR in between being affected at all. THEN the air will start to heat more.
2- The top air will heat FIRST. When the air on the top is hotter, now the air rising from below is cooling less. Eventually this effect propagates downwards until the bottom starts to heat up.
The answer again is obvious — 2 is what will happen. The bottom doesn’t magically “know” that now the top is warmer, the convection is due to the air temperature and the air temperature doesn’t magically update on the bottom right away just because you swapped the top out.
But in the experiment, 1 happened, not 2. How is it possible? The answer is because the effect was not due to convection, but due to the thermal light radiation emitted by the hotter glass, as this light travels instantly and does not have to propagate through the air. There is just no way to explain the result without this, and nothing you have written contradicts this or explains it.
As further proof the borosilicate glass in-between shows the effect was radiative, because if the effect was convective it would still happen even with an extra layer inbetween (why would it work with 2 layers in between but not 3 layers?), but it didnt because the borosilicate absorbed the extra radiation first.
Hope that clarifies
Best,
Dale
Reply
Jerry Krause
| #
Hi PSI Readers,
Lost in the debate about global warming is the original proposed magnitude of the warming by atmospheric greenhouse gases. The following is a common and accurate generalization. Earth’s atmosphere acts as a blanket, making our planet habitable. Without any heat-trapping greenhouse gases in our atmosphere, Earth would be a frozen ball of ice. For 100% Sun brightness and an albedo of 31%, Earth’s temperature as calculated from the simulation is 253 K, or -20°C. In reality, the atmosphere’s effect is powerful and warms our planet by an average of increasing the globally averaged temperature from -20°C to a much more habitable 15°C.
It is this magnitude, 33-35 K (33-35°C), that causes the idea of the GHE to be given serious consideration. and allows me to point to the fact that this ideal (reasoning) is absolutely wrong. For at weather stations two temperatures (atmospheric temperature. AT, and atmospheric dew point temperature, ADPT, wh are commonly measured side by side, the same distance above the earth’s surface. And it is an observed fact (Scientific Law) that the AT has never been measured to be less than the ADPT when these two temperatures are measured at the same time. So the AT cannot be even one degree less than the ADPR.
Have a good day
Reply
Dale Cloudman
| #
Hi Jerry,
That’s an excellent point. The alleged +33K is clearly the effect of the entire atmosphere, not just the so-called greenhouse gases. This includes the full effect of the adiabatic lapse rate, water with its latent heat, etc. It is certainly a clever trick to make the GHE seem more prominent than it really is.
Best
Dale
Reply
Jerry Krause
| #
Hi Dale,
Why no comment about the observed facts that the AT is never been measured to be less than the ADPT when both are measured at the same time?
Have a good day
Reply
Dale Cloudman
| #
Hi Jerry,
That fact (“that the AT is never been measured to be less than the ADPT when both are measured at the same time”) certainly is contained within me saying “That’s an excellent point”, along with “The alleged +33K is clearly the effect of the entire atmosphere” and “water with its latent heat” as well (specifically “latent heat of vaporization”, or more precisely the “heat of condensation”).
Best
Dale
Reply
Herb Rose
| #
Jerry,
As explained to you on numerous occasions the Dew Point is NOT a measured temperature. It is an expression of the water content of the air, just like humidity is. The reason the water content is expressed as a temperature is because as long as the level of water remains the same the value remains the same, while with humidity as the temperature changes so does humidity value, even as the water content remains the same. Your “law” simply says that the humidity is never over 100%.
The temperature does fall below the dew point producing clear sky precipitation (diamond dust) which is where the water in the atmosphere freezes as ice before it can condense into visible droplets.
I assume this explanation will again cause you to stop commenting until the next opportunity arises where you will again produce this “law” just as you continually repeat your inane quotes from famous people.
Reply
Jerry Krause
| #
Hi Herb,
The dew point temperature is temperature of a solid surface on which dew can be observed to form. The temperature of this surface can be measured by an instrument termed an infrared thermometer..
Herb, how is it that I know nothing and you know everything?
Have a good day
Reply
Herb Rose
| #
Jerry,
Thanks for the information.I thought they’d do it by heating a set amount of desiccant to see how much water it lost and at the same time resetting for the next measurement.
Reply
Jerry Krause
| #
Hi Herb,
Thank you for accepting my explanation.
Have a good day
Reply
Jerry Krause
| #
Hi Herb and Andy Rowland,
Please read this link (https://principia-scientific.com/ancestors-tracing-history-scientific-method/). I believe it may have been before either of your times here at PSI.
Have a good day.
Reply