Wednesday, June 23, 2010

Why conventional Greenhouse Theory Violates the 1st Law of Thermodynamics

Alan Siddons, a radiochemist, has forwarded a new essay explaining with simple examples why conventional greenhouse gas theory violates the conservation of energy demanded by the 1st law of thermodynamics. He writes, "Here’s another essay that deals with the silly heating-via-reradiation hypothesis, my critics be damned. It repeats the same point: If a body can heat itself by absorbing its own radiation, and thereby emit more radiation, then it is necessarily emitting more radiation than it’s receiving."


"One is disturbed each day by verifiably untrue statements touted as incontrovertible facts about hot-button issues." -- Richard S. Lindzen
"CO2 absorbs in the infrared and reradiates heat downward, thus heating the earth." -- Richard S. Lindzen

Okay then, let's examine that particular "incontrovertible fact." You know how a solar oven works.

By multiple reflections, the interior is exposed to more rays from the sun, so the food gets much hotter than it'd get otherwise. The operant principle is akin to how stage lighting works.

In the zone where the beams intersect, the photon density is greater so more light is delivered. It may be obvious, but it's worth pointing out that the two beams pass through each other — they do not clash like the Light Sabers in Star Wars movies. Worth noting too is that whereas more light makes the target brighter, i.e., increases the amount of light being reflected, it also increases the amount being absorbed, thus heating the target. Which of course is how a solar oven works.
Now, you’ve been told that terrestrial infrared is re-radiated back to the earth’s surface and heats it. So let’s test this notion by turning a spotlight off and seeing if we can mimic a second spotlight with a mirror, which will provide re-radiation. After all, a mirror has no idea of what it’s reflecting — it could be visible light or infrared 'heat rays.' It makes no difference to a mirror.
Well, is it any surprise that nothing happens? Reflecting light back to the bright spot doesn’t make it brighter. (You can try this at home on a wall.) 


Close as you hold the mirror to the bright spot, there’s no effect. You might notice, though, that offsetting the mirror a bit can illuminate a zone that’s in shadow. In this case, light reflected from the bright spot brightens a darker area. But the mirror cannot make the bright spot brighter.
The lesson I draw from this real-world example is that radiant energy can only light something that has less radiance. Brighter illuminates darker. You can get a sense of this by omitting the spotlight altogether and just imagining a surface radiating light on its own. Holding a perfect mirror directly above it, you'd see something like this:

There's no difference between them, and it shouldn't need explaining that the mirror image is not illuminating the very object that it's reflecting. But if the mirror isn't illuminating that object, the mirror isn't heating it either. To do that, the image in the mirror would have to be brighter, like the blazing image a solar oven provides. The same principle applies to conductive heat transfer: Higher temperature matter heats lower temperature matter. At equal temperature no heating occurs. Directing an object's own radiance back to it, then, doesn't do a thing and cannot make it more radiant. But from this proceeds another ramification.
A perfectly reflective 'white body' is the opposite of a blackbody. It can't be heated by radiative means. So suppose a mirror reflects a tiny amount of energy onto a white body.

In the scenario above, 1 unit of light hits the surface yet a total of 1.1 units are reflected, thus creating energy out of nothing and contradicting the 1st law of thermodynamics. Since this is impossible, it demonstrates that second-hand reflected light cannot be regarded as part of the object’s "radiation budget." Indeed, this reflection is only the object's own radiance in another guise, not an additional source of light.

Now consider a blackbody.

Without the mirror, a perfectly absorptive/emissive object would release one unit of light in response to one unit of irradiance. But if the blackbody absorbs another tenth from the mirror, then it's compelled to emit 1.1 units in all. To repeat, this is impossible. So this indicates that re-radiated infrared cannot be regarded as part of an absorbing object’s "radiation budget" either, or else the 1st law is wrong. This re-radiated infrared is only the object's own thermal emission in another guise, not an additional source of heat.

Putting these results together, one must conclude that "back-radiation" of any sort, whether visible or infrared, will simply not show up on an external radiometer because it represents no "extra" energy in the first place and it induces no illuminative or thermal effect. Re-radiated energy is neither reflected nor absorbed by a surface, therefore, because either outcome would point to additional energy that would register on a detector, signaling that more radiant energy is leaving the system than what is going in.

Summing up, re-radiated light has as much chance of enhancing surface energy as you have of standing in a bucket and lifting it off the ground. In other words, it can't be done. Following a principle similar to conductive heat transfer, and also similar to electrons needing a lower electrical potential in order to move, radiant energy can exert an effect only on something that is radiating less. Radiant energy is like water; it can only flow downhill. 

82 comments:

  1. Many thanks for this. Alan Siddons is again doing brilliant work to expose the nonsense that is the greenhouse gas theory. This latest analysis is a perfect addition to the ground-breaking paper, 'A Greenhouse Effect on the Moon.'
    http://www.ilovemycarbondioxide.com/pdf/Greenhouse_Effect_on_the_Moon.pdf

    Siddons and his co-authors, Dr. Hertzberg and Hans Schreuder, may be ignored but their detractors cannot credibly refute their study.

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  2. This is correct, of course, but radiating gases (most importantly water vapour) can reduce the cooling rate after sunset, as everybody has probably experienced. When there's a high air humidity, earth loses heat at slower rate when compared to dry air and so we can say that statistical average of day-night temperature cycle is somewhat higher, right? This can be false only in the case if the same GHG-s are reducing day-time warming rate, as some authors have decided.
    An important question is then how much raising CO2 concentration is affecting cooling rate, when compared to water vapour and what is the role of the GHG-s at daytime.

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  3. Richard GarnacheJune 25, 2010 at 7:54 AM

    I have been wrestling with this concept for some time. Thank you for the very simple demonstration.

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  4. Congratulations, Alan, by this excellent article, and thank you very much for sharing your knowledge with us!

    As we dig more and more on the AGW idea, we find more and more errors and flawed concepts.

    I wanted to investigate a little more on the power of the carbon dioxide for “warming up” the planet and I found something very peculiar. As the effective pressure of the atmospheric carbon dioxide (which depends on its density in the atmosphere) increases, its total emissivity decreases; this makes the carbon dioxide acts as a coolant, not as a warmer. See a graph on this coolant effect here:

    http://www.biocab.org/PE_vs_dT.jpg

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  5. Nasif,

    Yes - see post on Adiabatic Theory:

    http://hockeyschtick.blogspot.com/2010/04/adding-to-list-1-2-3-4-5-6-7-8-9-10-11.html

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  6. This is a fundamental misrepresentation of what the greenhouse effect is. The gases do not 'heat' the ground, since the ground is warmer than the radiating gases. However, the gases do indeed emit radiation, some of which is directed towards the ground, where it is absorbed. The ground doesn't know the source of each photon, so how could it choose not to absorb radiation emitted by gases (and clouds) in the atmosphere?

    This radiation, which has been observed even in the polar night (http://journals.ametsoc.org/doi/full/10.1175/JCLI3525.1) and reaches over 400W/m2 in the tropics (http://www.atmos-chem-phys-discuss.net/3/5099/2003/acpd-3-5099-2003-print.pdf), contributes to the total radiation budget. Globally the surface still emits as much as it absorbs, balancing the surface energy budget.

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  7. Dr. Siddons,

    Nicely illustrated, I have posted several times on the probability that a photon continues to be absorbed and remitted backs towards the surface indefinitely. This also shows that the “Greenhouse Theory” violates the Laws of Thermodynamics.

    Just a few clarifications;

    The “backradiation” does exist; it just does not warm the absorbing surface to a higher temperature unless the “greenhouse gas” is is warmer than the surface.

    My favorite example is a human standing in a cave, if the cave is made of “room temperature” stone your cooling rate is lower than if it is an ice cave. Both caves emit thermal ”backradiation” towards you, and this does “warm” you, but it cannot raise your temperature. As you have so nicely illustrated to raise your temperature would violate the First Law of Thermodynamics. Yes I know a human body “burns” food and creates heat but if you start right at the moment of death this analogy holds true.

    There are some counter-intuitive things about the thermal properties of materials when working in a system. Higher thermal capacity does not necessarily mean that something is “more effective” at heating or cooling something else. The important factor is the thermal diffusivity of a material; this considers both the thermal capacity and conductance of a material and is literally the “Speed of Heat” in a material. Increasing the Speed of Heat of one component in a system does not always increase the Speed of Heat through the same system. It turns out the material with the lowest Speed of Heat becomes the rate limiting step.

    My favorite analogy, if you are stuck in your family sedan behind a horse cart on a one lane road upgrading your sedan to a Ferrari isn’t going to get you anyplace any faster. Many plastics have thermal capacities close to copper, but “awful” conductivities. This is why you can’t heat a block of PVC very fast with a block of warmer copper.

    So in my final opinion; it turns out that increasing concentrations of “greenhouse gases” actually works to increase the response rate of the atmosphere since more heat/energy flows at the Speed of Light than at the Speed of Heat. This means the the atmosphere warms up slightly faster at sunrise, and cools down slight faster at sunset. Neither of these can cause any permanent warming or cooling.

    I do also believe that this effect is so small that we probably could not spend enough money to try to measure it with any believable accuracy. The official historic temperature record does not contain the necessary data and attempts to water board it until it confesses to Man Made Global Warming have just wasted 100 billion dollars and 2 decades.

    Cheers, Kevin

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  8. Dr. Siddons,

    I might suggest just one more observation about your Solar Oven Example. The oven does indeed “collect” or “concentrate” some solar energy to warm the food. As an engineer I find this a very elegant, simple, inexpensive and effective system. The interesting observation regarding the Laws of Thermodynamics is that the thermal energy that would have been absorbed by the Earth’s surface beneath the Solar Oven (and then transferred to the air around the oven) must be equal to the concentrated heat inside the oven. So a Solar Oven must deprive the surrounding surface of the Earth and Atmosphere of heat, OH, THE HUMANITY….

    Cheers, Kevin

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  9. Question: why then does the surface of the earth emit more infrared than is actually emitted into space?

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  10. Stu said...

    "Question: why then does the surface of the earth emit more infrared than is actually emitted into space?"

    Kevin said…

    This is one of the subtleties of heat transfer by radiation. A 255K blackbody emits from 5 microns out to well past 25 microns. This is termed the Midwave IR (MIR about 5 microns), Longwave IR (LWIR about 10 microns) and Far IR (FIR beyond 15 microns). Yes indeed the peak is at 11 microns, but only about 20% of the total is near the peak. Some of the other wavelengths do exit directly to space through “windows” in the atmosphere. So it is indeed possible for “the surface to emit more LWIR” while “the atmosphere emits less LWIR” to space. The “extra” is emitted out to space at different wavelengths, MIR or FIR.

    The point is that the energy budget is very complicated with lots of stuff flowing “forward” and “backwards” and at different wavelengths. The laws of thermodynamics tell us that this total budget must equal zero (unlike accounting which allows for profits, thermodynamics does not). The Earth (surface and atmosphere) cannot not create “EXTRA” energy/heat above that provided by the Sun.

    The Earth does indeed release a small amount of “EXTRA” energy/heat (volcanoes, geothermal, waste heat from human activities, etc.). But that energy/heat has nothing to do with the “greenhouse effect”.

    So the “greenhouse effect” cannot cause the “average” temperature of the Earth to rise (or fall). Other heat sources (volcanoes, etc) do release heat beyond the heat from the Sun and might possibly explain alleged increases in the temperature of the Earth. But we then need to consider the total amount of that energy with respect to the thermal capacity of the Oceans (HUGE), the Rocks (BIG) and the Gases (TINY).

    When we consider these thermal capacities I postulate that the possible increase from these sources is so small we lack the instrumentation to verify it.

    I know this is swimming upstream against the consensus of the climate scientists but this whole problem is just one huge heat transfer problem and many engineers are actually quite adept at analyzing those.

    Cheers, Kevin

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  11. "The point is that the energy budget is very complicated with lots of stuff flowing “forward” and “backwards” and at different wavelengths. The laws of thermodynamics tell us that this total budget must equal zero (unlike accounting which allows for profits, thermodynamics does not). The Earth (surface and atmosphere) cannot not create “EXTRA” energy/heat above that provided by the Sun."

    Correct (assuming a completely stable equilibrium - the total budget will equal close to zero in any real life long term averages), but yet the surface is warmer than is predicted by blackbody radiation. There isn't any 'extra' energy being created out of nothing - it's just that there is a (gross, not net) flow of energy from atmosphere to surface, which contributes to the net radiation. To stay in equilibrium, the surface temperature must increase when the energy absorbed and re-emitted by the atmosphere increases. This is why the surface temperature is higher than the blackbody temperature.

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  12. Kevin said "But we then need to consider the total amount of that energy with respect to the thermal capacity of the Oceans (HUGE), the Rocks (BIG) and the Gases (TINY)."

    Reply from Alan Siddons:

    That is very well-said. Here we are in the 21st century, looking up to the sky for answers to heat-retention... when we should first be looking down at our feet and out to the oceans. AGW skeptics are mad at me for swimming upstream too — Lindzen, Spencer, Singer, Monckton et al despise me — but to me their defense of shoddy physics (i.e., "the settled science") only enables the alarmist cause. Ask yourself: If you were uninformed and could only go by what people say, which side would you choose — the side saying that thermal pollution is endangering humanity’s future or the side saying that, sure, we’re warming the planet but it won’t be so bad? Warmists vs luke-warmists hardly makes for a good debate.

    People are still attacking me for the Greenhouse Effect on the Moon paper. Here’s a snippet of a reply that I just wrote.

    The blackbody need not have a 2-dimensional skin

    To me, addressing this point is largely a matter of logic. Since the conductivity of a theoretical blackbody is undefined in physics, one must regard it as zero. For if conductive transfer does cause any time delay prior to seeing a 100% emission (which is never obtained in reality anyway) then the body in question cannot be considered ‘black’ until that time. Thus, if only to avoid perplexity, the mathematical model employed for a spherical planet has to assume a responsive entity whose every zone is at thermal equilibrium with the available irradiance. And this requires a body which has no depth — something that’s paper-thin.

    Yet this is the very problem. The average temperature of this thin-skinned sphere won’t correspond to an averaged irradiance scenario unless the sphere is spinning quickly on two axes, so as to expose each part to the same amount of light. Otherwise, whether thin-skinned or fat, the 4th root relationship between light and temperature will necessarily cause thermal discontinuities, so to speak.

    Since the earth doesn't spin quickly on two axes, then, and a simple model of averaging the irradiance fails, you’re basically forced to return to a fat model with conductive features — which will constantly put it out of thermal equilibrium with irradiance as it spins! This yields the familiar diurnal pattern that we see, of surface temperatures peaking after peak irradiance. They never catch up.

    In short, assuming blackbody equilibrium and dividing irradiance by four to allow for surface area (as compared to a disc) is a lousy way to estimate a planet’s temperature to begin with. The root assumptions are wrong, and such a technique is little more than a wild stab in the dark. Exposed to the same intensity of sunlight, for instance, a spinning rock and a static rock will have very different average temperatures, yet both are likely to radiate the same "average" amount to space. Indeed, that’s why modelers use the sly term "effective temperature." By tying temperature to an average emission, you see, those two equally-radiating rocks can be said to have the same temperature — "effectively."

    Bottom line to me, then: Heat retention is a critical factor. Even for the moon, which endures a two-week night, researchers estimate that it never fully discharges its daytime-acquired heat and that it would take more than a thousand years to do so if the sun were abruptly "turned off." I’d call that very disequilibriated! So I do believe that 3 dimensions count. This complication is ignored in planetary temperature estimates.

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  13. It seems to me that most "greenhouse effect" theorists avoid violations of the first law of thermodynamics but violate the second law of thermodynamics. They violate the second law by postulating that the back radiation flows downward from the cold matter in the atmosphere to the warm matter in Earth's surface without a heat pump. To postulate this result is to imply the existence of a counter flow heat exchanger that transfers the heat in the downflowing heat flux to the heat in the upflowing heat flux and to imply that this heat exchanger is perfect in the sense of creating no entropy. If a real heat exchanger were employed, entropy would be created with the result that the downflowing heat would have to be pumped.

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  14. MS said:

    Terry,
    Yes correct - I just created a new post on this "Why greenhouse theory violates the second law of thermodynamics" and this is the main thrust of the Gerlich and Tscheuschner paper

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  15. Oh dear physics failure.
    I will concentrate on the first example as the rest just sets up straw men examples to knock down.
    "Well, is it any surprise that nothing happens? Reflecting light back to the bright spot doesn’t make it brighter. (You can try this at home on a wall.) "
    This is untrue, it does make it brighter though because the surface you might use does not reflect much light and the spot light is so bright the effect is hard to see. One everyday example would be a room painted white vs one painted black. The white room bounces the light around more and brightens the whole room making it appear lighter.

    Lets consider the example given where the mirror does not brighten the area illuminated by the spotlight. Any light striking the mirror are reflected as the laws of reflection dictate. Assuming the mirror is angled correctly some of those photons will be reflected back at the illuminated patch. Yet we assume that it is not any brighter. Therefore where did the energy go? Yep it would violate the conservation of energy!

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  16. mightydrunken,
    no it does not make itself brighter. If you believe light can brighten itself by reflecting from a mirror I suggest you try to get a patent on your perpetual heat engine. As Alan Siddons says, if light did NOT conform to the second law, we would be as gods, infinite power in our hands. And mirrors would be outlawed except for government projects.

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  17. mightydrunken,
    While you are working on your perpetual heat engine, I advise extreme caution if you use two mirrors instead of a mirror and a surface. You could create a laser of infinite power from the light of a match and vaporize the planet before AGW!!! DEAR GOD DO NOT TRY THIS AT HOME!!!

    Along similar lines, I advise extreme caution if you put a pot of water on a hot stove. The colder water will heat up the stove even more and once again create a planetary fireball even before AGW!!!

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  18. Regarding the Second Law of Thermodynamics, the AGW theory also violates this law. The second law simply paraphrased states: the NET (i.e. total) heat/energy flow is from the warmer location to the colder location. It places no restrictions on and provides no information about how each location arrived at its current temperature. It places no restrictions on the direction of the flow of electromagnetic radiation (i.e. infrared light, radiant heat, back radiation, microwaves etc.). It simply states that a colder location (the atmosphere) CANNOT RAISE THE TEMPERATURE of a warmer location (the surface). A colder location CAN slow the rate of cooling from a warmer location by any of the known heat transfer mechanisms, conduction, convection and radiation. How effectively it can slow the cooling rate is determined by the temperature differential and the “speed of heat” of the materials involved, the material with the slowest “speed of heat” determines the rate.

    Regarding the room with black paint versus white paint; if you put 1 watt of visible light into those two rooms the one with dark paint will absorb the light and convert it to heat in the walls. The room with white paint will bounce the light around and it will be absorbed by your eyes thereby also becoming heat. The watt of energy still gets absorbed someplace. Otherwise we could just heat our homes by painting them white inside and out. Then we could just chuck the whole furnace thing and eliminate our “carbon footprint” completely!

    Cheers, Kevin

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  19. Thank you Alan! There are some other laws of nature that can strengthen the argument further. As we know, absorbed radiation exerts a radiation pressure on the absorbed material or gas, likewise, if radiation is to be returned then by the law of conservation of momentum the gas must expand even more. I think that if the greenhouse effect was in action, the atmosphere would have to expand like a marshmallow in a cup of hot chocolate.

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  20. If I understand correctly, sir, I’d agree. You’re referring to the idea that a gaseous star assumes a globular shape, for instance, because its outward radiative pressure is countered by inward gravitation. Now, one can argue that a solid barrier above a warm emitter will slow down its cooling rate, and that would be true. A crust over the earth’s mantle does precisely that. But the real subject here is the role played by an atmospheric mass, which by its very nature cannot be made “stiff"! Its natural reaction to being heated is to fluff up like a marshmallow and hurry upward to release heat, while simultaneously displacing cooler gas downward to steal more heat from the surface. A solid mass hovering over an IR-emitter will retard its cooling rate, then, especially if that second mass conducts heat poorly, but a gas in contact with the emitter only promotes heat loss: the dynamic changes to that of an air-cooled car engine. If there is some way that an atmosphere can induce surface heating, it will not be found in convection or back-radiation.

    Still, all this pertains to the second law. My emphasis in this post is on the first. If reflecting an object’s light can make it brighter, then it’s obvious that this will generate an increasing illumination cycle, energy for free. People should pay more attention to how light really behaves instead of fantasizing about how they’d like it to.

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  21. Alan, I know the original post was about the First Law, but if I might still expound on the Second Law for just one more moment; the air cooled engine is a nice example. This type of engine is cooled by all three heat transfer mechanisms. Conduction happens right at the surface (aka the boundary layer), convection happens within a few feet of the surface, and radiation happens from almost all distances. Warmer air will slow the rate of cooling of the engine by all three mechanisms. As proof I offer the empirical observation that air cooled engines run warmer in warm weather.

    Now of course the AGW believers will latch on this example as PROOF of the “greenhouse effect”. Yes indeed if the air temperature is 10 degrees warmer the engine runs 10 degrees warmer as well. However if the engine is not running it is also 10 degrees warmer!

    Yes indeed the gases do not form a rigid barrier, but they still reemit infrared radiation backwards towards the engine. Perhaps I should have used the term “affect the rate of cooling” rather than “slow the rate of cooling”. I am still of the opinion that increases of “greenhouse gases” actually increase the rate of heat flow through the system since more energy/heat flows at the speed of light (very speedy) versus at the speed of heat (quite a bit slower).

    So the AGW hypothesis violates at least the First and Second Laws of Thermodynamics. I believe if you follow a “day in the life” of a “global warming” photon you will also discover that the photon will eventually escape the awful fate of being perpetually absorbed by a “greenhouse gas” and escape to the cold vacuum of space. It might be absorbed a few times or maybe even tens of times but eventually it escapes to space and cannot cause permanent warming.

    So we have a complete hat trick whereby the AGW hypothesis violates all three of the Laws.

    Cheers, Kevin.

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  22. This is rubbish. The first and second laws of thermodynamics only apply to systems that are "isolated" (An isolated system is one that cannot exchange heat with its surroundings.) The earth is not isolated and so is not bound by the laws.

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  23. Anonmymous,
    Your comment is complete rubbish. Physics 101: The 1st and 2nd laws apply to the entire universe and all isolated or non-isolated systems therein.

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  24. Alan Siddons email said...

    In thermodynamics, a "system" only refers to a conceptual category.

    For general purposes, then, the sun, earth, space system can be considered "closed." Infalling plasma and asteroids don't usually figure in. But in nature, everything is actually an open system, even black holes.

    1) Open system: The system in which the transfer of mass as well as energy can take place across its boundary is called as an open system. Our previous example of engine is an open system. In this case we provide fuel to engine and it produces power which is given out, thus there is exchange of mass as well as energy. The engine also emits heat which is exchanged with the surroundings. The other example of open system is boiling water in an open vessel, where transfer of heat as well as mass in the form of steam takes place between the vessel and surrounding.


    2) Closed system: The system in which the transfer of energy takes place across its boundary with the surrounding, but no transfer of mass takes place is called as closed system. The closed system is fixed mass system. The fluid like air or gas being compressed in the piston and cylinder arrangement is an example of the closed system. In this case the mass of the gas remains constant but it can get heated or cooled. Another example is the water being heated in the closed vessel, where water will get heated but its mass will remain same.


    3) Isolated system: The system in which neither the transfer of mass nor that of energy takes place across its boundary with the surroundings is called as isolated system. For example if the piston and cylinder arrangement in which the fluid like air or gas is being compressed or expanded is insulated it becomes isolated system. Here there will neither transfer of mass nor that of energy. Similarly hot water, coffee or tea kept in the thermos flask is closed system. However, if we pour this fluid in a cup, it becomes an open system.

    http://www.brighthub.com/engineering/mechanical/articles/3733.aspx

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  25. Kevin said, boy this is getting fun now!

    To reply to a few statements (sorry I don’t have the exact author/blogger);

    The laws of thermodynamics apply at every instant in time, they say nothing about “long term averages” and they are not restricted to a limited volume of the universe. At every instant in time, at all locations in the Universe the heat/energy budget MUST equal ZERO. Wow, I’d sure like to get funding to model that, only 2 decades until retirement, I might even get it done before I leave.

    The whole “Closed”, “Open”, “Isolated” characterization is a human invented separation of complex systems to make them more easily understood. An “Isolated” system simply means we are ignoring any other effects besides the primary effects of our system. A “Closed” system means we are considering our isolated system but we are also considering some simple external effects (i.e. how warm is it outside today?). An “Open” system means: “We are attempting to consider EVERYTHING that affects our system.

    In terms of heat flow through the Sun/Earth/Space system some people chose (for simplicity) to ignore the following additional sources of heat; volcanoes, radioactive decay of the rocks, waste heat from humans (potentially the largest source), geothermal energy (an underground volcano if you will), moonlight arriving on the surface, starlight arriving on the surface, frictional heat from rocks rolling downhill (don’t laugh the rocks got uphill in the first place from the formative processes of the Earth, not only by energy provided from the Sun), “warming” of the Sun by thermal radiation emitted by the Earth, a small portion of X-rays arriving from space that are absorbed by the Earth, and an even smaller portion of Cosmic rays arriving from space that are absorbed by the Earth, and probably a few more things as well.

    So, if my understanding of the Laws of Thermodynamics is “rubbish” then everything we have learned about the flow of heat/energy through a complex system in the last couple of centuries must also be “rubbish”. That probably explains why all those stupid engineers have not yet discovered that you can simply insulate your house with “greenhouse gases” and fill it up with heat once and just chuck the whole furnace thing altogether. Anybody want to invest in my new company: “Heat Trapping Gases Are Us” ?

    Really folks, this whole thing is the MOTHER (no offense intended against anybody’s maternal unit) of all heat transfer problems where the characteristics of all of the materials change with: time, altitude, albedo, salinity, asphalt coverage, crop coverage, emu flatulence, etc. etc. etc. (somebody could probably make a career of indentifying all of the causes, Oh wait, some people already have) . There are probably hundreds or thousands of inputs and our uncertainty of the outcome (i.e. +/- degrees F (or C if you please)) goes up exponentially with the number of inputs to the problem.

    Cheers, Kevin.

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  26. Kevin,
    Thanks again for your excellent commentary which Alan Siddons and I have both greatly appreciated. If you would like to do a "guest post" please post a comment anywhere with your email which I will NOT publish and then send you instructions, or if you wish please just keep commenting! Thanks again.

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  27. Ok, this is absurd on the face of it. In the "Now consider a black body figure," in reality total radiated energy would be 1 at equalibrium--same as the incident energy and the same as if no mirror were there. However, with the mirror, the energy flux density would be higher at all points elsewhere because no energy is radiated through the mirror. If you substitute a real absorber, it's temperature would rise as indicated by the increased flux density. If most of the radiated energy were reflected back by a domal mirror with a small hole, the temperature of the absorber would rise indeed, even though total energy emitted through the hole would remain the same once equalibrium is reached. The intensity of the emitted energy would be higher, but total energy would remain 1. On a cloudy night temperatures drop less due to reflection of heat by the clouds, this is obserable.

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  28. As I’ve indicated, a mirror supposedly increases “flux density” on the area it is reflecting. Yet no brightening or heating of that area occurs despite a prediction that it must. Stop thinking about photonic density and start thinking about how radiant energy is ACTUALLY transferred. (Hint: it has something to do with the second law of thermodynamics.)

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  29. I think that CO2 and other GHGs are thermally heated by the surface and other gases, then radiate net MORE back to space than comes in. It is water and thermal lag which allow the surface to warm above theoretical thinskin blackbody temperature. The water cycle (evaporation/condensation/cloud formation etc.) which act as a powerful governor/negative feedback system and stabilize the surface and climate.

    But as for the mirror experiment, there's a flaw in the numbers and diagram. The "1.1" figure is in error, since you have to subtract the proportion of outgoing radiation that the mirror intercepts from the total outgoing. If the mirror is perfect, and does not heat, that number is the same as the added re-reflection, leading to a net zero difference from the point of view of an observer above the mirror who collects all reflected photons. A few just took a wee detour. (Actually, if the white illuminated surface is "matte" and spreads its reflection widely, there would be a re-re-reflection declining sequence with the mirror, but it asymptotically shrinks very fast.)

    So any GHG back-radiated IR is by definition IR that wasn't allowed to exit normally, i.e. it was shadowed or blocked from the POV of the top-down observer--who in the end gets his normal quota of photons, no more or less.

    ReplyDelete
  30. But Alan...

    Wouldn't this also result in the 'This One Word' script being a blueprint to an edited version of the film?
    DEW

    ReplyDelete
  31. The greenhouse effect does not violate the first law of thermodynamics - your mirror does! How can the mirror transfer 0.1 units of heat to the surface if there are not 0.1 units of heat reflecting off of the mirror?

    In the correct version of this picture, 1.0 unit of heat strikes the surface from the outside source, 0.1 units of heat are radiated from the surface to the mirror, those same 0.1 units of heat are reflected back onto the surface, and 1.0 units of heat are emitted into space. In fact, this is an excellent model of the greenhouse effect for a young scientist -- because the surface is absorbing and emitting a total of 1.1 units of heat, it must be at a higher temperature than if the mirror weren't present and the surface were absorbing and emitting 1.0 units of heat.

    ReplyDelete
  32. David,
    No, you are quite confused. Note how initially you said 1.0 units are emitted to space and then you say the whitebody gets brighter and emits 1.1 units to space by reflecting off a mirror. This is clearly a violation of the 1st law.

    Try the experiment yourself using a light meter or your eyes. An object cannot - repeat cannot - make itself brighter by reflecting off a mirror. If that was possible, we would have an infinite energy source.

    ReplyDelete
  33. MS,

    I think you need to read my reply more carefully. In my correct version of the model, the surface emits 1.1 TOTAL units of heat -- 1.0 into space and 0.1 towards the mirror.

    The object is brighter (and hotter), and this is not an infinite energy source. If you look at all of the energy flows, you will see that energy is not created or destroyed at any point.

    A = Power in (from outside system) = 1.0
    B = Power reflected from surface to mirror = 0.1
    C = Power reflected from mirror back to surface = 0.1
    D = Power emitted from surface out to space = 1.0

    Total power created at mirror = C - B = 0
    Total power created at surface = B+D - (A+C) = 0

    ReplyDelete
  34. David,
    Once again, you are quite confused. If "total power created at surface = 0" then it is absolutely impossible for the "object to be brighter and hotter"

    I've got better things to do than go around and around with you on this. Comments by you are now closed on this thread.

    ReplyDelete
  35. In the drawing with the mirror and the white body... can't you see that if the mirror reflects back, say 10% of the energy, then the upward beam has 90% of the energy and therefore if you sum the first reflection and the second one you get 100% and not 110%?

    Best Regards

    ReplyDelete
  36. Giova Pelle,

    You have completely missed the point of this post which states that it SHOULD be 100% and nothing higher than 100%. The GHE claims it is higher than 100% due to backradiation:

    "But if the blackbody absorbs another tenth from the mirror, then it's compelled to emit 1.1 units in all. To repeat, this is impossible. So this indicates that re-radiated infrared cannot be regarded as part of an absorbing object’s "radiation budget" either, or else the 1st law is wrong. This re-radiated infrared is only the object's own thermal emission in another guise, not an additional source of heat."

    ReplyDelete
  37. Giova Pelle hasn't missed the point at all. The diagram shows TWO red arrows leaving the surface, 1.1 past the mirror, and 0.1 to the mirror. That's a total of 1.2 units. The 0.1 TO the mirror has been included in the 1.1 PAST the mirror.

    The mirror initially intercepts 0.1 from the surface, so only 0.9 can bypass the mirror. The mirror reflects 0.1 back to the surface, which (simple vectors) is exactly canceled by the 0.1 TO the mirror. There is NO net flow of energy from the mirror to the surface. There CANNOT be - the energy came from the surface in the first place. The mirror is not emitting radiation, it's reflecting it back.

    The mirror adds no net energy to the surface at all, which continues to emit 1 unit. No violation of the first law in this example.

    I see the article "Why 'Science of Doom' doesn't understand the 1st Law of Thermodynamics" has been withdrawn. This one needs correction.

    ReplyDelete
  38. MostlyHarmless,
    You also miss the point. The diagrams show the FALLACY of what the GHE claims, and state that what is shown in the diagrams is an IMPOSSIBLE FALLACY.

    Yes, of course the mirror would NOT add any energy to the surface and the surface would continue to emit 1 unit. That is the whole point of this essay! That backradiation cannot heat the surface!

    ReplyDelete
  39. The "black body" diagram misrepresents the situation EVEN IF the 0.1 units WAS somehow heating the surface. It shows 1.1 units from the surface bypassing the mirror, and a further 0.1 units striking the mirror. That adds up to 1.2 from the surface, NOT 1.1. Using a mirror in the example is a total red herring - the energy from the mirror can't affect the surface state, because the energy is simply reflected back where it came from.

    The earth/atmosphere system doesn't involve any mirrors, so the example here is irrelevant. If you want to prove a point about the greenhouse effect, then at least use a second body emitting radiation, rather than a mirror, especially one where the energy flows are misrepresented. This is a "straw man" argument.

    ReplyDelete
  40. MostlyHarmless,
    1. As already stated above a million times, a body cannot make itself hotter or brighter by reflecting off a mirror. You seem to agree, stating "the energy from the mirror can't affect the surface state, because the energy is simply reflected back where it came from." Correct and THAT IS THE POINT OF THIS ESSAY. Take a look at all the folks in the comments above who want to argue with that!
    2. OK fine - replace the perfect mirror with a second colder body re-emitting radiation from the warmer object - same end result- the surface temperature of the warmer object does not increase!
    3. Sorry you can't understand the analogy to backradiation from GHGs. Maybe Claes Johnson can explain it to you better:

    http://claesjohnson.blogspot.com/2010/07/mirror-example-by-alan-siddons.html

    The analogy is clearly not a "straw man" or "red herring" and does show why the GHE violates the 1st law.

    I've got better things to do than argue with you that this post shows what you agree it does show- i.e. that reflection off a mirror OR backradiation from GHGs CANNOT HEAT THE SURFACE, therfore comments by you on this post are now closed.

    ReplyDelete
  41. I find these thought experiments very difficult to think through to a strong conclusion. However I find convincing the argument that a blanket does not 'heat' the body under it but merely slows the rate at which the heat is able to be removed there-from, resulting in a warmer body than without the blanket.

    However, in the case of earth and the atmosphere, during the day the surface resists warming strongly with evaporation and convection overwhelming radiation in the entire troposphere.

    At night, there are very different conditions. Then with clear skies there is more rapid heat loss than with cloud cover. Under some conditions temperatures hardly change overnight.

    Surely, then, the effect of a small increase in atmospheric CO2 will be a very small change in the night-time heat loss?

    Paul

    ReplyDelete
  42. Why doesn't the roof of my mouth or my tongue burn as they both radiate at 310K at each other. Adding up to 620 K which is above boiling. :)

    ReplyDelete
  43. For the record folks, the arguments above are ALL wrong, wrong at many points, and I'll not waste the prose and the math to go through them. The fundamental demonstration that these arguments are wrong is really simple: EXPLAIN THE EXISTENCE OF THE TROPOPAUSE.

    Black-body radiation (and hence Stefan-Boltzman law) cannot do that.

    Yet very clearly a cold tropopause does exist, much colder than the atmosphere both above and below it. Nor does this violate thermodynamics, once you actually understand radiation transfer.

    Three big hints:

    1. Stefan-Boltzmann is NOT "the radiative transfer equation." Monckton started the habit of AGW denialists making asses of themselves with long-winded and completely ridiculous arguments from this conflation. Monckton actually claimed in print that the Stefan-Boltzmann law IS "the radiation transfer equation" ... which left all the physicsts who read that diatribe rolling on the floor.

    Monckton also entitled that "..... reconsidered," which instantly brings to mind the long history of utterly crank missives from the English peerage to the Royal Societies in the 18th and 19th centuries, where titles of this form were expected.

    2. Beer's Law really works, monochromatically. The problem is that you must do difficult integrals over the spectrum

    3. You must understand the vibrational rotational absorption/emission spectra of molecules, and particularly Lorenz collisional broadening, which effects itself in an atmosphere as "pressure broadening"

    Dr Siddons ... I'd recommend you take this blog down, you are being pointed out around the web as yet another blogger who is recapitulating Monckton ... (and not even giving him "credit") or paying attention to the scathing rebuttals of this line of argument.

    ReplyDelete
  44. Anonymous:

    The Tropopause can be explained by the lapse rate/pressure:

    http://hockeyschtick.blogspot.com/2010/09/shattering-greenhouse-effect.html

    and here:

    http://hockeyschtick.blogspot.com/2011/01/scientist-climate-changes-are-not.html

    You apparently didn't bother to read this post before criticizing it. This post shows that reflecting light off of a mirror does not increase it's intensity. Now I understand why you don't want to attempt to refute it with prose or mathematics and really have no idea what you are talking about.

    This post does not mention Stefan-Boltzmann once and the application or misapplication of that law has nothing to do with the point of this post. Same with Beer's law or emission spectra.

    You also didn't even read the 1st paragraph, which says this post is from an email I received from Alan Siddons. This is not Siddon's blog and it is not going to be taken down on your orders. BTW Siddons is highly critical of Monckton regarding the GHE.

    ReplyDelete
  45. MS,

    Your diagram with the mirrors is mistaken. The diagram violates the laws of thermodynamics because of an arithmetic error on your part. Look carefully at your diagram. How much energy is reflected from the whitebody without any mirror or reflection? The answer is 1.1. In your diagram, if we take the mirror away, then 1.0 units are reflected upwards, and 0.1 units are reflected diagonally toward where the mirror would have been. That adds up to 1.1 units of energy reflected from the whitebody even without any mirror, because of an arithmetic error. That is the violation of the laws of thermodynamics, not greenhouse theory.

    -tom

    (p.s. I'm an ecomonics student, not a scientist or engineer, but the point still stands).

    ReplyDelete
  46. MS,

    AGW theory doesn't violate the 1st law of thermodynamics. AGW posits that photons leaving Earth are absorbed and converted into heat. As a result, the Earth heats up, and more energy is emitted as heat, but less energy is emitted as light. The total amount of energy emitted is the same as had arrived, so there's no violation of the 1st law.

    Your mistake here was to consider heat and light separately. Heat and light are both forms of energy, and one can be converted into the other. In the greenhouse effect, light is converted into heat, thereby warming the planet, but the total energy radiated is the same as before (more heat, less light).

    GHGs in the atmosphere will increase the conversion of light into heat, leading to higher temperatures on Earth, but also less light emitted from Earth, and a dimmer Earth (please spare me any jokes) if viewed from the moon. The total amount of energy radiated is the same, with or without GHGs, so there's no violation of the 1st law.

    -tom

    ReplyDelete
  47. No Tom, that is incorrect.

    Without a mirror, 1 unit of energy in = 1 unit of energy out, not 1.1

    With a mirror, it's still the same: 1 unit of energy in = 1 unit of energy out, not 1.1

    Please read the post carefully. The diagram shows the fallacy that 1 unit of energy in can result in 1.1 unit out.

    ReplyDelete
  48. No Tom, your 2nd comment is also incorrect for several reasons, not the least of which is that outgoing longwave (IR) radiation has markedly increased over the last 62 years

    http://joannenova.com.au/2011/02/the-oceans-clouds-and-cosmic-rays-drive-the-climate-not-co2/

    ReplyDelete
  49. Tom, this post is also recommended reading showing your description of the GHE is falsified by the data:

    http://judithcurry.com/2011/02/09/decadal-variability-of-clouds/

    ReplyDelete
  50. Alan Siddon/MS,

    I love the simplicity of this very workable example. It forces AGW proponents to question their understanding...that is, if they are logical and not just AWG fundamentalist.

    My question, what happens to the radiation reflected by the mirror. I know it does not contribute to the spot's brightness but I find myself without a good explanation as to why. Thermodynamic laws are in fact well laws, but it does not offer specificity.

    Radiative transfer equations, those which I'm familiar, all specify vacuous surroundings. In such environment hot and cold, energize one another...or at least in theory no consideration is made for the 2nd law. My gut says this is incorrect, yet the equations say otherwise. And being I have no experience in an air free environment I must assume the math is correct.

    So when the discussion of atmospheric forcing enters the argument, it is easy to point why it can not happen...but doesn't answer what happens to the re-radiated IR. To this what "happens to re-radiated IR" question I fall back to the insignificance of forcing with the dominance of convection and conduction which do always follow the 2nd law. So my question is what happens to downward re-radiated IR.

    ReplyDelete
  51. LJ Ryan,

    The re-radiated downwelling IR is absorbed by the surface, but does not result in an increase in temperature/"brightness" since a lower (or equal) temperature/ frequency/ brightness/ entropy body cannot heat a higher (or equal) temperature/ frequency/ brightness/ entropy body (2nd law of thermodynamics).

    In the mirror example, we are assuming equal brightness from the mirror and the surface, ergo the brightness cannot increase. In the case of the Earth/atmosphere, the Earth (avg 15C) is considerably hotter than the atmosphere (avg -18C) and the 2nd law dictates that heat only flows from hot to cold.

    This is only a simplistic explanation - for an in depth treatment see multiple posts on Claes Johnson's site

    claesjohnson.blogspot.com

    and also on Charles Anderson's site

    http://objectivistindividualist.blogspot.com/2011/01/blackbody-radiation-and-consensus.html

    ReplyDelete
  52. "The re-radiated downwelling IR is absorbed by the surface, but does not result in an increase in temperature/"brightness" since a lower (or equal) temperature/ frequency/ brightness/ entropy body cannot heat a higher (or equal) temperature/ frequency/ brightness/ entropy body (2nd law of thermodynamics)."

    Brilliant. You just avoided an "apparent" violation of the 2nd law of thermodynamics by a "real" violation of the 1st (conservation of energy.)

    ReplyDelete
  53. No Anonymous- there is no 1st law violation. As shown in the example in the post, light cannot make itself brighter by reflecting off a mirror. Climate scientists effectively assume that it can.

    High potential energy falls naturally to a lower potential; low potential energy doesn’t spontaneously jump higher. Light illuminates darkness, not something that's shining brighter. Heat moves spontaneously to any lower temperature zone. A cool gas superimposed over a hotter background will be seen to absorb certain frequencies emitted by that hotter body -- but if the gas is hotter, instead, it will emit those same frequencies. A rattlesnake can only perceive its prey if its IR-detecting pit organs are cooler than the prey. So too, the detector of an infrared telescope must be kept cooler than the target in order to form an image -- otherwise it will merely report its own thermal output.

    ReplyDelete
  54. This is an intellectualmtrap that has periodically resurfaced in science since the late 1800's. Two things not accounted for in this model are that radiative efficiency of the atmosphere is not that of a blackbody and that system (planet) is not at thermal equilibrium.

    The empirical evidence that the first law violation argument presented above is wrong is found in the planets. Planets with infrared-active atmospheres (venus and earth) are hotter than those without (moon and mars) even taking into account the solar irradiance differences due to their distance from the sun.

    ReplyDelete
  55. wrong - and here's the calculations to prove it done by David Archer himself:

    http://hockeyschtick.blogspot.com/2011/08/professor-inadvertently-explains-why.html

    Then be sure to read & understand the link in that post to "Venus: No greenhouse effect"

    The adiabatic lapse rate alone fully & perfectly explains the temperature profiles on all 3 planets with no need for a GHE whatsoever.

    ReplyDelete
  56. With a PhD in optics, I can tell you for sure that your argument about the light reflected by the mirror back to the surface not contributing to its brightness is WRONG! The reason your 'experiment' 'proves' it is that the intensity in the bright spot changes by a small relative amount, and our perception to intensity is logarithmic - this is why you think the dark background becomes more intense that the original bright spot. However - if you would conduct a proper experiment with a calibrated detector you would notice that the bright spot becomes brighter by the same amount that the dark background does. If the rest of the 'proof' in this blog-entry is dependent upon this claim then it is wrong.

    ReplyDelete
  57. Well Mr. PhD in optics, I hope you will do the proper experiment with a calibrated detector yourself to prove that you are completely wrong. It is absolutely impossible for a light beam to increase its intensity/brightness by reflecting off a flat mirror. This is absolutely impossible due to the conservation of energy demanded by the 1st law.

    While you are at it, use two mirrors facing each other instead of a mirror and a surface. By your flawed logic, each time the light bounces off a mirror it will become brighter and result in a laser of infinite power. You better get to work on this right away as a Nobel Prize and unlimited military contracts await proof of your theory.

    ReplyDelete
  58. MS, In response to Anon PhD in optics, would it be fair to say that if the target area of the reflection by the mirror were brighter, this could *only* be because the mirror was reflecting *additional* light that was not in the original target area, which would make any area the reflected light would have illuminated had the mirror not been there, darker (based on the conservation of energy law)?

    The good question that I did see, is whilst a reflection of energy back to the source cannot add to the source's energy level, what happens to it? It is clear that if the source's emitted energy level is 1.0, and 0.1 is reflected back to it, the source's resultant level is still 1.0 and not 1.1. So what happens to the 0.1?

    ReplyDelete
  59. http://wattsupwiththat.com/2013/02/06/the-r-w-wood-experiment/#comment-1221185

    Clausius, a Thermodynamicist of impeccable credentials, also used the second law to derive what we call “The Optical Sine Theorem”, which basically says that no optical system can create an image that has a higher radiance than the source object. It also says that no optical system can form an image that is brighter than the image formed by an “Aplanatic ” system; that being a system that is corrected for both spherical aberration and coma, the two simplest of the Seidel aberrations of optical images.
    Clausius’ thermodynamic limit also applies to “non-imaging” optics, which is the main discipline behind illumination engineering.

    ReplyDelete
  60. Well, this post has a glaring problem insomuch as it states:

    "Re-radiated energy is neither reflected nor absorbed by a surface, therefore, because either outcome would point to additional energy that would register on a detector, signaling that more radiant energy is leaving the system than what is going in."

    Since, in back-radiation, we're talking about electromagnetic radiation we are talking about photons. Each of these photons have an energy given by E=hv. There's only three possible things that can happen to an incident photon: reflectance, absorption, or transmittance. In the case of back-radiation to the earth, transmittance is obviously absurd. You're statement that neither reflectance nor absorption take place removes the two other possibilities leaving you concluding that these photons simply cease to exist. Your conclusion actually violates the first law because each photon is possessed of energy. If you wish to get rid of it, you must convert it to some other form of energy, you can't hypothesize it away. But that conversion is what "absorption" is all about, which you exclude.

    The only retreat left with your logic is to propose that no back-radiation existed in the first place. If that were true, it would mean, for example, that the CO2 and H2O vapor in the atmosphere didn't re-radiate, which means that they remained in an exciting state, or that they re-radiated only in one direction. Both of these are counter to experimental evidence.

    ReplyDelete
    Replies
    1. "The only retreat left with your logic is to propose that no back-radiation existed in the first place. If that were true, it would mean, for example, that the CO2 and H2O vapor in the atmosphere didn't re-radiate, which means that they remained in an exciting state, or that they re-radiated only in one direction."

      Not true. All bodies above absolute zero radiate. Radiation transfer between 2 bodies is bidirectional, however, heat transfer between 2 bodies is always unidirectional from warm to cold, per the 2nd law. A lower energy/frequency/temperature body cannot add energy or heat to a higher energy/frequency/temperature body. The reason why is that all of the lower energy quantum states are already filled in the higher energy body. Thus, radiation from the lower energy source is not thermalized. This is referred to as the frequency cutoff for thermalization.

      Delete
    2. @MS

      You're begging the question. If you accept the claim that back-radiation of any sort is neither reflected nor absorbed (and obviously not transmitted) then what happens to the photons?

      Delete
    3. Your hangup appears to be in regard to only thinking about the particle nature of light while ignoring the wave nature of light.

      Basically, there is destructive interference between a lower frequency source and a higher frequency target. A simple analogy would be two people shaking a jump rope at a high frequency to create large waves in the rope. Then, a third person also grabs the rope and shakes it at a lower frequency or amplitude while the original 2 people continue at the same rate. Does the total energy in the rope increase due to the energy added by the 3rd person? No

      Delete
    4. Ah. Now you're just making it up.

      Photons have an energy, whether they are wave or particel. E=hv. You can't "destroy" energy, interference or not.

      Delete
    5. Re-read what I said above, a low energy/frequency photon reaching a higher energy/frequency body does not thermalize because those lower energy/frequency quantum states are already filled in the higher energy/frequency body.

      Your proposition is preposterous -that a flat mirror can increase the intensity of a spot of light. Why don't you use two mirrors instead and create a laser of infinite power?

      Delete
  61. I don't think that this post is a good analysis in any way. I have a extended comment on it which it too large to post as a comment here. You can find my comment at

    http://climatesciencereview.wordpress.com/2013/07/05/comment-on-the-hockey-schtick-why-conventional-greenhouse-theory-violates-the-1st-law-of-thermodynamics/

    ReplyDelete
    Replies
    1. Your post is refuted by my comments above as well as the comment from an optical engineer:

      Clausius, a Thermodynamicist of impeccable credentials, also used the second law to derive what we call “The Optical Sine Theorem”, which basically says that no optical system can create an image that has a higher radiance than the source object. It also says that no optical system can form an image that is brighter than the image formed by an “Aplanatic ” system; that being a system that is corrected for both spherical aberration and coma, the two simplest of the Seidel aberrations of optical images.

      Clausius’ thermodynamic limit also applies to “non-imaging” optics, which is the main discipline behind illumination engineering.

      Delete
    2. You're refutation is very non-specific. And the original claim was that it violates the first law. If you want to claim it refutes the second law, that's a different argument.

      Delete
    3. Not really, the 2nd law is derived from the 1st law. Anything that refutes the 2nd law also refutes the 1st.

      Delete
    4. The second law is not derived from the first law. There is a combined equation, but the second is not derived from the first.

      Delete
    5. What I should have said is the 2nd law places a restriction upon the 1st law regarding the direction of heat transfer. Thus, it remains true that anything that refutes the 2nd law also refutes the 1st.

      Delete
    6. Sorry. You're just hand waving at this point. There are many examples mentioned in thrmodynamics textbooks of processes which are consistent with the first law but not the second. None of this is convincing in the least.

      Delete
    7. Sorry, you are just hand waving, attempting to direct attention away from your perpetual motion machine/infinity laser that increases the intensity of light by reflection off a flat mirror. Why don't you demonstrate this device and earn a fortune instead of arguing about whether or not if something violates the 2nd law it also violates the 1st?

      Of course, there are examples mentioned in thermodynamics textbooks of IMAGINARY processes which are consistent with the first law but not the second. It remains absolutely true that any alleged REAL process that violates the 2nd law also violates the 1st. Whether you understand that or not is besides the point of this post. Apparently you misunderstood that those examples are IMAGINARY.

      Delete
  62. There's a simple experiment you can do which refutes the idea presented by this post that re-radiated energy of any sort is neither reflected nor absorbed. Again, the conclusion drawn in the original:

    ====
    Putting these results together, one must conclude that "back-radiation" of any sort, whether visible or infrared, will simply not show up on an external radiometer because it represents no "extra" energy in the first place and it induces no illuminative or thermal effect. Re-radiated energy is neither reflected nor absorbed by a surface, therefore, because either outcome would point to additional energy that would register on a detector, signaling that more radiant energy is leaving the system than what is going in.
    ====

    to perform this experiment you will need a bathroom with a mirror in it, some ambient light and a fair sized hand mirror.

    Stand about five feet from the bathroom mirror, hold out the hand mirror with the glass facing the bathroom mirror. Hold it at arms length. With a little manipulation, you will be able to see your own face - in the reflection within the reflection of the hand mirror.

    The ambient light landing on you, reflected off of you, into the bathroom mirror, into the hand-held mirror, back to the bathroom mirror, and into your eyes. Since you are seeing your own face within the reflection of the hand mirror, you know that this is light originally reflected from you. It is back-radiation for obvious reasons. It is or should be obvious that this back radiation can be reflected. In addition, since you are seeing that reflection within a reflection, since you are SEEING it, that back-radiation's light landed in your eyes, was absorbed by your retina, and transformed into the chemical-electrical impulses your brain recognizes.

    This simple experiment proves that back-radiation is both reflected AND absorbed. This post is simply wrong.

    ReplyDelete
    Replies
    1. No, once again Thor, it is you who fail to understand. For the 3rd time, so called "back-radiation" [a term only used by climate scientists] DOES EXIST, however since it is of lower energy & frequency than the hotter & higher energy & frequency body it returns to it does not thermalize. It does not increase the energy / frequency / temperature of the hotter body because those lower energy/frequency quantum states are ALREADY FILLED in the higher energy/frequency body. Therefore, back-radiation exists AND is both reflected AND absorbed but it DOES NOT THERMALIZE.

      I'm now finished with this discussion unless and until you demonstrate a working model of your infinity laser.

      I suggest, however, you read Postma's series of posts 1-12 on the fallacy of the GHE and explain to him how your laser works.

      http://climateofsophistry.com/2012/11/06/on-the-absence-of-a-measurable-greenhouse-effect-part-1-the-failure-of-ipcc-energy-budgets/

      Delete
  63. http://climateofsophistry.com/2013/09/19/fraud-17-watts-are-not-energy-ipcc-violates-conservation/

    ReplyDelete
  64. http://climateofsophistry.com/2013/10/26/greenhouse-fraud-20/

    ReplyDelete
  65. http://climateofsophistry.com/2013/11/10/greenhouse-fraud-21-carnot-engine-no-ghe/

    ReplyDelete
  66. Werner Brozek says:
    November 13, 2013 at 9:12 am
    We have all of this talk about huge numbers of Hiroshima bombs and about all of the 10^22 Joules the oceans are taking up. But when converting to degrees C, there seems to be almost nothing left as was pointed out in this article. I will illustrate the facts in another way.
    I will use Figure 5 and assume it is true.
    According to this diagram, the total heat increase is about 25 x 10^22 J over about 55 years.
    The total mass of the ocean above 2000 m is 48% of the total mass of the ocean.
    The total mass of the ocean is 1.37 x 10^21 kg.
    The specific heat capacity of ocean water is 4000 J/kgK.
    Applying H = mct, I get a change in t of
    25 x 10^22 J/(0.48 x 1.37 x 10^21 kg x 4000 J/kgK) = 0.1 K. Is that correct? If so, it would take over 100 years for the top 2000 m to go up by 0.2 C. Is that supposed to be a problem for us?

    http://wattsupwiththat.com/2013/11/13/comments-on-stefan-rahmstorfs-post-at-realclimate-what-ocean-heating-reveals-about-global-warming/#comment-1474228

    Vince Causey says:
    November 14, 2013 at 12:41 pm
    “Applying H = mct, I get a change in t of
    25 x 10^22 J/(0.48 x 1.37 x 10^21 kg x 4000 J/kgK) = 0.1 K. Is that correct? If so, it would take over 100 years for the top 2000 m to go up by 0.2 C. Is that supposed to be a problem for us?”

    I doubt it. This is the second law of thermodynamics at work – the entropy is increasing. Useful heat is being degraded into a more diffuse, low grade form.

    It’s like taking a red hot poker and dropping it in a bath of cold water. The heat of the original red hot poker can be used to do some work, but once the heat is transferred to the bath, the temperature becomes that of the bath, whose temperature has risen by only a tiny amount.

    The important point is that the heat, once diffused to a lower temperature, can never result in a subsequent rise in temperature, which would be contrary to the first law. Therefore, any heat, once sequestered by the oceans, can never reverse its thermodynamic pathway, and reheat the atmosphere to the temperatures that caused it to be absorbed in the first place.

    It is an admission that the problem (if it ever existed), is being solved by nature once and for all.

    http://wattsupwiththat.com/2013/11/13/comments-on-stefan-rahmstorfs-post-at-realclimate-what-ocean-heating-reveals-about-global-warming/#comment-1475425

    ReplyDelete
  67. http://climateofsophistry.com/2013/11/11/the-first-and-second-law-of-thermodynamics-debunk-the-greenhouse-effect/

    ReplyDelete
  68. http://www.principia-scientific.org/?option=com_content&view=article&id=404&utm_source=newsletter&utm_medium=email&utm_campaign=newsletter_November_25_2013

    ReplyDelete
  69. http://stevengoddard.wordpress.com/2013/12/04/shock-news-global-warming-is-a-fraud/

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