Guest Post by Willis Eschenbach
Once again, the crazy idea that downwelling longwave radiation (DLR, also called infra-red or IR, or “greenhouse radiation”) can’t heat the ocean has raised its ugly head on one of my threads.
Figure 1. The question in question.
There are lots of good arguments against the AGW consensus, but this one is just silly. Here are four entirely separate and distinct lines of reasoning showing that DLR does in fact heat the oceans.
Argument 1. People claim that because the DLR is absorbed in the first mm of water, it can’t heat the mass of the ocean. But the same is true of the land. DLR is absorbed in the first mm of rock or soil. Yet the same people who claim that DLR can’t heat the ocean (because it’s absorbed in the first mm) still believe that DLR can heat the land (despite the fact that it’s absorbed in the first mm).
And this is in spite of the fact that the ocean can circulate the heat downwards through turbulence, while there is no such circulation in the land … but still people claim the ocean can’t heat from DLR but the land can. Logical contradiction, no cookies.
Argument 2. If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass. If DLR were heating the air we’d all be on fire.
Nor can it be going to evaporation as many claim, because the numbers are way too large. Evaporation is known to be on the order of 70 w/m2, while average downwelling longwave radiation is more than four times that amount … and some of the evaporation is surely coming from the heating from the visible light.
So if the DLR is not heating the ocean, and we know that a maximum of less than a quarter of the energy of the DLR might be going into evaporation, and the DLR is not heating the air … then where is it going?
Rumor has it that energy can’t be created or destroyed, so where is the energy from the DLR going after it is absorbed by the ocean, and what is it heating?
Argument 3. The claim is often made that warming the top millimetre can’t affect the heat of the bulk ocean. But in addition to the wind-driven turbulence of the topmost layer mixing the DLR energy downwards into lower layers, heating the surface affects the entire upper bulk temperature of the ocean every night when the ocean is overturning. At night the top layer of the ocean naturally overturns, driven by the temperature differences between surface and deeper waters (see the diagrams here). DLR heating of the top mm of the ocean reduces those differences and thus delays the onset of that oceanic overturning by slowing the night-time cooling of the topmost layer, and it also slows the speed of the overturning once it is established. This reduces the heat flow from the body of the upper ocean, and leaves the entire mass warmer than it would have been had the DLR not slowed the overturning.
Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. The DLR is about two-thirds of the total downwelling radiation (solar plus DLR). Given the known heat losses of the ocean, it would be an ice-cube if it weren’t being warmed by the DLR. We know the radiative losses of the ocean, which depend only on its temperature, and are about 390 w/m2. In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.
But the average solar input to the surface is only about 170 watts/square metre.
So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?
Note that each of these arguments against the idea that DLR can’t warm the ocean stands on its own. None of them depends on any of the others to be valid. So if you still think DLR can’t warm the ocean, you have to refute not one, but all four of those arguments.
Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things …
w.
Tim Folkerts says:
August 23, 2011 at 3:41 am
they either measured the 324 W/m^2 (eg find the average of value from pyrgeometers around the world) or calculated it (from measured temperatures and compositions of the atmosphere around the world). I suspect the “thermals” component is tougher to get right than the IR, since it requires estimating the magnitude of updrafts and their intensity around the globe.
The worlds foremost expert on pyrgeometry says no change in atmospheric optical depth during co2 increase: http://www.warwickhughes.com/blog/?p=87 see comment 10 particularly.
The thermals also work through thunderstorm vortices as per Willis’ thermostat hypothesis. This means the atmospheric window is probably a good deal bigger then Trenberth thinks, and varies to deal with extra heat from radiation or albedo reduction.
Tim Folkerts says:
August 23, 2011 at 3:50 am
“The absorption around 1750 and 3500 nm is your incoming solar IR. It amounts to the ~67W/m^2 directly absorbed by the atmosphere.”
Presumably the absorbed UV also contributes to this total.
Yes, but the main climatic effect seems to be related to ozone destruction affecting plankton density.
the totals are pretty close to equal. with slightly more IR than visible energy. Depending on just where you draw the somewhat arbitrary lines between UV, visible, and IR, the numbers are ~ 10 UV, 40% visible, and 50 % IR
Well, ok, I was talking about the IR Myrrh is interested in and I lumped in near IR with visible. My bad.
Anyway, ~67W/m^2 absorbed in atmosphere, around 170W/m^2 absorbed in ocean: 2.5 times more.
Myrrh says:
August 23, 2011 at 2:45 am
Water absorbs longwave infrared and heats up, that is a basic bog standard real world physics known.
Microwaves at 122mm wavelength can warm water. IR at atmospheric and solar wavelengths can’t penetrate it.
From wiki’s page on microwave ovens:
http://en.wikipedia.org/wiki/Microwave_oven#Design
“More recently, some manufacturers have added high power quartz halogen bulbs to their convection microwave models, marketing them under names such as “Speedcook”, “Advantium” and “Optimawave” to emphasize their ability to cook food rapidly and with good browning. The bulbs heat the food’s surface with infrared (IR) radiation, browning surfaces as in a conventional oven. The food browns while also being heated by the microwave radiation and heated through conduction through contact with heated air. The IR energy which is delivered to the outer surface of food by the lamps is sufficient to initiate browning”
IR browns the outside, microwaves heat the inside:
Got that?
Tallbloke, I suspect we may mostly be agreeing, but often talking past each other.
“The worlds foremost expert on pyrgeometry says no change in atmospheric optical depth during co2 increase: http://www.warwickhughes.com/blog/?p=87 see comment 10 particularly.”
I was specifically talking about atmospheric IR and pyrgeometers, while this comment is about solar radiation and pyrheliometers, which is a side issue (i think, unless you have a connection in mind that I am missing). I was merely saying you can estimate the 324 W/m^2 IR by measuring the IR directly. Presumably you could also watch for trends in the pyrgeometers to see if there is indeed an increase in downward IR. I’d love to see reports/papers about that!
“Well, ok, I was talking about the IR Myrrh is interested in and I lumped in near IR with visible. My bad.
Anyway, ~67W/m^2 absorbed in atmosphere, around 170W/m^2 absorbed in ocean: 2.5 times more.”
I was talking about the “top” of the curve in the diagram you were referencing. The incoming solar radiation corresponds to ~ 10/40/50 UV/visible/IR.
The yellow part is the part absorbed by the atmosphere — that would presumably be the 67 W/m^2. I can’t easily estimate from the curve how much is due to different bands. For example, almost all of the UV is absorbed by the atmosphere, but it is a small amount to begin with.
The 170 W/m^2 that actually lands would presumably the red part of the graph. There may well me more visible energy hitting the surface than IR.
P.S. I have found from experience that Myrrh is very convinced of his position and on the authority of the websites he chooses to reference for us. No amount of evidence to the contrary is likely to change his view.
RE: Myrrh: (August 23, 2011 at 12:57 am)
“…water is a transparent medium for Light, the molecules do not absorb Light…
Only the empty vacuum of outer space does not absorb light. ‘Transparency’ has limitations. Destroyers and submarines use sonar instead of flashlights because the attenuation of the sounds they are using can be measured as a few dB/kyd (decibels per kiloyard). An attenuation of one dB indicates that the logarithm base 10 of the power of the signal has been reduced by 0.1.
On the Wikipedia ‘Electromagnetic absorption by water’ page, they are using a centimeter scale to measure EM absorption in water. This, I assume, is pure water uncontaminated by algae or plankton. Although undefined here, as far as I could find, the absorption coefficient usually indicates by what value the natural logarithm of the signal is reduced when passing through the specified distance (1 cm). An absorbed photon delivers its packet of energy to some ‘lucky’ molecule.
BTW, this page obviously appears to have been written by someone who has fully bought in to the greenhouse catastrophe theory. That does not mean that this largely unrelated data is incorrect or fudged—it also includes a cute water molecule vibration mode diagram.
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water
Myrrh :
you’ve got a weird notion about electromagnetic energy. best review physics 101 and get the basics right.
tallbloke: think about it = an object at the focal point of a parabolic reflector can NOT radiate more efficiently than it can without any reflector at all.
in fact, the extra chill that makes the ice is due to evaporation in very dry conditions.
back in the day of the raj, the brits had iced drinks in india – made the same way but with no reflectors anywhere.
http://query.nytimes.com/mem/archive-free/pdf?res=F20E1FF63E5F15738DDDA10894D9405B8284F0D3
Oops .. in my last reply to Tallbloke I left out the clouds and changes during the day! The red part is the sunlight that hits the surface at noon when it is actually sunny. Some of that Red part would be the reflection from clouds. Part of it would contribute to the atmosphere when it is not noon and more incoming light gets absorbed by the atmosphere.
http://tallbloke.files.wordpress.com/2011/08/spectral-content.gif
tallbloke says:
August 23, 2011 at 3:47 am
Question for Tim Folkerts:
O2 and N2 don’t absorb or radiate photons, but they must be excited by collisions with excited co2 and h2o molecules. Presumably they jiggle around more. What will the effect of this be? A priori, it seems likely they will more readily permit convection of more buoyant molecules. Won’t this speed cooling of the atmosphere to space?
All gases dissipate heat. We run our lives with this basic understanding. Hair dryers, radiators in cars, home heating systems, we perspire, dogs pant, etc. There are no examples of gases adding heat to a system on their own.
tallbloke- every bit of matter will absorb and emit photons.
http://brucegary.net/MTP_tutorial/MTP_ch5.html
gnomish says:
August 23, 2011 at 6:37 am
tallbloke- every bit of matter will absorb and emit photons.
http://brucegary.net/MTP_tutorial/MTP_ch5.html
Sure, but how much solar or Earth emitted E/M does O2 and N2 absorb or emit at atmospheric T’s and P’s?
P.S. Love the Raj ice-making story.
Gnomish says
“tallbloke: think about it = an object at the focal point of a parabolic reflector can NOT radiate more efficiently than it can without any reflector at all.
in fact, the extra chill that makes the ice is due to evaporation in very dry conditions.
back in the day of the raj, the brits had iced drinks in india – made the same way but with no reflectors anywhere.
http://query.nytimes.com/mem/archive-free/pdf?res=F20E1FF63E5F15738DDDA10894D9405B8284F0D3”
This is mostly wrong.
It is true that an object does not radiate better simply by putting it at the focus of a mirror. BUT the ground cannot radiate TO the object. The mirror blocks the warm IR from the ground and reflects in the cooler IR from above in its place. This allows the object to continue radiating the same energy but receive less energy, thereby cooling.
Evaporative cooling is also possible, but the article mentioned above only discusses cooling objects to 65 F by swinging wet objects, not freezing them. This chart http://mrsdlovesscience.com/realtivehumidity/pg12b.jpg shows how much cooling you you could get at various temperatures and humidities. For example, at 24 C and 9 % relative humidity, the max possible cooling would be by 14 C to 10 C. Google “wet bulb temperature” for more info)
Any further cooling was by endothermic reactions (“salts”) or refrigerators (“wonderful pneumatic machines”).
Konrad says:
August 23, 2011 at 4:56 am
Myrrh says:
August 23, 2011 at 2:45 am
“Water absorbs longwave infrared and heats up, that is a basic bog standard real world physics known.”
The fact that Myrrh writes this statement after I have taken the time to design a simple, repeatable empirical experiment and reported the methodology and results to this thread proving that statement false should indicate that responding to Myrrh is not worth the wear on your keyboard.
Myrrh, follow the instructions, do the experiment and learn.
Why don’t you re-read what I wrote and the particular point I was making. I thought your experiment was good, but what do I know..
Now do something really useful to help bring down All the AGWScience Fiction Inc scam, design something that will measure just how much blue visible light from the Sun heats water..
gnomish says:
August 23, 2011 at 6:04 am
Myrrh : you’ve got a weird notion about electromagnetic energy. best review physics 101 and get the basics right.
Can’t watch it, but you really should put away your ego when talking to me..
It’s very simple. Visible light is transmitted through water without being absorbed because water is a transparent medium for visible light. That is a technical description of what happens to visible light travelling through water. Because, water is a transparent medium for visible light. Transmitted means specifically it is not absorbed. Absorbed in the specific physics meaning of the word and not in the general ‘disappeared somewhere in the deep and so much have been absorbed’ nonsense.
If the ‘greenhouse glass is transparent to visible light, it passes straight through without being absorbed’ as per the AGWScience Fiction telling of the story, which happens to be a physical fact, then the FACT that water is a transparent medium for visible light must mean that it is also not absorbed in water.
You can’t have it both ways.. Water is a transparent medium for visible light, just as glass is.
PS There is mention at the very bottom of Gnomish’s article about freezing the water by evaporation, but only when the air temperature is 42 F or lower. I had missed that earlier. So, yes, when the temperature is quite close to freezing, then evaporation can get you the rest of the way (which actually agrees quite well with the table I linked to).
The link I had posted earlier in the thread (http://solarcooking.org/plans/funnel.htm) was for IR cooling in a sealed container, so evaporative cooling was not an option. Here they could freeze water even when the overnight temperature got no lower than 47 F, so this can work at higher temperatures than the evaporative method.
MKelly,
both O2 and N2 absorb UV. It is part of the atmospheric chemistry so important to our ground life. O2 additionally is a minor absorber in the visible range.
MKelly,
Here is a source: http://www.coe.ou.edu/sserg/web/Results/results.htm
Hey Myrrh,
you might want to take a look at this data also. Pure Water is NOT TOTALLY transparent as it DOES absorb and emit at very low levels. Please note the exponents in the numbers.
Here is another H2O data source: http://www.martin.chaplin.btinternet.co.uk/vibrat.html
tallbloke,
“Energy in = energy out. ”
I can’t argue with that!! 8>)
Nope, I am primarily talking about the flux “measured” above the surface. The fact it is a flux means it is not resident in the water or the atmosphere. I am wondering about the residence at the water end myself which is why I am only willing to accept “slowing the cooling”. Whether it is the same energy or not, the fact is the surface loses more than it receives so CAN’T be “warming” due DLR.
Tim Folkert: i’ll buy that the foiil is insulating the contents from external energy except from space.
now i read the article more carefully and see, as you point out, that the jar was sealed. i was guilty of skimming instead of paying proper attention.
heh- so i guess the dlr makes good ice of water…lol
myrrh – the video is a kid burning stuff around the house with a blue laser.
lasers are monochromatic. visible light lasers can be used to burn stuff.
visible light, when absorbed, becomes heat.
visible light does not reach the bottom of the deep ocean because it is absorbed.
when it is absorbed, it becomes heat.
glass does not pass IR . when canadians use a magnifying glass in the sun to burn ants, it’s visible light delivering the heat energy.
dude- everybody has been trying to help you get right. my ego can’t be harmed by your stupid, so forget about that. i don’t suppose i’ll continue to be charitable, though.
“heh- so i guess the dlr makes good ice of water…lol”
It is an interesting concept, isn’t it?
The extension, of course, is that if you could somehow aim the device so that it also “missed” the atmosphere’s IR (for example by removing all the greenhouse gases), the water inside the foil cone would mostly “see” the 3 K of outer space. Then the water would get MUCH colder yet.
Good thing we have all those GHGs and their IR helping to keep us all warm!
kuhnkat says:
August 23, 2011 at 11:22 am
Hey Myrrh,
you might want to take a look at this data also. Pure Water is NOT TOTALLY transparent as it DOES absorb and emit at very low levels. Please note the exponents in the numbers.
Here is another H2O data source: http://www.martin.chaplin.btinternet.co.uk/vibrat.html
You’re a comedian right? You’re defending that Visible is the main heating wave to convert to heat all the land and oceans of our planet, to raise the temperature of all the Earth as claimed by AGWScience Fiction Inc’s takeover of the education system, and you give me a page which says:
“Water is almost perfectly transparent to ‘visible’ light, a property which is made good use of by photosynthesis and allowing production of both biomass and oxygen.”
Even if this work is replicated, all it is doing is confirming that water is transparent to visible. What isn’t being used in photosynthesis which is a chemical conversion and not a heat creating conversion, is transmitted through. What can’t you understand about this? The graph shows that water is even in this work, if this work is correct, for all practical purposes not absorbed by water. Not that I can make sense of it, the UV looks very odd.
Why don’t you understand that you’re confirming what I’ve been saying here?
Spector says:
August 23, 2011 at 5:47 am
On the Wikipedia ‘Electromagnetic absorption by water’ page, they are using a centimeter scale to measure EM absorption in water. This, I assume, is pure water uncontaminated by algae or plankton. Although undefined here, as far as I could find, the absorption coefficient usually indicates by what value the natural logarithm of the signal is reduced when passing through the specified distance (1 cm). An absorbed photon delivers its packet of energy to some ‘lucky’ molecule.
BTW, this page obviously appears to have been written by someone who has fully bought in to the greenhouse catastrophe theory. That does not mean that this largely unrelated data is incorrect or fudged—it also includes a cute water molecule vibration mode diagram.
Yes, obviously written by someone wanting to minimise what the graph is saying.. 🙂 “This water absorption occurs preferentially at certain characteristic wavelengths while the balance of the spectrum is transmitted with minimal effects.” That is, zilch absorption for Visible, it is transmitted with zilch effects, and, all the absorption is by the bulk of thermal infrared; the ‘peaks’ here he points out are for the most part below the bulk ‘peak’ of thermal infrared which is a table top mountain kind of peak..
Visible light does not work on a molecular level, it can’t move molecules as does thermal infrared. Visible is tiny. It gets lucky if it hits an electron.. That’s the best it can manage in the atmosphere, where absorbed briefly by an electron it gets bounced back out the way it came, no heat is being created. That’s what reflection and scattering is.
Visible is not powerful enough to move the molecules of oxygen and nitrogen in the atmosphere to convert them to heat. Water is completley transparent to Visible, not proven in any way that it isn’t, you can see straight through clear water.. Visible doesn’t have the mechanism to interact even on an electron scale with water, is really transparent, it is transmitted through unchanged. Unlike the atmosphere which isn’t really transparent, because of absorption by electrons to produce reflection and scattering.
Here, look at this in the real world where applied scientists who actually do things that have to work: “That sunlight can penetrate water is a well known phenomenon. In fact, it is an essential requirement to sustain the life of aquatic plants like algae that grow in water.” But do read the next sentence. http://almashriq.hiof.no/lebanon/600/610/614/solar-water/unesco/29-34.html
..just fades away…
See though, UV is not absorbed, this is all about targetting nasties in the drinking water supply where it’s important to get UV to them (because UV is more energetic and works on DNA levels).
Here, for UV not being absorbed by water which remains unchanged, question 8 http://www.iuva.org/iuva/faqs
‘Oh, we’ve seen a little blue’ said some ascientists digging around in some ice somewhere, we think water is really blue’ Nanas. Blue is more energetic and smaller so travels slightly differently, it’s more easily scattered (so the blue sky).
OK, enough of this (y’all generally). Water is well known to be a completely transparent medium for Visible. This means that it is not absorbed by the molecules of water, but passed through. Visible cannot heat water. It has no method in place to do so.
It cannot move the molecules into the vibrational states necessary to convert to heat. It’s a bloody physical impossibility for the tiny little tikes.
Visible is used for photosynthesis primarily, that’s its role in life.
Unless you can show that real science, tried and tested and used in countless applications worldwide in a huge variety of fields, doesn’t understand this, then you must accept that the AGWScience fiction you’re promoting ‘that visible heats the Earth’, is just that, science fiction.
Strange how oceans get darker with depth if they are transparent to visible light.
Myrrh, I see you have not yet explained why car steering wheels get hot after long parking exposure to summer sunlight through the windscreen. Odd really, considering IR cannot penetrate the glass.
The colour of the car has an effect too. White reflects visible light a great deal more than black, (white cars are cooler) but IR absorption is not affected by visible body colours.
Stephen Wilde, Re:
To be fair, I think Myrrh did say early-on that visible light gets tired, and just slows down in the oceans. He/she did not explain what energy is involved in decelerating those naughty photons from ~300,000,000 M/sec to zero though, and I‘m not convinced.
Just sayin’ ….
http://math.ucr.edu/home/baez/crackpot.html
REPLY: and 100 points for being Tim Folkerts – Anthony
Tim and gnomish,
before worrying about how to sneak between those hot DLR’s, you might consider exactly how much IR is radiated to the water by the device itself!
Myrrh,
“Water is almost perfectly transparent to ‘visible’ light”
OK, I think I am starting to see where your problem comes from. That ALMOST is all that is needed to absorb most of the visible energy even without the contamination in the water. You apparently have even less of a concept of the magnitudes that we are dealing with than I do. It really helps to have a LITTLE math skill. Even though that absorption is positively miniscule, the number of layers of molecules the light has to pass through adds up. Do you have any idea of how many molecules of water it takes to make a column of water 500 ft. deep??? Neither do I. It is a rather large number comparable to the attenuation provided by those large negative exponents on the absorption charts. In other words, even though each layer takes away an amount so small it is difficult to measure, after that many layers it has all been nibbled away. Notice that the least absorptive bandwidth is actually in the UV close to your BLUE light you don’t think absorbs or carries energy.
Repeating that water is TRANSPARENT is a childish tool to reinforce your own misunderstanding of the issues. Until you stop it you will simply continue to look silly.
You still haven’t explained to me how the tiny amount of THERMAL or FAR IR from the sun that makes it to the upper atmosphere manages to warm the earth. Want to take a stab at it now?? About 50% of the sun’s output is in the Infrared. Over 90% of that is in the near infrared you don’t think heats anything. So, you are left with less than 5% of the sun’s output attenuated by distance and a small cross section heating the earth. Please explain how this works??
Very troubling that one of my favorite bloggers decided to abandon all dissenting views and ignore challenges to his “integrity?.” The cowboys that raised me would label this behavior as chickenshit. WTF, Willis?