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.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
You lost me when you quoted that insolation is reduced to 170 W/sq m at the surface while DLR is more than 4 times the 70 W/sq m you say is accounted for by evaporation – ie ~280 W/sq m.
I guess reality is dead.
How can DLR exceed insolation ? That is impossible !
“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.”
This beats me. The big ocean with its enourmous heat capacity warms the small atmosphere with its little heat capacity and then the big ocean is heated by the small atmosphere with its little heat capacity, again and again. Like if I should be pounding with my own straight lefts on Mike Tyson.
I kinda sniff that something is seriously wrong here…
Or maybe, if you take the radiation approach, you always have to consider radiation in two directions, like Kiehl and Trenberth. Isn’t this correct? Then the net radiation is outgoing (sum of two vectors) and the DLR is negative so it doesn’t heat the ocean.
Simple as that. From a layman’s perspective, that is.
Sun Spot says:
August 15, 2011 at 5:05 pm
re: Alexander Duranko says: August 15, 2011 at 3:28 pm
“Anyone ???, I’m hearing crickets on this item.”
It’s so wrong it’s not worth a response would be my guess. That’s certainly why I didn’t bother. A cheap remote IR thermometer aimed upward on a clear night with air temperature exactly the same with the only difference being humidity will register a higher temperature on the more humid night. A rather expensive gadget costing in the five figure range can measure DLR day or night under any conditions. It’s called a pyrgeometer. Doesn’t every grizzled old engineer have one of those puppies on his bench sitting between his ohmmeter and oscilloscope? /sarc
Judith Curry posts another round of Slaying the Greenhouse Dragon: Part IV by Vaughan Pratt at Climate Etc.
(Willis is clearer)
As we all know, heat is transferred by conduction, convection and radiation. If the top 50 micrometers of the ocean is warmed by IR, wouldn’t those molecules slough off their extra energy almost instantly via conduction, warming the adjacent molecules? And so on, transferring the heat deeper into the ocean. Am I missing something?
Smokey says:
August 15, 2011 at 5:36 pm
=============================================
Me too Smokey…..I know I’m missing something
We don’t plant tomatoes until we can dig down a good foot and a half and the soil has warmed up…
I cannot see why it is considered valid to reduce the insolation to Earth by a factor of 3/4 because half the earth is in darkness at any one time and a disk has half the area of a hemisphere.
Almost nowhere on earth therefore receives the 234 W/sq m used to calculate the fictional minus 18/19 C.
Surely the whole of the tropics is subjected to an insolation 2-3 times more than this figure during a significant portion of any day and then it commences cooling at night.
The mean is meaningless and ought not be used. I refuse to believe DLR heats the earth more than the sun – it defies logic.
The real question we ought to consider is given space is ~3 K why is it Earth isn’t even colder ?
Perhaps then we may perform better analyses than being done at present.
Even IPCC documents quote higher insolation figures than the oft quoted 235 W/sq m but they trickily often lead with the Kiehl & Trenberth diagram which defies logic or science.
You can’t argue with someone who is trying to deceive you and using a model which in no way reflects reality is deception.
How do they calculate the work of the earth’s climate system or is radiation the only game in town ?
gnomish says:
August 15, 2011 at 1:08 pm
“http://en.wikipedia.org/wiki/Solar_pond”
Solar ponds are shallow, need a dark bottom, and have to be shielded from turbulence so that saline stratification can occur. The mode of operation is that visible light from the sun is absorbed by the dark bottom and that conductively heats the water from the bottom up. The high salinity of the bottom water changes the temperature/density relationship such that the hot brine can’t rise off the bottom and mix with the low salinity layer on the top.
The ocean is nothing like that. It is effectively bottomless with regard to light from the sun and no signifcant saline stratification occurs in the mixed surface layer which absorbs the solar radiation.
.
First, should it not be an extremely simple task (for a scientist) to measure how much DLR there is? That way it can be determined how much it heats. Say for instance, you cover the ground in a substance that can measure the DLR to some precision out in the open air. Then you move a DLR reflective surface on top and ULR absorbing surface on the bottom over top of the measuring substance, and measure it again without most of the atmosphere shooting the DLR downward.
I hear all the thermodynamics arguments of whether radiative energy can warm an item, but is that really the question we are looking for? I just have not seen any evidence that much radiation ever returns to the surface, if it really increases based on the amount of greenhouse gas present, even with the experiment being done in a laboratory where you can limit the number of contributing factors to come out with a realistic starting point.
Dave Springer says:
” Most the remaining heat loss is blocked by the reflective coating. That’s why laying tin foil over non-reflective attic insulation doesn’t help a whole lot but does help some.”
on the floor, where little heat is lost by conduction and absolutely none by convection, a layer of foil, shiny side up, will provide quite excellent insulation – that floor will be the warmest one in the house. next time you tile a floor, try it. it works great.
if you consider co2 functioning as a mirror for IR, then it’s obvious that it is preventing heating from above exactly as much as it is preventing cooling below – except at night, of course.
now, if you could bring that agw ocean.in.a.bottle out of the virtual world inhabited by alarmist witch doctors, the computer models might apply. but not in real life.
no real ocean is anything like that. there is a constant flow of water gas from surface to stratosphere (and back). any co2 gets swept along in a refigeration cycle by the hadley heat pump. it does not sit like a reflective shield over the sea.
you can add all the shiny crap you want to the freon in your air conditioner – it will not alter the efficiency of the refrigeration cycle through radiation physics.
similarly, the heat pump of the hadley cell- IR radiation exchange within the flux has zero effect on the transfer efficiency.
adding heat capacity, however (and co2 is a bit more dense than elemental gasses) improves the efficiency, not the opposite as claimed by the fetishist rent seekers.
Phytoplankton are packed with chlorophyll and other chromophores. The IR spectrum of different chlorophylls is well known,
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC540707/pdf/plntphys00384-0069.pdf
Plants are only green because we can’t see very far into the red.
The classical way to cool a white light source, for doing photolysis, is to place a round bottomed flask full of water in the beam. This stops sample heating.
Willis
First of all and importantly, we are not talking about mm but rather microns. Due to the wavelength, DWLWIR cannot penetrate more than a few microns into water. The problem is that nearly always these first few microns of the ocean are divorced from the ocean being in reality wind swept spume and spray. These few microns are therefore, in the limit, not connected with the ocean and thus even if there was sme physical process (which no one on the warmist side has yet sought to explain) whereby heat can in effect be conducted downwards, it beggars belief that there can be any transport of the warmer micron layer into the bulk ocean. The most likely scenario is that the DWLWIR absorbed by these first few microns are heated and evaporate thereby causing cooling not warming. As you know, the ocean is constantly evaporating and in your earlier article, you discuss the cloud formation which is the result of this evaporation. The DWLWIR cannot overcome the evaporative and convectional forces that are at play (which forces have principally been driven by solar energy). If you like the DWLWIR cannot swim against the tide.
Your dismissal of the first argument is with respect complete and utter rubbish. The DWLWIR absored by the first few microns of the ocean is burnt off and evaporates thereby preventing any transport of heat into the deeper ocean. There is no equivalent process with respect to the land. Rocks or sand or tarmac or whatever does not get burnt off and evaporate away. Further, I suspect that there is a large number of people who consider that backradiation cannot heat the land. Whilst the warmists struggle to describe the basics of their conjecture, many armist acept that backradiation does not heat but rather it acts in some way akin to a blanket and reduces the rate of heat loss, ie., because of backradiation the land cools slower than would be the case if there were no backradiation.
As regards the second argument. DWLWIR is not heat. We all know that the DWLWIR cannot effectively heat anything and cannot do sensible work. That is why no one is seeking to utilise the alleged 333 mw per sqm of backradiation and use this to cure the world’s energy needs. After all, according to Trenberth, the backradiation is nearly twice the solar energy and if it was truly a source of heat or if it could truly do work, man would exploit this valuable resource.
As regards argument 3, I dealt with that at the outset. We are taliking about the first few microns which is wind swept spume and spray and which if anything is evaporated (eventually giving rise to the clouds in your earlier post) and never mixes with the bulk of the ocean.
As regards argument 4, I believe that I have had that argument with you before. One of the biggest problems in the AGW debate is that climate scientists seek to deal with averages. However, this hides what is going on. The reality is that huge amounts of solar energy are being input into the tropical oceans such that these oceans would never freeze. The heat being absorbed by these oceans is then transported towards the poles with the ocean current conveyor belt acting as a huge heat pump distributing the heat absorbed by the tropical oceans. Of course, by late summer/winter there is not quite enough energy being absorbed by the tropical oceans to offset the reduced solar energy being received by say the Artic ocean such that that ocean begins to freeze over.
I had enjoyed your previous recent articles but this latest one is very light weight and nothing more than conjecture.
If you want to suggest that the oceans would be frozen but for the absorption of backradiation, you need to do 2 things. First, prove that DWLWIR is absorbed to a significnt degree by the oceans. Second, you need to carry out a calculation for each kilometre square of ocean based upoj the amount of solar energy received by the ocean at that point and ignoring DWLWIR and detailing at what point in time that part of the ocean would freeze.
Presently, I am extremely sceptical of the points that you raise and consider that they fall well short of discharging the burden of proof that lies on someone claiming that DWLWIR heats the oceans and prevents them from freezing. .
Merrick says:
August 15, 2011 at 1:39 pm
Small issue of heat capacity notwithstanding.
1DandyTroll says:
August 15, 2011 at 1:22 pm
“So a IR lamp directed at a one square meter tub of water of say one meter in depth to radiate the surface at 170W wouldn’t warm that body of water if it was circulating?”
Not if it’s longwave infrared. An “IR” lamp emits shortwave infrared and a significant amount of visible red light which will penetrate and heat both the water and the walls of the vessel. The infrared coming from the air above the ocean is emitted by a source that’s about the same temperature as the water. DLR cannot possibly warm the water warmer than the air from which the DLR comes from. The argument is over whether it can slow down the rate of cooling. So the experiment to try would be taking two vessels of warm water, say 100F, and suspending a third vessel with water at the same temperature over one of them. The third vessel is your “heat lamp”. If the DLR from the third vessel is capable of slowing down the cooling rate then as all the vessels cool down to room temperature the one without the LWIR “lamp” over it will cool more slowly and you should be able to see this with a therometer in each of the test vessels after some period of time.
Actually duplicating what happens over the ocean makes it sound a lot less likely to the lay person that the “lamp” hung in the air over one of the vessels is going to make any difference in the cooling rate of the vessel beneath it once they realize that the “lamp” is an object the same temperature as the water. Otherwise they imagine the kind of heat lamp commonly used for home heating purposes which has a heated element hot enough to glow cherry red.
“If it only heats the first mm why then are several inches usually warm when going swimming? Or is that an odd question?”
Visible sunlight is heating it far deeper than 1mm with decreasing effectiveness over increasing depth so more heat is added nearer the surface. Without significant turbulence a warm surface layer develops. This discussion isn’t about visible light. It’s about longwave infrared light. There’s very little LWIR in sunlight. Virtually all the LWIR that shines on the surface is LWIR emitted from the surface where a portion of it is reflected back downwards from the atmosphere above it. More or less is reflected back down depending on the exact composition of the gases.
“”””” Dave Springer says:
August 15, 2011 at 4:46 pm
Yet another general misconception is that all the water molecules in any arbitrarily thin layer are at the same temperature. That isn’t how it works. Some of those molecules are boiling hot and some are ice cold. The average of many of them is the temperature. “””””
Do you want to take a Mulligan on that one Dave ?
At any given Temperature, the average kinetic energy per molecule has some specific value, but the individual molecules have energies which plot as some Maxwell-Boltzmann distribution, which puts the most probable velocity at some value, and tends to cluster the velocities at the low energy side of the peak, giving a long tail on the high energy side; but theoretically having no upper bound on the maximum energy.
For any sample that is at some fixed Temperature, this distribution of velocities (energies) is constant. But any individual molecule can at any time adopt any possible value of energy, as a result of collisions with its neighbors. The thermodynamic Temperature is defined in terms of that distribution and the mean energy per molecule.
I have argued that since the M-B distribution is always fully populated, and any molecule can at any time be anywhere in that distribution, that for any individual molecule, the time averaged distribution of energies , must be the same M-B distribution of the whole population, so one can argue that any individual molecule, has a Temperature that is the same time averaged distribution of kinetic energies as the population as a whole has.
It is not true that near by molecules can have greatly different Temperatures, simply because they have greatly different instantaneous energies.
For example, there is absolutely no tendency for an individual molecule that currently has a high KE (velocity) to seek out and target a neighboring molecule, that has a much lower energy and velocity. The directions of two such molecules are entirely random; and whereas heat would tend to flow in the direction of the highest Temperature gradient; that is only true over time, and nothing would stop the slowest molecule in the bunch from colliding with the fastest; but that is just a random event; not a directed result.
I have expressed this before, I consider that at least some of the water vapour released from the ocean skin is condensed well before it drifts magically upwards in a pristine state to condense only at a certain altitude.
If the air at the ocean surface is humid that means it has already condensed and released its heat; there are plenty of nuclei available, salt, DMS, dust particles near land.
This could be a source of DLR.
When considering the point raised by Willis, it is important to bear in mind that due to the wavelength of solar light, solar light can penetrate well to a 10 a depth of metre or even more ( indeed, it can penetrate to a limiyed extent as far as about 100m). If you can see the bottom of the sea bed, you can see how far solar light can penetrate.
However, the DWLWIR being re-radiated by the GHGs is of a wavelength that can penetrate no more than a few microns into water. Crispin in Waterloo (August 15, 2011 at 2:07 pm) gives a good explanation of the physical processes that are going on and details why the DWLWIR does not make its way to any significant extent into the ocean.
Accordingly, when one goes diving or snorkelling one can feel warm layers of water (sometimes due to currents) and one can feel the warmth of the sunlight. It is solar energy that is absorbed and which heats the water.
The so called greenhouse effect works differently over land than over water. There is a significant diurnal temperature range over land, but very little diurnal range over the deep oceans. It may be that over land, DWLWIR helps the keeping of night tme temperatures up, by reducing the rate of heat loss from the land. However, over the oceans, night time temperatures do not drop off because in the limit (by which I mean for the period of darkness), the ocean itself is an all but limitless heat source having the ability to constantly replenished the air temperature above it. Further, over oceans there is high humidity such that the effects of CO2 if any are substantially dwarfed.by the effects of water vapour and the water vapour being evaporated from the ocean beneath it acts so as to impede DWLWIR coming from the atmosphere high above.
@Willis Eschenbach says: August 15, 2011 at 2:04 pm
“…You are right. To be accurate, DLR means that the surface is warmer than if the DLR weren’t there. So you are technically correct, but in common parlance we don’t usually say “It slows the cooling so it ends up warmer than it would otherwise”. We just say “it warms it”.
///////////////////////////////
It is only in climate science that such a statement could be made!!!
It is akin to a bioligist claiming that he has created life by not killing a laboratory rat. Sure, by not killing the rat, the net effect is that you have a living rat, but not killing something is not the same as giving life to something. Likewise slowing the rate of cooling is not the same as warming. They are quite different processes. No wonder climate science is so off the rails.
One obvious reason why Trenberth cannot find his missing heat in the oceans is that it never made its way into the oceans because DWLWIR does not heat the oceans.
@Willis
Point by point flaws:
“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.”
No, it isn’t. Land doesn’t evaporate. Neither does land allow sunlight to warm it uniformly down to any significant depth. That’s why the surface of a lake doesn’t get as hot during the day as a blacktop parking lot and why the lake won’t get as cold at night as parking lot. The solar heating of the parking lot is concentrated on the surface. It gets far hotter during the day and gives up the heat far quicker at night.
“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.”
The energy is going into latent heat of vaporization. That is the very large amount of energy that water absorbs going from liquid phase to vapor phase with no change in temperature. No change in temperature is why it’s called “latent” heat. The energy needed to turn a pound of water into a pound of water vapor at the same temperature is about 1000 times as much as it takes to raise the temperature of a pound of water by 1F. Water vapor of course rises until adiabatic cooling lowers its temperature below the dewpoint but I’m presuming you already knew how clouds form even if you didn’t know how much energy is transported in the form of latent heat from surface to cloud.
“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.”
Wrong and wrong. Wind driven turbulence doesn’t mix down the top 10 micrometers. That’s a film thinner than the wall of a bubble where viscous forces overwhelm other forces. There’s very little diurnal turning in deep bodies of water and again viscosity is the overwhelming force in the top few micrometers preventing it from mixing deeper. Major turning of deep bodies of water is seasonal not diurnal. If you had much experience with deep freshwater lakes year round (I live on the shore of one) you’d know that from experience – you can smell it when it overturns in the springtime. The ocean is no different in that respect which also explains why so much summer heating is retained and released in the winter causing far less seasonal temperature variation than that of land surfaces at the same latitude.
“Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. ”
Wrong again. Liquid water has the same properties as water vapor when it comes to being transparent to visible light and opaque to infrared. The same properties that make water vapor a greenhouse gas make liquid water a greenhouse fluid. The difference is that liquid water is like water vapor on steroids since there’s more water in the first meter of the ocean than there is water vapor in the column of air above it and sunlight penetrates far beyond the first meter.
You have really failed to think through any of these arguments. You experience sailing and surfing and diving isn’t serving you well at all in understanding the physics of water.
Tallbloke @ur momisugly August 15, 2011 at 1:46 pm seems to have kicked off the concept that DLR is absorbed in what is a nano-skin of the water from where it is rapidly reemitted.
Others have added supportive arguments, to which I’d like to add a point:
I think an important aspect of this concept is that it is an inconceivably rapid process, so fast that wave action and water turn-over etc is not a consideration. Since it all happens so fast in a very thin skin, those naughty photons and evaporating molecules radiate hemispherically back into the air where their free path lengths are very much longer and things are more complicated.
Dave Springer,
Since the ocean doesn’t glow in the infrared, it seems to me that when a water molecule receives an IR photon, it immediately sloughs off the extra energy through conduction to adjacent water molecules. Some of the energy is probably contributing to vaporization, but the air even right above the ocean surface is usually not 100% R.H., so there’s not a lot of vaporization going on. And of course, something is warming the oceans.
Trenberth et al are looking in the wrong direction for the “missing heat”. After making a quick pitstop in the cloud deck the missing heat from the last 50 years is now unformally distributed radiation in a sphere surrounding the earth with a radius of 50 light years. They aren’t going to find the missing heat looking down into the ocean. It just ain’t there.
Often I find these threads when all others are long since gone, but this one looked like beating up a straw man to me, so I am now glad to have missed it. I have always assumed that when people say “LW does not warm the ocean”, they mean LW does not propagate in the bulk ocean. To have LW warm the bulk ocean requires absorption at the surface and transfer to depth by other means. It does not behave like SW, which I think is the idea most people are trying to convey.
One is not going to illustrate much about heat transfer in a fluid like water using arguments that exclude some modes of heat transfer.
Anyone who has a pool knows full well that the top few inches gets warm first. That goes down quite deep, even in a pool where there is little turbulence (ie not being used) by the end of the day.
If the pool is being used, that warm water is spread pretty much all through.
I would imagine the sea would behave in the same way, and my experience when out surfing makes me believe it is.