Guest Post by Willis Eschenbach
One of the best parts of new tools is new discoveries. So the tools to calculate the heat constants of the ocean and land as described in my last post, Lags and Leads, reveal unknown things to me.
A while back I wrote a post called “Wrong Again”, about a crucial mistake I’d made that was pointed out by a commenter, might have been Mosher. In any case, what I’m wrong about this time is that I always thought that over the course of a year, much more energy per square metre of surface was stored in and subsequently released from the ocean than was stored in the land. I figured that in part this was a result of the difference in heat capacities (specific heats) of water and soil, which are in a proportion of about three to one per cubic metre. That is to say, it takes about three times the energy to warm a cubic metre of ocean by one degree C that it takes to warm a cubic metre of soil. I also figured that because the heat can penetrate the ocean more deeply, it would store more of the heat than the land would. Finally, I figured that the differences in albedo would favor the ocean over the land.
One of the joys of writing for the web is that commenters set off whole trains of thought about what I’m investigating. It’s like having a set of global colleagues. Sometimes rambunctious colleagues, to be sure, but well worth it. In this case a comment from Bruce Ploetz led me to look into the question of the implied heat storage in the thermally lagged system.
In my last post I used the example of putting an aluminum pan on the fire versus putting a cast-iron pan on the fire. The aluminum pan has a small heat constant “tau”, while the cast-iron pan has a large heat constant tau.
An oddity of this is that we can calculate the relationship between tau and the actual size of the thermal mass. For the earth system, it works out to a thermal mass per square metre of 7.9 metres depth of water. This is the amount of thermal mass that is involved in the annual temperature swings. I usually call that 8 metres for quick first-cut calculations. (I’m not sure where I got that number, 7.9 metres thermal mass per month of tau in the earth system, although it was from a trusted source of some kind. Any assistance in backing up that number would be appreciated.)
As a result of my newly-gained ability to calculate the time constants tau for the ocean and the land (on the order of 3+ months and 1 month respectively), we get about 24+ metres and 8 metres of thermal mass for the ocean and land respectively.
So … given that we have that much thermal mass for the land and the ocean, how much energy goes into and out of the thermal mass per year?
Now, I’m a great fan of rules of thumb, which I keep in my head for back-of-the-envelope calculations. One such rule of thumb is that a watt per square metre of incoming energy applied over a year (1 W-yr/m2) will raise a cubic metre of seawater by 8°C.
This means that for the land, with an involved thermal mass of 8 metres depth, 1 W-yr/m2 changes the temperature of the thermal mass by 1°C. And similarly, for the ocean’s involved thermal mass of 24+ metres depth, it takes about 3+ times that or 3+ W/m2 to change the temperature by 1°C.
Here’s the key graph, from my last post:
Figure 1. A comparison of the annual temperature swings of the northern hemisphere ocean and land. Solar variation in both cases is the same.
The land is easy, because the involved thermal mass warms and cools at one degree C per watt-year/m2. It swings 28°C, but remember that is in half a year. So the rate is 56°C per year. And conveniently, this is also 56 watts/m2. So every year, in back-of-the-envelope terms, there is a flux of about half a hundred watts first into and then out of the land.
Next, the ocean. The swing of the ocean temperatures is 7.7°C per half-year, or a rate of about 15°C per year. The involved thermal mass of the ocean requires 3+ watts per square metre per degree. How nice, the thermal flux in W-yr/m2/°C is equal to tau. That works out to 45+ watts/m2. Now, the “+” I carried through the calculations was plus 10%, as the true tau for the ocean is 3.3. So we need to add an additional 10% to the 45+ watts, giving us 50 watts first into and then out of the ocean … versus 56 watts for the land. Half a hundred either way.
Of course, now that I think about it, it makes perfect sense. The smaller involved thermal mass will heat up more, the larger involved thermal mass will heat up less, and they both actually store about the same amount of energy per square metre. Of course, in global terms, the 70:30 ratio of the sizes of ocean and land comes into play … but per square metre, they each have a flux of about fifty watts per square metre, first into and then out of the thermal mass each year. Good to know.
Always more to learn,
w.
My Usual Request: If you disagree, do us all the favor of telling us exactly what you disagree with. There’s only one way to do this, which is to quote the exact words you disagree with. Anything else is your interpretation. We can only be clear what you think is incorrect if you quote the precise words they actually used.
Math Note:
Here’s the actual calculations in comma separated format.
Type, time constant tau, depth heat mass, W-yr/m2/°C, Swing °C, Equiv W/m2
Land, 1, 7.9, 1.0, 28.3, 55.8
Ocean, 3.3, 26.1, 3.3, 7.7, 50.3
Note that this is only valid with tau expressed in months.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
We’ve seen, maybe, 0.9C increase in GAST over the last 150 years (which goes back before the Wright Brothers. While I can’t speak to the other topics you cover, I can state that, in my 40+ years in aerospace engineering, we designed our aircraft to be sufficiently robust that 0.9C wasn’t going to make a serious difference.
Sir, you raise some interesting points. But I think you have it wrong. Those parties should be screaming. Screaming the opposite: That they aren’t finding what was projected. Never mind the ancient refrain about theoreticians vs the engineers – both have their place in the mobius conveyor that is science.
The tragedy is that we have darn near every research scientist, engineer, and data handler remaining utterly quiet about the factual status of climate change.
Keck for example has specific forecasts for nightime temps. but only since 1998 or so.
fundamentally you are looking for forecast of good weather, the specialized forecasts predict nightime
temp within a degree, windshake for the telescope and atmopheric turbulance.
You only care about the temp to match the mirror temps as closely as possible.
http://www.gb.nrao.edu/~nradziwi/Starcasting.pdf
People still use Refractors. No mirror.
michael
Largest refracting telescope is 40 inches. Largest reflecting telescopes are around 10 meters.
Refracting telescopes are opaque to certain wavelengths, and suffer from chromatic and spherical aberration problems. Larger retractors suffer from lens sagging.
Serious scientists don’t use refracting telescopes.
PA
True but that is not the issue. The issue is how an increase in heat would effect telescopes..
And Serious Astronomers sometimes do use refractors. They have for hundreds of years.
CO2 does rule the night, at least in deserts. How much of the BEST increase is the night component?
It does rule the night, but it hasn’t reduce nightly cooling in any measurable way.
Here’s a graph of the difference between yesterday’s warming and last night’s cooling.
But they have rolled out artificial stars and fast mirrors to reduce the effects of scintillation, one might wonder if they collect the correction logs to aid future debugging of issues?
Willis,
You wrote: “This means that for the land, with an involved thermal mass of 8 metres depth”
That is not correct. Go even a meter down in the ground and the temperature hardly varies. But the temperature of the ocean varies throughout the mixed layer, to a depth of tens of meters. Combine that with the factor of 3 for relative volumetric heat capacities and the thermal mass of the ocean is at least 100 times that of the land, per unit area.
I think that your mistake is that you seem to be treating the ocean and land as isolated systems. But there is a large exchange of heat between them, largely in the form of latent heat. So you can not get a meaningful result for the part that constitutes 99%.
The last bit of my message seems to have been garbled. It says “So you can not get a meaningful result for the part that constitutes 99%.” But what I wrote was more like “So you can not get a meaningful result for the part that constitutes less than 1% of the whole without considering the exchange with the part that constitutes greater than 99%.”
But I used the symbols . I wonder if that caused the problem.
“But I used the symbols . I wonder if that caused the problem.”
Yep. The symbols and all that was between them was deleted by the software.
Howdy Willis,
Interesting articles. I just wanted to point a bit of a confusing paragraph in the first part of your “Lags and Leads:
“Second, and this is an important point. In general, the shape of the exponential lagged response (the pan temperature) is NOT the same as the shape of the impulse (the on/off of the fire). In the red lines in Figure 1 above, you can see how the square-wave on/off thermal impulse of the flame (black/yellow line) is transformed into a kind of shark-fin shaped thermal response in a heavy cast-iron cook pot.”
When you say “transformed” it sounds like you are stating that the input is transformed rather than what I think is your actual meaning, which should be “appears to be transformed”, because the 2 are completely separate. I know it’s nitpicking, but it can be confusing to some, as I had to read it a couple of times.
Thanks
Dahlquist
Drive on, you magnificent Bastard!
Entertaining and interesting posts as always, Willis, but I think there is a mistake in the calculation of the heat capacities C for ocean and land in the previous post. Without explanation it is said to be just 8 times tau in appropriate units while in the beginning of the paragraph you claim that it can be calculated from lambda. The correct relation is C=tau/lambda. This gives your number for the sea but a factor of two less than your number for land. Probably you found the factor 8 from your calculation for the sea and just applied it also for land?
To people who have had an elementary course in electronics, the model will be very familiar as an RC circuit. Just replace Joule by Coulomb, Kelvin by Volt, and lambda by R [Ohm]. The time constant for discharging is tau=RC.
Here is comparison (without comment) of the CET daily max/min (low pass filter) to the insolation with one month delay
http://www.vukcevic.talktalk.net/CET-dex.gif
Willis asked: “I’m not sure where I got that number, 7.9 metres thermal mass per month of tau in the earth system, although it was from a trusted source of some kind. Any assistance in backing up that number would be appreciated.”
The trusted source who wrote this article calculated a mixed layer of 60 meters. Others usually get 30-50 m, but they use different methodology. http://wattsupwiththat.com/2015/06/06/can-we-tell-if-the-oceans-are-warming/
Willis wrote: “One such rule of thumb is that a watt per square metre of incoming energy applied over a year (1 W-yr/m2) will raise a cubic metre of seawater by 8°C.”
However, as that cubic meter of seawater warms it begins to emit more thermal radiation though its surface. If we treat the seawater as a blackbody at 288 degK, a 1 degK rise in temperature produces a 5.4 W/m2 increase in thermal radiation. So after approximately, 0.2 degK of warming, your 1 m^3 of seawater will be emitting an additional 1 W/m2 of thermal energy – negating the incoming 1 W/m2 and preventing further warming. It is more appropriate to say that 1 W/m2 radiative imbalance can raise the temperature of a 50 m mixed layer at an INITIAL RATE of 0.16 K/yr. However, within about half a year, Planck feedback will have negated about half of this forcing and the rate of warming will have dropped roughly in half. This produces an exponential decrease in the warming rate.
Why would you treat the oceans as a blackbody? Do they have an absorbancy of 1? Are they isolated by a vacuum? No. So the ocean isn’t a blackbody, so you can’t treat it like one.
Wicked, water is indeed very close in its emissivity to a black body. If it is warmed slightly, it emits more IR. We can’t perceive things in IR like some animals. If we could and could not see in the ‘visible’ range, water would look like Texas Tea.
And what becomes of the increased radiation from the warmed water? Ignoring the substantial energy diverted from radiation to evaporation at the surface as evidenced by the “cool skin”, that radiation is mostly immediately absorbed by water and CO2 in the atmosphere and flung back. Cycling, constantly cycling over short distances at the speed of light in…the photon food fight.
Gymnosperm asked: “And what becomes of the increased radiation from the warmed water?”
Great point. (I was trying to keep things simple.) A little of the radiation from warmer water escapes directly to space. The rest is absorbed by the atmosphere, warming it. A warmer atmosphere radiates more LWR to space (OLR) and towards the surface (DLR). The increase in DLR arriving at the surface does negate most of the increase in emission of thermal IR. As you note, the warmer water also transfers more energy to the atmosphere via evaporation. A 1 K increase in SST increases equilibrium water vapor by about 7%, potentially a 5.5 W/m2/K increase in latent heat – bigger than the Planck feedback I discussed above. Since air very near the surface of the ocean is saturated, the rate of evaporation depends on turbulent mixing (followed by convection) and can’t be calculated from simple principles of physics. So I chose to illustrated the problem with this post only by bringing up the increase in thermal emission with warming. That is easy to calculate.
One way to avoid this problem is to consider the surface and atmosphere as a single unit which can gain and lose energy only by radiation. Since the whole planet doesn’t emit like a blackbody, we use the climate feedback parameter: the increase in net radiation to space associated with a 1 degK rise in surface temperature. The climate feedback parameter is the sum of Planck, WV, LR, cloud and ice-albedo feedbacks in units of W/m2/K. CERES provides some information about these feedbacks and shows that AOGCMs don’t get feedbacks right. ECS is the reciprocal of the climate feedback parameter. If the earth were a blackbody at 255 K (our blackbody equivalent temperature), it would emit about 4 W/m2 more radiation to space for each degK of warming and our ECS would be about 1 K for 2XCO2. If ECS were 2 K, the planet would emit (or reflect) only 2 W/m2 more radiation for every degK of surface warming. If ECS were 4 K, only 1 W/m2/K. And if the planet could warm without any addition radiation being emitted or reflected to space, we would have a runaway GHE.
Does Willis understand that when a radiative imbalance produces a temperature change, the Planck feedback associated with the temperature change starts reducing the radiative imbalance? This produces the negative exponential rate of approach to a new equilibrium temperature that appears in his equations. Willis mistakenly converts a radiative imbalance into a temperature change using dW = lambda*dT, an equation that is only valid for a change in radiative forcing – dF – after many decades when equilibrium has been reached.
“One way to avoid this problem is to consider the surface and atmosphere as a single unit”
Luv it. What choice do we have? It IS effectively a single unit.
“Does Willis understand that when a radiative imbalance produces a temperature change, the Planck feedback associated with the temperature change starts reducing the radiative imbalance? ”
Not sure but I think this generally plays into what Willis is saying. There must be negative feedbacks to all temperature forcings or this planet would have gone willy nilly a looong time ago.
I personally need to dig into your Plank aspect further, and I bet Willis will as well.
Thank you.
Gymnosperm wrote: “Not sure but I think this generally plays into what Willis is saying. There must be negative feedbacks to all temperature forcings or this planet would have gone willy nilly a looong time ago.
I personally need to dig into your Plank aspect further, and I bet Willis will as well.”
You (and Willis) are correct that the sum of all feedbacks – the climate feedback parameter – can’t be positive (or even so close to zero that it might have become positive in the past). Willis believes negative cloud feedback limits local daily warming in the tropics.
I’ve left several comments for Willis on Planck feedback, but they don’t appear to have changed his thinking. Radiation is power, energy per unit time; while temperature is proportional to energy. If there were no Planck feedback (increased emission of thermal radiation with increasing temperature, W = eoT^4), all illuminated objects would continue to warm indefinitely. A radiative imbalance always produces an initial warming RATE, not a temperature change. As the object’s temperature rises, its thermal emission rises, diminishing the radiative imbalance and slowing (via a negative exponential) the warming rate. Eventually, a new EQUILIBRIUM temperature is reached, which gives us ECS. Only when equilibrium has been reached does dF = lambda*dT. Willis (and many others) improperly apply this equation to non-equilibrium situations and don’t discriminate between a radiative forcing (which generally persists at equilibrium) and a radiative imbalance (which goes to zero at equilibrium). This flawed approach avoids the need to deal with heat capacity and Planck feedback.
If the earth were a blackbody at 255 K, the S-B equation predicts that it would emit 4 W/m2 more OLR if the temperature rose to 256 K. Since this is power lost, it is a Planck feedback of -4 W/m2/K. (Sophisticated calculations accounting for the actual temperature of the emitting GHGs, clouds and surface give 3.2 W/m2/K). Increasing water vapor in the atmosphere of a planet that is warmer absorbs about 2 W/m2/K (+2 W/m2/K since absorption keeps heat in) of that increase in OLR before it can reach space. Increasing water vapor also reduces the lapse rate so that the surface warms less than the bulk of the atmosphere. For every 1 degK rise in surface temperature, the atmosphere warms a little more and emits a little more, about -1 W/m2/K. The sum of these three feedbacks is -3W/m2/K (or more accurately 2.2 W/m2/K). That leaves the big source of uncertainty, cloud feedback. There are also feedbacks associated surface albedo, the reduction in seasonal snow cover and ice caps. All of these feedbacks except ice caps develop during the seasonal cycle. The sum of all feedbacks is called the climate feedback parameter, which might be negative. The reciprocal of the climate feedback parameter is ECS in K/(W/m2) and is usually multiplied by 3.7 W/m2 per doubling of CO2 to give ECS in K/doubling.
With all these new learnings about sensible heat and the laws of thermodynamics that you’re having, are you ready to discard “back radiation” and “greenhouse effects” yet?
‘Backup’ a little Backup to your 8 metrrs annual temperatur swing is the temperature of ground water and springs. Year round springs mostly have water temperature close to yearly average trmperature of the place. Melt water run off or hot springs are discernable by temperature.
Greetings
Very good. Now, if we could just model what the clouds are doing over the seas, plains and mountains we might be able to understand what is really happening .
Willis,
I have often made the point that land thermal mass is greater than it appears because the rain water carries energy directly into the ground, often to great depths. Unlike the ocean, the heat/cold is always being pulled downward by the direction of water flow from the surface. It is very difficult for the heat to escape the surface of the ground if a warmer rain transfers its energy and a couple days later another warmer than average rain is flowing more water through the same rock.
It is very interesting to me and it is something climate models don’t appear to handle well.
Jeff Id,
More complexity regarding temperature in the ground (shallow solid earth). For example, in the young volcanic mountains of Oregon’s Cascade Range, replete with active volcanoes, one might guess that the geothermal gradient is high, that is, that the temperature increases rapidly with depth because of the active volcanism. Not so because young volcanic rocks are greatly fractured and fine aquifers, generally. Subsurface temperatures in the Cascade Mountains are depressed significantly for a great depth (thousands of feet, if my memory is correct) because of the permeability that permits cold rain water to circulate deeply and cool the rock mass.
Not to mention that heat flow varies greatly. I recall “normal” heat flow values of 45-60 W per Squ M.
Bill Rocks
Jeff Id, a very cogent factor! It would be interesting to measure heat flux at the surface of an area of very low rainfall with one of high rainfall, say east and west of the rockies at the same latitude and altitude and probably rock formation (igneous or sedimentary) or between the dry part of South Dakota and Wisconsin.
the heat/cold is always being pulled downward
=================
I was surprised to learn years ago that rain delivers more heat to the surface than does sunlight.
Thanks guys, your replies made me want to blog a bit on what seems to be an endless cycle of reading paper after paper.
https://noconsensus.wordpress.com/wp-admin/post.php?post=14865&action=edit&postpost=v2
Well, Nick Stokes left a comment on my link which I found fairly convincing. Compared to other factors, there simply isn’t much heat which can be carried by rain to the ground. I just wanted to leave this here because wrong information isn’t very helpful.
Well that was pretty cool. Thanks for putting those thoughts out for us. I fly and what you say in that field resonates with my own experience.
Several years ago, I enquired (many times) whether anyone knew whether the increase in DWLWIR causes for infra red telescopes, the equivalent of light pollution for ordinary astronomical telescopes operating in the visible light spectrum.
If it does, then all IR telescopes would have required some form of re-calibration or would be less effective since the mid 20th century (due to the increase in DWLWIR coming from increased levels of ghgs and/or due to the warmer atmosphere) and one would have expected to see articles written in the astronomy journals on this subject.
Indeed, even optical telescopes may be affected by an increase in temperature. As one knows from Hubble, the tolerances are very small, and just a very small expansion of the metal frame supporting the mirror could lead to distortion of the mirror and hence have a significant effect on image quality/focus.
As far as I am aware, I have never seen an article discussing any of this, and that raises the spectre of whether there has been any significant increase in DWLWIR, or significant increase in temperature. It may well be the case that the land based thermometer record showing warming is very much exaggerated.
.
While I was skeptical, it wasn’t until I started paying attention to how quickly it cooled after sunset while logging night time temps when I was doing astrophotography. Since I don’t use a cooled camera, to reduce sensor noise from temperature you create dark frames, and you have to have darks for the temp you capture lights.
My question was is it cooling slower now than it use to, and I’ve found no evidence in over 120 million surface station records. I posted a chart of the difference between yesterday’s rising temps and last night’s falling temp.
A little further down below this reply.
“Willis wrote: “One such rule of thumb is that a watt per square metre of incoming energy applied over a year (1 W-yr/m2) will raise a cubic metre of seawater by 8°C.” ”
Watt is not energy.
A watt is a power unit, energy over time, Btu/h or kJ/h. So 1 W-yr/m^2 is 3.412 Btu/h for 8,760 hr or 8,760*3.412 = 29,890 Btu/m^2 per year.
Is that a spherical or circular m^2? Btu per land m^2 (heat capacity Btu/lb-F) or Btu per ocean surface m^2 (sensible heat capacity 1 Btu/lb-F or latent heat of 970 Btu/lb)?
Nicholas, a Watt-yr IS a unit of energy. A kilowatt-hr is a unit of energy. A watt is power but when measured as application of this unit over time – it is energy. Please, unless you have some science background, don’t try to correct peripheral details. One watt-yr is close to 30,000BTU and it would take about 4BTU to heat a kg of water by 1C. 1m^3 is 1000kg, so it would take about 3,900BTU to heat it one degree and 30,000/4000 is 7.5C, close enough for this calculation to 8C.
Nicholas did not say watt-yr is not energy.
I like your attempt to quantify this but it’s so much more complex in reality. Much more thermal energy DOES go into and out of the oceans than land, and I think it is quite apparent in temperature observations.
There are only 3 points that matter:
1) IPCC can’t assign blame to industrialized mankind for the increase in atmospheric CO2 between 1750 and 2011 because they have no reliable numbers for the natural sources and sinks. (& neither can anyone else)
2) The 2 W/m^2 additional RF due to CO2 between 1750 & 2011 is of no consequence considering the magnitudes and uncertainties in the overall global heat balance especially the water vapor cycle.
3) IPCC admits their GCMs can’t explain the pause/lull/stasis/hiatus and are essentially useless.
All the rest of the discussions and debates are interesting academic sideshows, but simply a lot of distracting noise.
Cool
I know that Willis is busy, but there are a number of points raised in the various comments upon which it would have been good to have had the benefit of his views.
The same happened, in the recent earlier threads. I am not criticising, but it is frustrating for this to happen, and much could be learned by continuing the debate..
richard verney,
I share your frustration. But knee-jerk defense is easy, while reconsidering one’s position is hard. I am hoping that Willis is carefully reconsidering his calculations in light of the comments here and will treat us to an improved version in the near future.
Think you are wrong now ?, wait 1-200 years.
Wish I could see all the new discoveries 🙂
Hi guys,
I am new in this area, so here is a stupid question that bothers me. Related to:
Khwarizmi August 25, 2015 at 2:48 am: Good question. Temperature increases by ~30°C per kilometer below the sediments.
Ok, this is a geothermal temperature increase with depth. Make sense. But here comes the question:
Why does the temperature DROP below the water surface in the ocean, reaching +4 oC in the bottom, while it increases by ~30°C per kilometer below the surface on land? If there is geothermal energy flow, shouldn’t it equally heat the ocean water as it does the solid matter on land?
OK, there is a difference in thermal conductivity, but this should only affect the GRADIENT, but not the direction of change. In other words, the temperature below the layer affected by solar radiation and thermal conductivity with the atmosphere (say, 200 m) should ONLY INCREASE with depth in the ocean as it does on land. Right?..
Will be thankful for any insight.