Wrong Again, Again

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:

scatterplot NH ocean land temps vs solar spline360Figure 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.

108 thoughts on “Wrong Again, Again

      • Excepting Gaia. Just as well, because our concerns are trivial to her. As George Carlin said, we are just a surface nuisance like a bad case of fleas.

    • Now vary the incidence angles of solar radiation on the wrinkled land masses and variation of vegetation cover and consider the churning action of ocean waves to bring cooler water into the solar radiation and so on. Very complex and not easily measured.

      • Evaporation, condensation, atmospheric circulation, storms….
        Oh, and if you want anything close to a realistic solution, do all of that to a spatial resolution of 1mm.

      • Hmmm, in fact what Willis says – if I understand correctly – is that the amount of energy he specifies moves in and out of the respective thermal masses each year. He does not specify how. How would include radiation, conduction, evaporation, etc. All of those are routes the outgoing energy can take in their convoluted paths away from the planet.

      • Matt, Michael, Kim,
        And so on, etc., etc. The complexity boggles the imagination. Like many, though, I believe the oceans have much more to do with long term climate and its changes on our 70% water covered world than any other single variable. However it is made up of and impacted by a multitude of other intervening variables not the least of which is TSI from which it stores and circulates and disperses heat in unknown ways. Add to this the unknown amount of geothermal energy it absorbs and the clouds mentioned above and it is questionable if the equation to describe the system could even be written were the values of the variables known quantities.

      • Jim G
        Absolutely.
        The problem is very complex.
        Our data – especially over the oceans (which are big) – is sparse.
        This blog once listed a shed-load of variables.
        Auto

  1. One difference between ocean and land is that heat transport from the surface to below the surface is done almost solely by conduction in land, where convection dominates in the ocean for all but the shortest time sales.
    Another difference is that radiant energy at optical wavelengths (sunlight) will penetrate a fair distance into the ocean as opposed to being stopped at the surface on land.
    Having said that, knowing the Tau for land and water does give some nice information.

    • “…the shortest time sales.” Would that be an instant discount, or as we say in Asia, “For you, special price”?

    • Convection does not transfer enthalpy from the surface to depth in the ocean. Quite the reverse. Convection insures that enthalpy at depth from the deep penetration of light quickly moves to the surface.

  2. Willis, a man who admits his error makes me think on Burns:

    What though on hamely fare we dine,
    Wear hoddin grey, an’ a that;
    Gie fools their silks, and knaves their wine;
    A Man’s a Man for a’ that:
    For a’ that, and a’ that,
    Their tinsel show, an’ a’ that;
    The honest man, tho’ e’er sae poor,
    Is king o’ men for a’ that.

    (The whole can be ).
    Willis, you are a king o’ men for a’ that. Would any other Mann admit to a’ that?

  3. “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.”
    A watt is a power unit, not an energy unit, i.e. Btu/h. 3.412 Btu/h per W.
    The heat capacity of air is 0.24 Btu.lb – F, water, 1.0 Btu/lb-F. No watts involved.
    Three Legged Stool of CAGW: 1) Anthropogenic 2) Radiative Forcing 3) GCMs
    Leg the 2nd
    Radiative forcing of CO2 warming the atmosphere, oceans, etc.
    If the solar constant is 1,366 +/- 0.5 W/m^2 why is ToA 340 (+10.7/- 11.2)1 W/m^2 as shown on the plethora of popular heat balances/budgets? Collect an assortment of these global energy budgets/balances graphics. The variations between some of these is unsettling. Some use W/m^2, some use calories/m^2, some show simple %s, some a combination. So much for consensus. What they all seem to have in common is some kind of perpetual motion heat loop with back radiation ranging from 333 to 340.3 W/m^2 without a defined source. BTW additional RF due to CO2 1750-2011, about 2 W/m^2 spherical, 0.6%.
    Consider the earth/atmosphere as a disc.
    Radius of earth is 6,371 km, effective height of atmosphere 15.8 km, total radius 6,387 km.
    Area of 6,387 km disc: PI()*r^2 = 1.28E14 m^2
    Solar Constant……………1,366 W/m^2
    Total power delivered: 1,366 W/m^2 * 1.28E14 m^2 = 1.74E17 W
    Consider the earth/atmosphere as a sphere.
    Surface area of 6,387 km sphere: 4*PI()*r^2 = 5.13E14 m^2
    Total power above spread over spherical surface: 1.74E17/5.13E14 = 339.8 W/m^2
    One fourth. How about that! What a coincidence! However, the total power remains the same.
    1,366 * 1.28E14 = 339.8 * 5.13E14 = 1.74E17 W
    Big power flow times small area = lesser power flow over bigger area. Same same.
    (Watt is a power unit, i.e. energy over time. I’m going English units now.)
    In 24 hours the entire globe rotates through the ToA W/m^2 flux. Disc, sphere, same total result. Total power flow over 24 hours at 3.41 Btu/h per W delivers heat load of:
    1.74E17 W * 3.41 Btu/h /W * 24 h = 1.43E19 Btu/day
    Suppose this heat load were absorbed entirely by the air.
    Mass of atmosphere: 1.13E+19 lb
    Sensible heat capacity of air: 0.24 Btu/lb-°F
    Daily temperature rise: 1.43E19 Btu/day/ (0.24*1.13E19) = 5.25 °F / day
    Additional temperature due to RF of CO2: 0.03 °F, 0.6%.
    Obviously the atmospheric temperature is not increasing 5.25 °F per day (1,916 °F per year). There are absorbtions, reflections, upwellers, downwellers, LWIR, SWIR, losses during the night, clouds, clear, yadda, yadda.
    Suppose this heat load were absorbed entirely by the oceans.
    Mass of ocean: 3.09E21 lb
    Sensible heat capacity: 1.0 Btu/lb °F
    Daily temperature rise: 1.43E19 Btu/day / (1.0 * 3.09E21 lb) = 0.00462 °F / day (1.69 °F per year)
    How would anybody notice?
    Suppose this heat load were absorbed entirely by evaporation from the surface of the ocean w/ no temperature change. How much of the ocean’s water would have to evaporate?
    Latent heat capacity: 970 Btu/lb
    Amount of water required: 1.43E19 Btu/day / 970 Btu/lb = 1.47E+16 lb/day
    Portion of ocean evaporated: 1.47E16 lb/day / 3.09E21 lb = 4.76 ppm/day (1,737 ppm, 0.174%, per year)
    More clouds, rain, snow, etc.
    The point of this exercise is to illustrate and compare the enormous difference in heat handling capabilities between the atmosphere and the water vapor cycle. Oceans, clouds and water vapor soak up heat several orders of magnitude greater than GHGs put it out. CO2’s RF of 2 W/m^2 is inconsequential in comparison, completely lost in the natural ebb and flow of atmospheric heat. More clouds, rain, snow, no temperature rise.
    Second leg disrupted.
    Footnote 1: Journal of Geophysical Research, Vol 83, No C4, 4/20/78

    • Are you not joining their (false) assumption that you can “average” heat transfer energy over the entire year by “averaging” the insolation at the equator, mid-latitudes, and poles into a single “flat earth” value valid for both day and night?

    • Although it’s a bit of a bass ackwards sentence ” . . . watt per square metre of incoming energy applied over a year . . . ” But W-yr is an energy unit and a W-yr/m2 is an energy flux across an area. That was my understanding of the sentence.
      Cheers, Mark

    • <i?“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.”
      A watt is a power unit, not an energy unit
      But a watt-year is.
      Please read carefully before jumping in feet first.

      • as stated above by other, a Watt-year is energy ; 31.5 MJoules. Which indeed will rise a cubic meter of water by (a little less than) 8K.
        I one saw a greeny very proud of his energy conservation system for his house : a huge pool in the cellar. The poor guy had no physical ideas of how few energy his costly and inconvenient (thanks to humidity everywhere) system actually contained.

  4. The land is easy, because the involved thermal mass warms and cools at one degree C per watt-year/m2. It swings 28°C,
    Are you saying the ground temperature swings 28°C (82 F). Or are you getting air temperatures above the ground? How deep do you have the 28 C going into the mass of the earth?

    • I live in the state of Wisconsin. The frost line here is 6ft, which means over the course of the winter the ground will freeze to a depth of 6 feet and will thaw to that depth over the spring/summer. My understanding is that summer ground temp averages somewhere around 68 F, and since if will freeze to 6 ft in the winter, an 82 F swing over the course of a year is not that unbelievable.
      P.S. This is why most buildings here have full basements, since you have to sink the foundation below the frost line anyway, once you are below 6feet, there isn’t that much extra cost to putting in a full basement.

      • Rockford, Il, just 15 miles South of the Wisconsin border had, according to It’s Water Department, 6 feet of ground freeze this year. This is near a record as it froze mains that had not previously frozen and residential pipes more than they could remember. The City issued a warning on pipe freezing and offered to reduce water bills for those who left water trickling to prevent mains freezing. Here water mains are under the streets and the ground density and lack of sod covering would allow deeper cooling and freezing. Grave diggers generally found frost just below 4 feet under grass.
        I wonder how deep the temperature difference goes? Does ground cover make a big difference everywhere? In heavily wooded areas would both increase and decrease be reduced? Has paving changed the heat exchange with the atmosphere much? Finally, is the estimate for heat gain and loss pretty close?
        Our well water is way colder than 68 degrees from a 130 feet deep well in the yard. Doesn’t change much by season. Mighty cool watering the garden in August. Does not freeze quickly if the hose supply system is opened and used in March, something I have done.

      • “swings 28degrees C” = delta C. 28C delta = 50F delta.

        Average daily range since the 50’s is ~17.8F (specifically 17.83143F rising temp, 17.83556F falling temp).
        Cooling slightly more than warming.

  5. if the surface of the water and the surface of the land had radically different rates wouldn’t we have expected to see very distinct and unusual weather patterns at the shorelines of every ocean ? you would have had a very distinct line of demarcation of differing thermal behaviors that would have led to interesting weather patterns everywhere the oceans touched the land …

  6. This seems to imply there’s a significant difference between the Northern and Southern Hemispheres.
    There’s far more water down under.
    So these lags ought to cause systematic differences between the N and S temperatures. But the Jet stream and the Gulf Stream both cross the Equator,
    If that’s right, there ought to be repeated patterns in accelerations and decelerations of the Jet stream and the Gulf Stream.
    It may not be periodic in frequency. But, if this is right, it should be repeated in sequence.
    And that would be a useful thing to learn.

    • It’s interesting that despite the large difference in NH and SH land:water, that their albedo is the same.
      ”This symmetry is achieved by increased reflection from SH clouds offsetting precisely the greater refl
      ection from the NH land masses. (ii) The albedo of Earth appears to be highly buffered on hemispheric and global scales as highlighted by both the hemispheric symmetry and a remarkably small interannual variability of reflected solar flux (~0.2% of the annual mean flux). We show how clouds provide the necessary degrees of freedom to modulate the Earth’s albedo setting the hemispheric symmetry. We also show that current climate models lack this same degree of hemispheric symmetry and regulation by clouds. The relevance of this hemispheric symmetry to the heat transport across the equator is discussed.”
      http://webster.eas.gatech.edu/Papers/albedo2015.pdf

    • The jet streams are normally restricted to northern and southern hemispheres and do not cross the equator. There are polar jets and subtropical jets in each hemisphere. Also, I doubt that any part of the Gulf Stream is considered to be from south of the equator. I know there is a flow of water from off the coast of Africa near the Congo that feeds into the Caribbean Sea but that is not normally considered part of the Gulf Stream.

    • Don’t forget oceanic currents that transport the solar energy imparted in the tropics and transport it in 3 directions, particularly pole words.
      For example, the reason why the ocean off Iceland or the West coast of Norway does not freeze, but the Baltic Sea at the same latitude does freeze, is down to oceanic currents. One cannot look at how much solar is being received at any one location at any given moment in time, because energy is being transported through the system to different places by different means and at different rates. The energy budget at any one location is made up of a number of different factors.
      I explained this to Willis in his Radiating the Oceans article. There is so much solar being inputted into the tropical oceans that the tropical oceans never freeze. The excess energy is transported in 3 directions (some being distributed to depth, and some polewards) at different speeds. Some seas in some places are the recipient of this warm water currents and they do not freeze; other areas are not and they gradually freeze when the input from their direct local solar insolation falls below critical levels, and thaw, once direct local solar insolation picks up.

  7. i would love to see that posted on the skeptical science site.even if it did not last long,it may even provoke some proper thought among he worlds leading warmists.

    • A very strong and rapid decarbonisation of the global economy could stabilise the climate system below 2°C,
      =============================
      It would sure as shooting destabilize the global economy. The stock market swings of the past week would be remembered as the “good old days”.
      Funny how we are told that fossil fuels are not sustainable, yet every time prices rise there is a huge glut of oil on the market and prices fall. This suggests that supply is a function of price, not fundamental scarcity.

  8. Willis – I’m not at all sure about this, but you may have a logic error. It looks to me like the way in which you have calculated the relative thermal mass for land and water necessarily means that you will find the same stored energy per square metre. IOW your conclusion was initially assumed. For example, does your relative thermal mass calc cater for possible difference in albedo? I repeat : I am not at all sure about this, you are much more likely than me to be correct, but I would appreciate it if you could check.

    • Willis, logical slip it seems:

      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.

      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.

      Accepting your 3:1 ratio that should be 3°C.

  9. I would just add that you need to keep in mind the internal heat that comes from radioactive decay from within the earth, (mainly U, K and Th) which in this solar system is unusual, and which ultimately and largely drives plate tectonics and volcanoes. The mid ocean ridge system is quite extensive and certainly effects chemistry within the deep oceans.
    It might not be much of a percentage in terms of heat in the discussion above, but over long periods of time and in terms of chemistry and other effects in can be quite significant. Many scientists have made the mistake of ignoring internal earth processes, such as ascribing the origin of the ocean water from comets (when magma solidifies it expels water, which is more than enough to explain the origin of the ocean water), ignoring the buffering capacity of ocean sea floor with ocean acidity, and mistaking the age of the earth by not taking into account the effect of radioactive decay with heat loss calculations. I’m sure there are other examples.

  10. Willis
    Have you taken into account that the temperature gradient between land and ocean is fundamentally different, and that they are heated in different ways from different sources?
    As one descends through the oceans the temperature at 24 metres is considerably cooler than the temperature at the surface (the SST is very slightly cooler than the top mm,) Also there is a fundamental difference between day and night temperature profiles due to the lack of solar input.
    During the night, a typical ocean has much the same temperature extending over a depth of between about 1mm and say about 8 metres (although anyone who has dived/snorkelled will be aware of how there can be significant temperature fluctuations over very small distances), but during the day, the temperature drops off significantly from 1mm and below. Most of the solar is absorbed within the first few metres. See generally (where (a) is the night time profile and (b) is the day time profile):
    http://disc.sci.gsfc.nasa.gov/oceans/additional/science-focus/modis/MODIS_and_AIRS_SST_comp_fig2.i.jpg
    However with land the temperature at 8 metre depth is higher than that at say a few centimetres. This is why in Northern countries, plumbing has to be buried below the perma frost depth.
    There is a fundamental difference between how the oceans are heated (top down by solar which can penetrate to a depth of about 100 metres, although most is absorbed within 3 to 5 metres). On land the vertical penetration of solar is negligible. Obviously solar heats the very top surface, but below that it is geothermal heat that is heating the ground below the very top surface, ie., it is being heated from below by a different heat source. And there is no equivalence of ocean overturning inland. Heat is transferred by conduction, not by physical mixing.
    Incidentally, if you look at the NASA plot (a) of the ocean temperature profile, you will note the steady temperature profile between about 1mm and 8 metres. This suggests that ocean overturning does not effectively mix the very top layer of the ocean. This, of course, has a bearing on whether DWLWIR which is fully absorbed in the top microns, can effectively be mixed and dispersed/dissipated into the bulk ocean below and hence whether it plays an effective role in keeping the oceans warm and preventing their freezing.
    .

  11. Thermal mass is one parameter, thermal conductivity is another. Experiment with a stainless steel teaspoon and with a Silver, Aluminum or Copper one. In a hot cup of tea or coffee, the handle of the SS one will stay very cool, compared to the others. Little to do with thermal mass. What is the temp profile in degrees difference vs. depth for seawater and for land (whatever that is)? In my quick skim, I didn’t trip over this relationship, so pls accept my apologies if it’s covered. / mark F

  12. It’s a good point. I wonder what percentage of so called climate scientists ever read Verhoogen “Energetics of the Earth.”

  13. Willis,
    You state:
    “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.”
    //////////////////////////
    It is nearly 3 am and it may be that I am not thinking straight, but is this a comparison without real meaning?
    The ocean is uniform in its nature and characteristics but the land is not, having vastly different albedos and varying heat capacities. For example how much of the of the land surface of the planet is covered by forest? There is all but no solar reaching the land/forest floor. The trees are more than 8 metres high. In (tropical rain) forests, it is the high humidity (water vapour) that is keeping the temperature stable, nothing to do with the land. And where is the distribution of these forest, since over the equatorial area, there is far less seasonal variation in solar insolation (indeed, they have less seasons)
    PS. Per Wikipedia: “According to the widely-used[5][6] United Nations Food and Agriculture Organization definition, forests covered an area of four billion hectares (15 million square miles) or approximately 30 percent of the world’s land area in 2006.[4]”

    • Richard, I think you are on the right track.
      The vegetation prevents warming in two ways: evaporation, which is how it grows, and absorbing solar energy and converting CO2 into biomass, which is how it grows.
      In the oceans, incoming energy evaporates water, lots of it. There is also a considerable amount of biomass growing in the ocean which converts energy into chemical stored energy.
      The emissivity of water is higher than nearly everything on land, as well as more reflective. The idea that water is a powerful emitter of IR is not easily apprehended because we see it as transparent. Water looks clear and reflective, but in IR it is jet black. Irradiate it, it evaporates. Leave it alone, it sheds heat.

    • Should have added that the difference in heat capacities of the land and ocean end up being equalized in the NH and SH because clouds are generated that reduce the insolation that other wise would have seen vast differences and probably killer currents and winds from the difference in land:ocean ratio. A remarkable balance that I hadn’t even heard about until I read it in one of your articles as a passing thought.

  14. 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.

  15. 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.

  16. 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

    • 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.

    • There are a LOT of websites that discuss the effect of scintillation and as everyone notices,
      not a soul in one of the most documentation-obsessive, and instrument potential aware atmospheric fields
      the world has ever seen has ever uttered a SINGLE solitary PEEP.

      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?

  17. 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.

  18. 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

  19. 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.

  20. 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.

  21. 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?

  22. ‘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

  23. 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 .

  24. 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.

    • 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.

  25. 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.

  26. 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.
    .

  27. 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.

    • 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.

  28. “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.

  29. 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.

  30. 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.

  31. 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.

  32. 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.

Comments are closed.