The Ocean, By Other Means – Air Temperatures

Somewhere along the line I heard the quote that “The air heat is just the oceans by other means” (or something close to that). Well, it’s pretty obvious if you have an instinctive grasp of specific heats, but I got to wondering about how to make it clear to folks without 3 or 4 years of chemistry classes…

So it occurred to me, one could show how much ocean equals all the air above it, clear to space, in terms of heat capacity. That ought to give an intuitive grasp.

So the atmosphere weighs about 14.5 pounds (or 1 BAR) per square inch. (Yes, I’m mixing weight and pressure units. It doesn’t matter for this purpose). How far down in the ocean is the same weight (pressure) of water? In this case my old scuba diving units make it easy. 33 feet or 10 meters.

https://oceanservice.noaa.gov/facts/pressure.html

Yes, I’m ignoring the tiny bits past the decimal point. We’re finding an intuitive rule of thumb, not measuring gold ingots… I’m also going to ignore the small change of specific heat with air density…

Ok, but water holds more heat per unit weight (or more precisely, mass) than does air. How much? (Or what are their specific heat values?) Well, air is about 1, and water is about 4.

http://www.engineeringtoolbox.com/air-properties-d_156.html

http://www.engineeringtoolbox.com/sea-water-properties-d_840.html

So we need about 1/4 that depth of water to hold the same heat as one atmosphere of air, or about 8.25 feet or 2.5 meters. Hardly enough to float a small boat.

Now the Earth is only 2/3 oceans, so that would be about 12 feet of the ocean to equal ALL of the air, even that over land too. Call it obout 3.75 meters or 375 cm of ocean. Just about two people deep.

So, to offset a 1 C rise in air temperatures, only a little more cold deep ocean needs to be overturned by wind and tides. Conversely, even a minor change to lighter winds or tides would let the air warm a degree C.

Consider just how deep the oceans are, thousands of meters, and you can see it would take a very precise knowledge of exact ocean temperatures and depths to prove any air temperature change was not just due to minor variations in the ocean.

But Wait, There’s More!

As I remember it, the heat of fusion of water is about 80. (Someone can check me on that)

https://en.wikipedia.org/wiki/Enthalpy_of_fusion

So about 1/80th that much water turning to ice would be the same. Call it 1.8 inches or about 5 cm.

Now that seems pretty small to me. Maybe I’ve blown a calculation somewhere… but I don’t think so…

So when a few feet of ice form over the arctic ocean, that is one heck of a lot of air equivalent heat getting radiated out to space.

The heat of vaporization is even bigger. The Specific Heat of Vaporization for water being about 540.

So our 375 cm of ocean divided by 540 = 0.7 cm of rain. So to completely remove 1 C of temperature rise in the air would require 3/4 cm of ocean to evaporate, rise to the top of the troposphere, dump that heat to space (it can radiate out through the stratosphere just fine) as it condenses to a small cloud, then fall back to earth as rain. Call it 1/4 inch of rain.

So that’s all it would take to cause 1 C global increase in warming (1/4 inch less rain) or to reduce the globe 1 C (1/4 inch more rain). Do we even know how much global rain changes year to year or decade to decade?

Now I’ve done this all “off the cuff” and only one pass, and those numbers look mighty small to me, so definitly needs a cross check. Still, I think it isn’t too far wrong.

33 feet /4 / 540 ×12 = .183 inches.
Water ft./ air equiv. ht. / Sp. Ht. V. X ft. Per inch.
Seems to cross foot.

So provided I didn’t mess up too badly, those are some human sizes for water vs air on Earth. Measuring air temperature is exactly wrong for finding global heat. It’s the ocean and the rains…

A technical managerial sort interested in things from Stonehenge to computer science. My present "hot buttons' are the mythology of Climate Change and ancient metrology; but things change...
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15 Responses to The Ocean, By Other Means – Air Temperatures

1. erl happ says:

The rate of overturning of the ocean is obviously a critical determinant of surface water and air temperature. That would seem to depend upon the strength of the pressure differentials that drive the winds. Lowest pressures by far are on the margins of Antarctica. Anybody know why?

2. E.M.Smith says:

I’d guess it is because of the giant, perpetual, cyclonic wind circulating around the continent (along with a circumpolar current).

There’s a huge descending vortex there. Where all the tropical air goes once it reaches the tropopause. Look up Polar Night Jet.

3. oldbrew says:

Various stats in the link below but the summary says:
‘The ocean is the most dominant component of the Earth’s heat balance, and most of the total warming caused by climate change is manifested in increased ocean heat content (OHC).’
– European Environment Agency

http://www.eea.europa.eu/data-and-maps/indicators/ocean-heat-content-1/assessment

4. Graeme No.3 says:

Well, it is obvious that we must ban evaporation to save the planet. Quite how I don’t know but no doubt an expensive scientific program should be set up immediately to provide a refuge for victims of Trump’s refusal to believe faked data.

5. erl happ says:

Re ‘There’s a huge descending vortex there. Where all the tropical air goes once it reaches the tropopause.’

Although there is indeed descent of a very gentle nature over the continent where high pressure and anticyclonic winds prevail, more so in winter than summer, the margins of Antarctica experience cyclonic winds and rapid ascent in all seasons but especially in winter. The wind reaches maximum velocity between 400 hPa and 50 hPa where there are marked differences in the density of different air parcels in the horizontal domain, the less dense being the ozone rich portions of mid latitude origin. Adding to the density contrast (in part due to the latitude of origin of the air and its temperature) is that fact that ozone absorbs outgoing radiation from the Earth itself. There is generalised movement of air parcels in a west to east direction in mid and low latitudes towards the margins of Antarctica.Over the last 70 years surface pressure on the margins of Antarctica has fallen by about 15 hPa while it has risen in the mid latitudes of the southern hemisphere and all points north. As a direct result there has been a marked increase in the velocity of the north westerlies at the surface.These are the strongest winds on the planet and they are responsible for the circumpolar current.
In the late 1970’s there was a sudden increase in the temperature of the upper air (especially from 100 to 10 hPa) over Antarctica and this increase affected the southern stratosphere generally. Surface temperatures rose to a smaller extent, probably due to a reduction in ice cloud density.
It’s interesting to note that parts of the Pacific ocean are cooler at the surface today than back in the 1970’s.
The steepest increase in surface temperatures has occurred in the Indian Ocean to the west of what I call home. However, over the last several years surface temperatures in this region have fallen to the point where they are now anomalously low.
The point to note is that the Jet Stream is a manifestation of polar cyclone activity that owes its intensity to the nature of the air parcels drawn into the ascending vortex that manifests from about 400 hPa to the upper limits of the atmosphere and ozone plays a part in conditioning that activity.
The question then arises: What determines the ozone content of the air and changes therein.

6. cdquarles says:

Ozone content is a function of synthesis rates that makes it and reaction rates that destroy it. There are only a few synthesis reactions (UV and higher energy light incoming, though mostly UV; electrical discharge in air, which also emits UV, and catalytic formation) and a lot of destructive reactions. Final concentrations in any one defined sample will be a function of these plus net bulk transport.

7. erl happ says:

That’s the theory.

Operationally, its more important is to explain the distribution by altitude, latitude and season and the flux in ozone content over time.

The zone that we call the ‘stratosphere’ is only in small part, affected by short wave radiation that is energetic enough to split the oxygen molecule. Consider the winter hemisphere where ozone proliferates to a variable extent from year to year.

8. cdquarles says:

@erl,

Chemistry is still chemistry. and the amount found in any column in three dimensions will still be determined by local synthesis rates, local destruction rates and local bulk transport, integrated over whatever sized ‘parcel’ you want to examine. Even in the winter hemisphere and even at night, there is still incoming radiation. It might be small relative to the local star’s output on the lit side, but it is still not zero. The presence of noctilucent clouds and other aerosols will vary by local conditions. Surface chemistry is way too often insufficiently considered, even if considered at all.

9. erl happ says:

In the ionosphere O2 is broken down by short wave radiation. O3 in the stratosphere exists via transport and in that location is relative immune to photolysis because the wave lengths required are in the main already exhausted.

If O3 in the stratosphere is largely a product of transport from the ionosphere how is it that O3 does not proliferate all the way to the surface?

Why is the tropopause relatively elevated in the northern hemisphere and yet that hemisphere is relatively ozone rich by comparison with the southern?

There is massive ascent of ozone rich air on the margins of the polar caps all the way to the top of the atmosphere. Where is the balancing descent?

10. David A says:

EM says… So to completely remove 1 C of temperature rise in the air would require 3/4 cm of ocean to evaporate, rise to the top of the troposphere, dump that heat to space (it can radiate out through the stratosphere just fine) as it condenses to a small cloud, then fall back to earth as rain. Call it 1/4 inch of rain.”

Which raises the question of the warming equivalency of an equal amount of downwelling LWIR vs disparate SW radiation.

Does not increased LWIR radiation on the ocean result in the work of increased evaporation and convection?

Not to mention the correlating reduction in SW entering the oceans.

OTOH an increase in SW penetrating the ocean surface can, depending on the unknown current ocean equalirium, directly increase OHC for a very long time.

BTW, has anybody calculated how much volcanic energy the oceans hold? It would of course be a WAG on the input, and a WAG on the ocean residence time.

Same question regarding non volcanic geo-thermal ocean energy.

11. E.M.Smith says:

@David A:

I think you see the problem! A shift of solar activity between EUV / UV / Blue and Red / IR can have a big effect in heat distribution and rainfall. I find it interesting that since the shift to more Red / IR happened, we’ve had a lot more news of massive floods all over the planet.

More IR, more evaporation, more rain, cooling planet. More UV, less evaporation, warming air temperatures, warming ocean, heating planet.

Don’t know how long it will take for the evaporation / flooding to cool things off a lot, but do note in passing that the Little Ice Age famines were caused more by incessant rain, muddy fields, inability to harvest, and ‘lodging’ where grains were blown down than by the actual cold.

12. David A says:

If I may make a slight modification to this…” A shift of solar activity between EUV / UV / Blue and Red / IR can have a big effect in heat distribution…

add to “heat distribution” total heat content of earth atmosphere, land, oceans.

Question; does anybody wish to make a WAG on my questions regarding volcanic and geo-thermal ocean heat content?

13. E.M.Smith says:

@David A:

I thought distribution between ‘deep ocean’ vs ‘prompt evaporation / rain’ implied changes in total heat content; but I guess stating it is better than assumptions…

Per volcanic WAG: I have no clue. One would need to start from some data. Maybe find heat output of a ‘black smoker” then try to size the length of ridge it represents then multiply by total ocean ridge length and a SWAG as to % of ridge that’s active. I think the ridges vastly out heat the individual volcanoes, but a bit of data on them would help too.

So I think I have a clue how to approach it, but right now I’m already a few hundred percent over committed on other projects…

14. David A says:

Thanks em, ya no worries on quantification of volcanic and geo thermal. I think the kicker is residence time ( of a significant percentage) from the ocean bottom is likely measured in years, decades, centuries?bottom is likely meadure