Layman struggles with Science

[Richard111]

This clueless pensioner layman is much heartened to note the rising level of hard science in the posts and comments on the sceptic blogs these days. Sadly, this results in a decline of my own ability to keep up with these discussions so I must limit myself to basic basics.

Be that as it may, I still have a question to which I cannot find an answer. The question is: "What percentage of radiation from the surface is absorbed by the CO2 in the air above and how much is "backradiated"? I have attempted to work this out before Post: #146 on Page 8 but I need the opinion of someone who knows what they are talking about. Note I ask about "percentage". I will need the answer in "percentage" as I simply cannot get my head around "watts per square metre" from the atmosphere which can can absorb and radiate over the total VOLUME from the surface to space.

To try and explain my thinking I want to conduct a thought experiment. I have a mild steel plate with dimensions of 1 metre by 0.5 metre by 3 millimetres thick. This has a total volume of 1000 x 500 x 3 = 1,500,000 cubic millimetres. The mass, from my bathroom scale, is 11.5 kilograms and the total surface area is 1 square metre. I'm ignoring the 9,000 square millimetres of the edge (less than 1/100th of a square metre), you'll see why shortly.

If we melt that sheet down and recast it as a cube, we will get a metal block of 11.5 kg, and six sides measuring 114.47 millimetres squared which gives a total surface area of 78620 square millimetres or just 0.07862 square metres or if you like just 7.862% of a square metre. You can see where I'm going?

We have two equal masses of identical material with rather different surface areas, 1 square metre as against 0.07862 square metres. If we heat both masses to the same temperature, say 16.8C, both masses will radiate at 400 watts per square metre. Now 1 watt is 1 joule per second so after 1 minute the plate has lost 400 x 60 = 24,000 joules but the cube will have lost (400/100)x7.862 x 60 = 1,886.88 joules in that 1 minute.
So in one minute the cube lost just 7.862% of the heat energy lost by the plate. Surface area has a MAJOR effect on the rate of cooling.
Lets split the plate into three, each 1mm thick, and we will have 3 square metres of radiating surface that will lose 72,000 joules in one minute. Split the plate into 30 x 0.1 millimetre slices and we will have a radiating area of 30 square metres losing 720,000 joules per minute.

The specific heat of mild steel is about 500 joules per 1kg per K (or C if you like) (LINK). We have 11.5 kg so to change the temperature by 1C we need 11.5 x 500 = 5,750 joules so our 30 square metres losing 720,000 joules per minute divided by 5,750 equates to a loss of 125.22C ! ! ! ! !

We started off at 16.8C and in one minute our block lost 1,886.88 joules but we need to lose 5,750 joules to reduce the temperature by just 1C so our block lost 33% of 1C so the block will now be at 16.5C. But when that block is flattened out to a plate of 15 square metres just 0.1 millimetre thick it radiates from 30 square metres (both sides) and loses 720,000 joules in just 1 minute thus cooling to MINUS 108.42C.

Actually, this WON'T happen! Because the rate of radiation from the plate drops off as the temperature drops and if the local air temperature is already at 16.8C then the plate will be in thermal equilibrium and no further heat loss will occur. As I said above this is a thought experiment. IF the local air temperature was below say -60C (as found up at the tropopause) then our plate WOULD cool very quickly indeed.

We are still dealing with SOLID metal here even if it is just 0.1mm thick. My guestimate is that this thickness is about 5,000 molecules deep. Just think, if we had 30 x 5,000 plates just one molecule thick they would radiate from 150,000 square metres! That ain't going to stay warm very long!

If our 11.5kg of mild steel was reduced to individual molecules and thoroughly mixed into the air then cooling is going to be EXTREMELY rapid.

Lets talk about carbon dioxide, CO2, at 400 parts per million evenly mixed through the atmosphere. Over any 1 square metre of the earth's surface there will be about 6kg of CO2 (0.04% by volume times factor for mass of CO2 molecule) spread evenly through out that column of air. Let us assume our column of air is over the ocean. Then the first 6 millimetres of water will have the same mass as all the CO2 in the air above. Now consider the heat capacity of the water at 4,000 joules per kg per degree C and CO2 gas at 844 joules per kg per degree C. [LINK]

The water, with the emissivity of better than .95, will be emitting almost at black body levels, thus the energy contained in that 400 watts per square metre is spread through a large range of electromagnetic frequencies, from say 3 microns to beyond 100 microns. How much energy then is carried by the 15 micron band? My own thumb suck guess is about 8% or say 32 watts (or 32 joules per second).

Just from curiosity let's assume the CO2 in our column absorbs all the available energy provided by that 8% of 400 w/m^2, or 32 joules per second. Therefore over twelve hours, (no sunshine) 12 x 60 x 60 x 32 = 1,382,400 joules, divided by 844 x 6(kg) = 273 degrees. Gosh and golly! I've found the missing hotspot!

Not really. :-) Remember that 6kg of CO2 is well mixed into over 10,000kg of nitrogen and oxygen with an average heat capacity of 1,000 joules per kg per degree C. So, assuming all that energy absorbed by the 6kg of CO2 was passed into the 10,000kg of the atmosphere, what temperature increase can we see?

Over a period of 12 hours 1,382,400 joules was pumped into 10,000kg of air. I make that 1,382,400 divided by 10,000 = 138.24 joules into each kilogram of the atmosphere. To raise the temperature of 1kg of air by 1C we need 1,000 joules but we only have 138.24 joules, therefore the whole 10,000kg of air warmed up by just 0.14C degrees!

Must stop now. This post is getting too long. What I want to talk about next is the way energy is absorbed into a molecule and how that molecule releases or emits that energy to change to a lower energy state. I hope to be able to explain why it seems to me that CO2 in our column of air will hardly absorb any of that 15 micron radiation leaving the surface. Stay tuned.