Now we can calculate how much energy it would require to raise the temperature of the troposphere by a single degree Kelvin:

1.012 J/g·°K = 1.012 kJ/kg·°K

1.012 kJ/kg·°K · 1.2 kg/m³ = 1.2144 kJ/m³·°K

1.2144 kJ/m³·°K = 1,214,400,000 kJ/km³·°K

You are assuming a constant density of 1.2 kg/m^3 for the atmosphere up to 17 km. This is entirely incorrect. At 17 km, atm density is 0.16 kg/m^3. You can find a rough average density by integrating density by height from 0 to 17 km altitude and then dividing by 17 km. So your real value you should be using is roughly:

(1.22 - 0.16) / 2 = 0.53 kg/m^3

Also the atmosphere extends out to roughly 60 km (depending on time/temp/atm conditions), with the density changing as a function of all these things (it gets a lot lower).

To do this properly, you need to use the absorptivities (throughout the optical spectrum) of CO2, and all other greenhouse gases to find heat input for each greenhouse gas.

Integrate over the absorption spectrum/irradiant solar energy like this:

Qtotal=ʃα(λ)Q(λ)dλ

where α is the absorptivity of CO2 at a given wavelength of light, and Q is the irradiant power from the sun as a function of wavelength (you essentially assumed this was a constant based on 1366 W/m^2*Earth cross sectional area).

This is probably the most important issue. You have neglected the fact that excessive CO2 and other greenhouse gases can affect the total absorptivity greatly enough to cause a temperature rise.

[edit on 1-12-2009 by erkokite]

[edit on 1-12-2009 by erkokite]

Originally posted by neo5842
This is what we need, though in my simple understanding of the mathematics here, i would like to see, how sunspots would change things when added to the equation, though i understand its not possible to see with any degree of accuracy because of how random they can be from year to year, though i do understand the 11, 80 (and so on) year cycle, but still not predictable. As far as i am aware they contribute the largest amount of energy attributed to the heating and cooling of the planet, and for that reason it can take up to 800 years for the CO2 to follow with its increase. Leading to many to conclude that the earth is in a state of cooling rather than warming as a result of the sun's relative inactivity for the past 9 years or so. Please correct me if i am wrong. However like i said i would like to see how things look after sun spots are taken into consideration.

IT IS NOT CARBON DIOXIDE (CO2)
IT IS CARBON MONOXIDE (CO) !!!!!!
will you people please!!!!! learn your table of elements!!@!!!!!

Originally posted by DjSharperimage

Originally posted by neo5842
This is what we need, though in my simple understanding of the mathematics here, i would like to see, how sunspots would change things when added to the equation, though i understand its not possible to see with any degree of accuracy because of how random they can be from year to year, though i do understand the 11, 80 (and so on) year cycle, but still not predictable. As far as i am aware they contribute the largest amount of energy attributed to the heating and cooling of the planet, and for that reason it can take up to 800 years for the CO2 to follow with its increase. Leading to many to conclude that the earth is in a state of cooling rather than warming as a result of the sun's relative inactivity for the past 9 years or so. Please correct me if i am wrong. However like i said i would like to see how things look after sun spots are taken into consideration.

IT IS NOT CARBON DIOXIDE (CO2)
IT IS CARBON MONOXIDE (CO) !!!!!!
will you people please!!!!! learn your table of elements!!@!!!!!

Thank you for pointing that out, however, no need to be so rude, just a simple point in the right direction would have sufficed. :0)

Oh-Oh. a concepual problem - the sort my physics students have when they haven't studiedfor the final . You multiply Radiant energy times the % increase in CO2 per Wikipedia. That's a big fail. It's like painting a 150 mm thick window with5mm of black paint and saying it's only going to cut down on 3% of the light coming in. It ignores totally the absorption spectrum of the substance. Think about it. The sun warms one=half of the earth 's atmosphere by 15 degrees K in 4 months every spring and summer.
Your post is going to be a good exercise for my class this weekend. I'll let them dissect the wikipedia references and double check the caculations.
Offhand, it appears that you have used a specific heat value for fresh water. That is enough to get you a reject from a refereed journal. You need to use the correct value for the correct salinity. And specific heat also varies with pressure. What Q are you using? ISA? And what values for the albedo from all surfaces?
Yor approach is actually conceptually correct for the first part of the problem = find out the total rnergy budget. But it's only a start. Next look at absorption spectra The absorption spectra must be applied And the specific heat and relevant albedos. Then you look atthe real world and see if it fits your model.

i'm afraid your work is not quite ready for prime time. Or a refereed journal. But keep working. I like the effort.But before your next effort you should study the concept of significant figures and computational accuracy. 8,694,154 km, huh. How sure are you that it's not 8,694,153 or 8,694,155? There is a difference between precision and accuracy. To the layman, the purported exactness seems impressive. To a scientist, it screams poseur. I'm not going to tell my class where this little homework exercise in critical analysis came from. Unless you want me to.
EYou should see some of the crackpot stuff we get at the journals for which I review. Yours is not crackpot, just fatally flawed. For now. Keep working, tho.

Originally posted by 4nsicphd
Oh-Oh. a concepual problem - the sort my physics students have when they haven't studiedfor the final . You multiply Radiant energy times the % increase in CO2 per Wikipedia. That's a big fail. It's like painting a 150 mm thick window with5mm of black paint and saying it's only going to cut down on 3% of the light coming in. It ignores totally the absorption spectrum of the substance. Think about it. The sun warms one=half of the earth 's atmosphere by 15 degrees K in 4 months every spring and summer.
Your post is going to be a good exercise for my class this weekend. I'll let them dissect the wikipedia references and double check the caculations.
Offhand, it appears that you have used a specific heat value for fresh water. That is enough to get you a reject from a refereed journal. You need to use the correct value for the correct salinity. And specific heat also varies with pressure. What Q are you using? ISA? And what values for the albedo from all surfaces?
Yor approach is actually conceptually correct for the first part of the problem = find out the total rnergy budget. But it's only a start. Next look at absorption spectra The absorption spectra must be applied And the specific heat and relevant albedos. Then you look atthe real world and see if it fits your model.

i'm afraid your work is not quite ready for prime time. Or a refereed journal. But keep working. I like the effort.But before your next effort you should study the concept of significant figures and computational accuracy. 8,694,154 km, huh. How sure are you that it's not 8,694,153 or 8,694,155? There is a difference between precision and accuracy. To the layman, the purported exactness seems impressive. To a scientist, it screams poseur. I'm not going to tell my class where this little homework exercise in critical analysis came from. Unless you want me to.
EYou should see some of the crackpot stuff we get at the journals for which I review. Yours is not crackpot, just fatally flawed. For now. Keep working, tho.

You should also FAIL everyone for using Carbon Dioxide CO2 instead of Carbon Monoxide CO

Far more accurate than the rubbish the governments have taken as their model, and now using to try to get some more money. I think it would be far more useful if anyone with knowledge to even consider picking this apart should be the ones trying to show the world leaders the real calculations instead of the lies they have. :0)

ooh boy... first let me say that i am not disagreeing with your conclusions, as i am a proponent that the earth is warming outside of human and industrial influence. the issue that i have here is some flaws in your thinking and such that i would like to point out to help strengthen your case and remove holes.

Originally posted by TheRedneck
It has been theorized that the use of anthropogenic (man-made) carbon dioxide is the reason for the recently observed warming trend from ca. 1960-1998. The present level of CO2 in the troposphere is stated by multiple sources as being on the order of 380 ppmv. This represents an increase, based on the most liberal estimates I have uncovered for pre-industrial levels of 280 ppmv, of 100 ppmv or 0.01%. Since this base point is considered to be 'safe and natural', it would logically follow that any warming would have to be associated with the 0.01% increase and it alone.

ok sounds good so far, although i am not sure where the 0.01% is coming from. is this a ratio to the whole volume of the troposphere or a ratio of initial ppmv to current ppmv, which would be an increase of nearly 36%. be clear here what you are saying.

All heat energy reaching the earth is from the sun, in the form of solar irradiance. Heat reflected back into space is a result of this solar irradiance, and can therefore be considered the same in energy calculations.

Again, this is not entirely true. You have to talk about the interaction with the solar radiation and the earth/atmosphere. a large portion of the suns energy is actually filtered in the thermosphere. The earth is considered to be a black body, and for all intensive purposes will emit all of the energy that is absorbed. The heating is due to the absorption of the shorter wave energies coming in and the emitting of the longer wave terrestrial radiation. It is the longer wave energy that is emitted that is absorbed by the atmosphere which is NOT a black body and has absorption windows cause them to trap certain wavelengths.

Solar irradiance can and has been quantified. The amount of energy reaching the planet is on the order of 1366 W/m². The planet presents a more or less circular profile to the sun, so the area of the earth normal to solar irradiance can be calculated as this circle. The earth is an average of 6371 km, with a troposphere layer surrounding it that averages 17km in height, which also must be included since it is the location of the atmospheric carbon dioxide.

Woah woah! Slow down! Important note here! You are calculating for a flat disk. Not to mention you are not taking into consideration the VOLUME of the troposphere, or the atmosphere’s above it. The earth is a spherical shape, and the angle of incidence of incoming solar radiation has a HUGE impact on the amount of absorbed solar radiation. Why do you think it can be daylight for 6 months at the poles and be cold as all get out? Any figure that you propose to be the quantified amount of solar energy that is absorbed by the earth at this point is about as accurate as me throwing a dart at a board blind folded and dizzy! It is not as simple as you try to make it seem.

That result is in Joules (or kiloJoules) per second. Since most climate predictions are based on much longer time intervals, I will now calculate how much energy would be available during such a longer time interval such as the commonly used 100-yr. period:

100 yr = 36,525 days = 876,600 hr. = 52,596,000 minutes = 3,155,760,000 s

We can now multiply this time interval by the rate of energy influx to obtain the total energy that the planet will receive from solar irradiation over the next 100 years:

175,117,838,274,000 kJ/s • 3,155,760,000 s/100yr =
552,629,869,311,558,240,000,000 kJ/100yr

again you forget to correctly unit your statistic in volume, so now you have an inaccurate and debatable value that is not given in the proper units. You need to be dealing with joules per kilogram or per meters squared and then per second. Atmospheric energy is unitized in j/kg, or j/m^3.

Now we must calculate exactly how much of that energy will be affected by the increase in the amount of carbon dioxide in the troposphere. Remembering that the increase from pre-industrial levels is 0.01% of total atmospheric volume, we multiple this total energy by 0.0001:

552,629,869,311,558,240,000,000 kJ/100yr • 0.0001 =
55,262,986,931,155,824,000 kJ/100yr intercepted by anthropogenic CO2

you are now assuming a homogenous atmosphere with a homogenous ppm for co2. this is not the case. Co2 levels vary, and greatly, depending on where in the globe you are. And then you have to take into consideration how much energy that part of the globe is receiving, what the surface coverage type is and its albedo, then find out the average amount of water vapor is available and the average height of the tropopause and the average volume of the column of atmosphere you are looking into. You cannot simplify such a complicated process!

Now let us turn to the question of how much energy is needed to increase global temperatures. Of course, the first and most obvious area to be heated is the troposphere itself.

There are multiple layers of the troposphere and each one is heated differently. For instance the molecular boundary layer (that being the point where the air “touches” the ground, is heated through convection, while the planetary boundary layer is heated mostly by mixing (that is the “dirty” layer that interacts with all the terrain and what not) and the free atmosphere is heated through radiation. Each process requires different circumstances and cannot be married so simply as you state.

Air under average atmospheric conditions has a specific heat capacity of 1.012 J/g•°K and an average density of 1.2 kg/m³. The troposphere itself can be calculated by using the information presented earlier (average radius of earth = 6371 km and a troposphere extending 17 km above the surface). Thus the area of the troposphere can be determined by calculating the volume of a sphere of 6388 km radius and subtracting a sphere of 6371 km radius from it:

V(tot) = 4/3 π r³ = 4/3 π • 6388³ = 1,091,901,171 km³

V(earth) = 4/3 π r³ = 4/3 π • 6371³ = 1,083,206,917 km³

V = V(tot) - V(earth) = 1,091,901,171 km³ - 1,083,206,917 km³
= 8,694,154 km³

Now we can calculate how much energy it would require to raise the temperature of the troposphere by a single degree Kelvin:
[align=center]1.012 J/g•°K = 1.012 kJ/kg•°K

1.012 kJ/kg•°K • 1.2 kg/m³ = 1.2144 kJ/m³•°K

1.2144 kJ/m³•°K = 1,214,400,000 kJ/km³•°K

again, not area… but volume. The volume of the troposphere is what you calculated, but that still does not take into effect that the troposphere is denser at the poles, despite being a thinner layer there.

While I give you a star and flag for effort and overall importance of the message, there as you can see many many holes that under scientific scrutiny would be torn to shreds.

Regards-

Wx4cAsTeR