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Originally posted by ElectricUniverse
How about you give us ONE research paper that proves atmospheric CO2 absorbs as much radiation as it is CLAIMED by the AGW proponents?...
Originally posted by Animal
Actually if you payed attention to what I was saying you would realize it is the OPPOSITE of your above assertion hat environmentalists strive for...
Originally posted by Animal
Yawn. . .Nope, not even close, just more of the same old fanatical opposition. . .
Originally posted by Animal
OH MY GOD, REALLY?!?! Well that just puts a nail in the old 'lets take care of the environment coffin' now doesn't it?
Originally posted by Animal
If you say it it must be true, right?
Australian TV Exposes Stranded Polar Bear Global Warming Hoax
Remember that wonderful picture of stranded polar bears on an ice floe that were used by folks like soon-to-be-Dr. Al Gore to demonstrate how dire the man-made global warming issue is?
Well, ABC television in Australia, on a show called “Media Watch,” recently debunked the entire issue (video available here, h/t NB member dscott).
It turns out -- as NewBuster Jake Gontesky reported on March 20 -- the picture was taken in August, “when every year the fringes of the Arctic ice cap melt regardless of the wider effects of global warming.”
The photographer, Australian marine biology student Amanda Byrd, didn’t think the bears were in any jeopardy:
They did not appear to be in danger…I did not see the bears get on the ice, and I did not see them get off. I cannot say either way if they were stranded or not.
Denis Simard of Environment Canada agreed:
You have to keep in mind that the bears are not in danger at all. This is a perfect picture for climate change…you have the impression they are in the middle of the ocean and they are going to die...But they were not that far from the coast, and it was possible for them to swim...They are still alive and having fun.
How delicious. Think this kind of broadcast would ever happen in America?
Originally posted by Animal
Greenpeace is much like Al Gore to you and yours, a straw-man.
Originally posted by Animal
New to me, how about a link to these 'leaders' . . .
Originally posted by Animal
Well like i already said, if you said it, it must be true.
Originally posted by Animal
The most pathetic, weak, ignorant, and spun out argument against protecting hte environment EVER told.
Given the present composition of the atmosphere, the contribution to the total heating rate in the troposphere is around 5 percent from carbon dioxide and around 95 percent from water vapor.
It would appear that China is not the energy utopia after all.
Originally posted by mikelee
A Kelvin is -457.87 degrees Fahrenheit.
Originally posted by wx4caster
reply to post by TheRedneck
to further clarify for all on kelvin and celcius-
zero celcius is the freezing point of water.
zero kelvin is the temperature at wich all molecular motion stops.
the incrementation is the same.
Originally posted by TheRedneck
However, I do not agree there is a problem with the climate.
Yes, we have seen some warming in recent times, ..
We also know that global temperatures tended to rise and fall during historical times, even within the human timeframe.
If the Earth has survived earlier warming episodes, what is so bad about this one even if it is not natural? The problem is that our civilization—where cities are located, where we grow food, where we get fresh water—is all based on the climate we have experienced for the last 10,000 years. So are many of the world’s ecosystems. If the climate changes, many of those things will suddenly find themselves in the wrong place.
The difference is that in the natural cycle CO2 lags behind the warming because it is mainly due to the Milankovitch cycles. Now CO2 is leading the warming. This is clearly not natural cycle. The earths natural cycles, if human industrial output had not been involved would have us near or slight below thermal equilibrium, possibly slightly cooling.
In other words, if were were in the natural cycle without human influence, the forcing levels would likely be around 0W/m2 to -0.1W/m2. We are currently experiencing a positive forcing of around 3.6 to 3.8W/m2 and a human induced negative forcing of around 2W/m2. The resultant forcing, depending on current levels and the Schwabe cycle is around +1.6W/m2 above natural cycle as estimated.
Originally posted by TheRedneck
The specific heat capacity of water by volume is 4.186 J/cm³·°K[6] at 25°C. Thus, in order to raise the temperature of the oceans by a single degree Kelvin:
4.186 J/cm³·°K = 4,186,000,000,000 kJ/km³·°K
4,186,000,000,000 kJ/km³·°K · 1,347,000,000 km³
= 5,638,542,000,000,000,000,000 kJ/°K
Originally posted by truthquest
reply to post by TheRedneck
I will admit TheRedneck that I don't understand global warming physics, so your maths simply did not make sense to me.
The part I don't understand is this: How does more CO2 cause global warming?
First, please help me rule something out. Is it the thermal insulation properties of CO2 that is causing the increase in warming? I imagine not since even if it were a perfect insulator as a gas it only takes up a small fraction of the total gasses in the atmosphere. So the thermal insulation of CO2 is then mathematically insignificant, correct?
So what then is it about CO2 that causes an insulating effect? Apparently you refer to something about CO2 interacting with photons to create heat, but I can't quite understand what that is exactly. CO2 is not an optically clear gas and therefore absorbs light? Is that what it is?
Originally posted by truthquest
Originally posted by TheRedneck
reply to post by truthquest
Thanks, that pretty much explains everything I was wondering about.
In your equations then, how do you determine the percent chance of light that will hit a CO2 atom? It seems you are saying in your OP that if CO2 was say is 2% of all gasses in the atmosphere that it would intecept 2% of the light. Is that correct?
In my opinion, given your description of how global warming works, the way to determine how much heat will be captured would be to determine:
- The area (from a cross-section-wise perspective) of an individual CO2 atom.
- The area within that area which if hit would result in absorption of a photon... I would guess 100% but don't know the physics.
- How many CO2 atoms there are per area unit in the atmosphere.
So, by multiplying the cross-section area of CO2 by the number of atoms per unit area you can calculate what the maximum area is of CO2 per unit area that would be that is vulnerable to interaction with light rays. Using a statistical function (not sure which one) you could cancel out CO2 atoms who's area overlaps.
The next step would then be to determine the area (again as a cross-section perspective) of a photon. Since I don't really know too much about thermodynamics there is probably no point in me going on much further but the point is my idea would be to match up the area of the photon.
You are absolutely right.CO2 has nothing to do with acid rain. Except that it is an ingredient in Coca-Cola, which, when shaken and poured on a windshield, works great to clean off the acid rain residue.
Originally posted by TheRedneck
reply to post by 4nsicphd
although they spew enormous amounts of CO2, and SO2, which combines with O2 and H2 to form H2SO4
I just want to make a clarification to that. From the wording, it could be believed that you are saying carbon dioxide (CO2) is an ingredient in the manufacture of sulfuric acid (H2SO4), which it is not. Sulfur dioxide (SO2) is the main ingredient in atmospheric production of sulfuric acid; carbon dioxide does not enter into that equation.
I know you didn't intend for it to come across in that way, but it could to someone without chemical knowledge.
Kudos on your explanations, nonetheless.
TheRedneck
Please amend your calculations.
Originally posted by TheRedneck
reply to post by TheProfessional
Please amend your calculations.
I'll be happy to. I need to know a couple things first, however.
- You state your values in J/g·°K, whereas I was calculating based on volume in J/cm³·°K. As I am sure you are aware, the difference in density of seawater as compared to 'fresh' water will offset the differences in value (although quite possibly not completely), so this will have to be converted into volumetric units instead of mass units for the new value to be applicable. Was the use of mass units instead of volumetric units correct, or a typo?
- The values I have been using came, admittedly, from Wikipedia and are therefore subject to correction. However, this source agrees with the value I have commonly used for some time, which is the value for volumetric specific heat capacity of 'fresh' water of 4.186 J/cm³·°K, and mass-based specific heat capacity of 4.183 J/g·°K. Can you provide a link to this new value for verification?
Thank you for the additional information. I look forward to your reply so this re-calculation can be included in the new calcs I am performing now, or at least can be corrected in the original calcs.
TheRedneck
[edit on 12/4/2009 by TheRedneck]
3.993 J/g·°K · 1.030 g/cm³ = 4.113 J/cm³·°K
This is equal to 4.113·10^12 kJ/km³·°K
4.113·10^12 kJ/km³·°K · 1.347·10^9 km³ = 5.540·10^21 kJ/°K
This compares to the previous result of 5.639·10^21 kJ/°K.
5.540·10^21 kJ/°K ÷ 55.26·10^18 kJ/°K = 100.3
This compares with the previous result of 102.03.
with the further notation of
(ρ/kg m^−3 − 1000)*
* In oceanographical work these data are always expressed in terms of 1000(S − 1) where S is the density relative to water at 4 °C. Values for this can be obtained by adding 0.03 to the values in the table.