Let's see if this is right:
Weight of the atmosphere: 5 x 10^15 metric tons
CO2 emitted in 2007: 2.93 x 10^7 metric tons
So the 2007 CO2 added 1 part in 1.6 x 10^8 to the atmosphere.
An Olympic sized swimming pool contains at least 2.5 x 10^6 liters of water.
There are about 20 drops in a milliliter or 20000 in a liter, so there are 5 x 10^10 drops in an Olympic sized pool.
So as far as relative mass is concerned, the CO2 addition in one year is more like 300 drops from an eyedropper in an OSSP, or about 15 ml or 1 Tablespoon of water in an Olympic sized pool
Lot of time on your hands huh, LOL? OK so maybe I read the analogy wrong an eyedropper full.
I think if Chris asked the question, these scientist advocates would freeze up.
I just took another trip through the central valley of California. The area is a moonscape. To save a small freaking fry fish they have turned the area back to nature.
These scientists don’t think about the other side of the cost equation. How many people will die due to decreased crop yields that the CO2 would have given us? How many people will die from cancer caused by the Mercury in the CFLs?
They keep telling us that the other side of the equation isn’t a concern but refuse to quantify it.
I don’t trust them. I think they need to be exposed. For ever benefit they claim there is an offset. Trees cut to make room for windmills and solar panels. Water required to clean the solar panels. etc...
5.512 X 10^15 is the Weight of the Atmosphere
The CO2 can’t be right though, because 0.039% of this is CO2.
Remember, my question for Christi to ask is how much has persisted since the beginning of the Industrial Revolution?
If we are emitting it and it gets absorbed as quickly as it is emitted, then it is of no concern.
My understanding is that something on the order of 0.0013 x 10^-58 is the correct answer. I saw a video somewhere that laid this all out and I think it would be devastating to anyone arguing AGW.