If 2 kilograms (70.5479239614 ounces ) of mass is converted totally to energy the quantity of energy will be as follows: M2=4 K0=2 U=4*299792458=1199169832 U/k=1199169832/2=599584916 E=599584916*299792458=1.7975103575e17 joules .........According to R.L.’s estimation it takes about 6*10^24 J to heat the Earth’s atmosphere and ocean by one degree, then according to R.L. all we need to do is convert 2 kilograms of mass into energy every sixty thousands of a second for a duration of one second to heat the atmosphere by one degree and have some energy left over.....
Ok nimrod... ignoring your stupid rewrite of e=mc^2, since when does (1.8*10^17)* 60,000 equal 6*10^24?! again you are off by orders of magnitude! You dont understand a damn thing about basic math or fission- as others have already pointed out...

You are absolutely right, I made an error, but the error was not the type that you are referring to.
What I did was; I assumed that the quantity of joules that is represented in the calculation is the heat equivalent of watts per second, in other words, if that quantity of energy is released in two seconds then the effect during one second is only half of that amount. If on the other hand that quantity of energy is released in half a second, then the effect that is created is twice the original amount.
Or to explain it differently, if we have a reference power of one watt per second, and that power is distributed over a period of two seconds, then during one second the power that is released is .5w. If that one watt of power is released in .5 sec, then it would take 2w of power to maintain that effect for a full second.
So the power equivalent of 1.8*10^17 joules distributed for a period of 60,000 th. of a second represents the power of 1.08e19 joules. Now if we released 1.08e19 joules every sixty thousandth of a second until a full second has passed then we would have accumulated 6.46e23 joules of energy. The error that I made was that I calculated the needed energy for a period of one minute but stated that it was for one second.
The one minute equivalent power is 3.888e25 joules.

But don’t worry, the grand children are teaching me arithmetic so I should improve eventually :-)

You asked how much uranium must be consumed in fission to convert 2kg of mass into energy...Ok here we go: Energy released from 1 atom of U235 going fission: 2*10^8 eV= 3.2*10^-11 J mass to energy in one U235 fission: 3.5*10^-28 kg mass of U235 atom= ~ 3.9*10^-25 kgnumber of fusion events to turn 2 kg of mass into energy: 5.7*10^27 eventsmass of U235 atoms that must fission to turn 2 kg into energy (and remember the amount of U235 needed will be MUCH greater than the number of atoms that actually split) is :(5.7*10^27)*( 3.9*10^-25)=2.2*10^3 kgNow to heat the earths ocean and atmosphere by 1 degree, I calculated you need 6*10^24 J (actually the true amount is ORDERS OF MAGNITUDE MORE since I did not include the thermalization of the earth itself, or the energy radiated to space)... the mass of U235 that must fission is (2.2*10^3 kg)*(6*10^24 J)/(1.8*10^17 J)=2.2*10^3 kg *(3.3*10^7)= 7.26*10^10 kg to fission!!!!!!that is SEVENTY _BILLION_ KILOGRAMS!!!!!!!!!!NOW WILL YOU DROP THIS NONSENSE?!

Why do you consider it “nonsense” to speculate that since the fission reactors can heat many tons of water that they might also be able to heat the atmospheric molecules to some extent?
I know that you and many others would not want it to be that way, but you can not change the laws of physics by putting big numbers on a piece of paper.
Show me FACTUAL measurements indicating that the Earth’s atmosphere has not heated up, increased in radioactivity or reduced in density to ANY degree in the last 70 years and I will believe your numbers.

you are wrong- demonstrably so- you just advertise your ignorance and stupidity when you post this crap.... You are wrong by MANY MANY ORDERS OF MAGNITUDE!!! As Nick said "GET AN EDUCATION!"YOU are too lazy and stupid to run the numbers yourself, and when we do them for you, you put your fingers in your ears and close your eyes and keep posting garbage...

You are behaving in the same manner by arguing that radioactivity can not increase the kinetic energy of the atmospheric molecules to any extent, You should talk to the people that were living close to the Three Mile Island nuclear power plant when the accident happened there.

In _all_ my back of the envelope calculations I made only assumptions that helped your case, and effects I ignored, if included, would only help -my- case and still we get 70 billion kg!!! Go play in another sandbox, kid- the adults here are tired of changing your mental diapers...

My calculation shows 120,000*60 kg. or 7,200,000 kg.

Thank you for changing my “mental diapers”. I’d do the same for you if you wanted it.

But then naturally you are ignoring the “stupid rewrite of e=mc^2,” and have very emphatically stated that I can never teach anything to you, so I suppose you will have to figure out how “stupid” the re-written Einstein’s formula is. Although I do not expect you to bother with such triviality since you all ready know everything that there is to know about the universe.

Thanks to everyone else also for their helpful comments.

I have learned a lot about the ‘ignorance’ of scientific thinking.

Keep smiling anyway (And burying your head in the sand)

Cheesh !!……Brings to mind the warning about throwing pearls before swine.

Johannes