How Much Material Is Needed to Meet the USA's Annual Energy Requirement?

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Homework Statement


The annual energy requirement of the USA is of the order [tex]10^20[/tex] J. If we could find a 100% efficient process that could change matter into energy, how many kilograms of material would be needed to meet this requirement?

Homework Equations


[tex]E = mc^2[/tex]

The Attempt at a Solution



Well, i simply sub [tex]10^2^0[/tex] into [tex]E = mc^2[/tex]
And i obtain, [tex]\frac{10^2^0}{(3.0*10^8)^2}[/tex]
then [tex]m=11,111.1111[/tex]

Is this correct? It looks to me like too simple a question, since this is the last question of my tutorial.
 
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It might have just been a question put into make you think about the possible consequences. But there's a slight miscalculation - you're off by one decimal place: ~1,111 Kg. Whats so special about this? Dividing by 365 to get the daily rate, you arrive at the conclusion that to support the energy needs of the USA on a daily basis would require 3 Kg of matter to be completely transformed into energy. Maybe you could compare it to the magnitude of fuel used nowadays? Just food for thought.