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

[Delzac]

Homework Statement

The annual energy requirement of the USA is of the order $$10^20$$ 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

$$E = mc^2$$

The Attempt at a Solution

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

Is this correct? It looks to me like too simple a question, since this is the last question of my tutorial.

 

[koofle]

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.

 

[Delzac]

K, thanks for the help.