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An incorrect but illustrative answer

Most energy sources we are used to involve chemical reactions which convert one compound into another, and energy is given off in the process. This is true of all burning processes such as burning gasoline, burning coal, etc. In chemical reactions we have something like :

equation18

In this we have two molecules come together on the left hand side of the equation, produce a new molecule + energy on the right hand.

A measure of ``bang for the buck'' is how many Joules of energy you get for each kilogram of reactants on the left hand side.

As it says in the book, a typical number for chemical reactions is about a million Joules per kilogram. Higher for nitroglycerine, lower for peat moss, but in this ballpark. Therefore tex2html_wrap_inline67 .

Let's do an interesting calculation, which was first done in the middle of the last century. Assume that the Sun obtains its energy by some sort of chemical reaction. This reaction proceeds at a rate that releases tex2html_wrap_inline69 W by converting the Sun from reactants to reaction products. How long could it keep this up?

Use the equation E = PT, where E is the energy (Joules), P is the power (Joules/sec) and T is the time (seconds) you can keep this up. Therefore the time (which is what we want) is:

equation25

What is P? We know that, it is the luminosity of the Sun = tex2html_wrap_inline69 Watts. What is the total energy content of a big ball like the Sun? tex2html_wrap_inline83 Joules.

Therefore tex2html_wrap_inline85 seconds.
Question for audience; Is that good enough? How long does that represent?



Steve Spangler
Tue Aug 31 16:29:42 CDT 1999