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SOLUTIONS TO EXERCISE 23

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energy: TOC for Knowledge Concepts, Exercises, and Solutions


(a) The Sun's luminosity is 3.9 * 10^26 J/s. (1 J = 1 kg m^2/s^2)
If this luminosity is generated by converting matter into energy, then each
second, 3.9 * 10^26 J worth of mass must be converted.  Using E = mc^2,

  m = E/c^2 = (3.9 * 10^26 J) / (3 * 10^8 m/s)^2 = 4.3 * 10^9 kg

There are about 8 million people in New York City, with average weight
about 40 kg (from babies to big-and-tall people included).  That's 

  m(people) = 8 * 10^6 * 40 = 3.2 * 10^8 kg

So each second, the Sun consumes matter equivalent to about 13 times the
entire population of New York City.  Good thing the Sun is so massive.


(b) The proton-proton chain describes four protons coming together
to form a helium nucleus.  A proton's mass is about 1.673 * 10^(-27) kg.
The difference, then, between the mass of four protons and one helium
nucleus is ((4 * 1.00794) - 4.00260) * 1.673 * 10^(-27) kg, and that
converts to

  E = mc^2 = (4.03176 - 4.00260) * 1.673 * 10^(-27) * (3 * 10^8)^2

    = 4.39 * 10^(-12) J


(c) For every four protons - that is, hydrogen nuclei - used in fusion,
we get 4.4 * 10(-12) J.  so produce 3.9 * 10^26 Watts, the Sun must use

       3.9 * 10^26 J/s
 ----------------------------- = 3.55 * 10^38 hydrogen nuclei per second.
 4.39 * 10^(-12) J/(4 H-nuclei)

There two hydrogen atoms in each water molecule, and the molecular weight
of water is 18 (H-two-O).  So one mole of water weighs 18 grams, and 
contains two moles of hydrogen atoms (1 mole = 6 * 10^23).  A gallon of
water contains 4 quarts; 1 quart is slightly less than 1 liter, which
is 1000 cm^3; and water's density is 1000 kg/m^3 = 1 gram/cm^3.  So, 
approximately, one gallon of water contains about 4000 grams of water, or

  (2 * 6 * 10^23 H-nuclei/mole) * (4000 g) / (18 g/mole) = 2.67 * 10^26

  hydrogen nuclei.  So one would need (3.55 * 10^38)/(2.67 * 10^26)

  = 1.3 * 10^12 gallons.  Over a trillion gallons of water each second!