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SOLUTIONS TO EXERCISE 23
(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! |