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C WWW sapling learning.com /ibiscms/mod/ibis/view.php?id-2568677 Question 10 of 12 Map UNIVERSITY PHYSICS presented by sapling learning According to the U.S. Energy Information Administration, the United States generated approximately 3.97x1012 kWh of electrical energy during 2004. If we could generate all this energy using a matter-antimatter reactor, how many kilograms of fuel would the reactor need, assuming 100% efficiency in the annihilation? Number kg Suppose the fuel consisted of copper and anti-copper, each of density 8940 kg/m3. If the copper and an copper were each stored as a cubical pile, what would the side-length of each be? Number Previous 3 Give Up & View Solution Check Answer Next Exit Hin 5/15/2016 11:55 PM Due Date: Points Possible 00 Grade Category: Graded Description: Policies: Homework You can check your answers. You can view solutions when you complete c up on any question. You can keep trying to answer each questior you get it right or give up. You lose 5% of the points available to each a in your question for each incorrect attempt a answer O eTextbook O Help With This Topic O Web Help & Videos O Technical Support and Bug Reports 8:45 PM 5/11/2016Explanation / Answer
You are indicated efficiency is 100% the rest energy of 0.511 MeV electron is its antiparticle the positron and has the mimma energy.
The energy of positronium atom (electron + positron) is
2 me c = 1.02 MeV
1.02 106 eV (1.6 10-19 /1 eV) = 1.632 10-13 J
The total energy is 3.97 1012 Kwh = 3.97 1015 Wh (3600 s/1 h) = 14.292 1018 W s
P = E/t
[W s] = J
1 atom positronio ------ 1.632 10-13 J
a ------ 14.292 1018 J
a = 14.292 1018 /1.632 10-13
a = 8.757 1031 atom positron
This amount of weight atomos
1 atom ----- 2 me = 2 9.1 10-31 Kg
8.757 1031 a ---- w
w =18.2 10-31 8.757 1031
w = 1.59 Kg
Part B)
= 8940 Kg/m³
the mass of each part is half of the total
m = 0.795 Kg
= m/V
V = m/
V = 0.795 / 8940
V = 8.89 10-5 m³
V = L3
L = (V)1/3
L = (8.89 10-5 )1/3
L = 4.46 10-2 m
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