Question

Burning 20 tons of coal provides about 5. 1011 J, which is approximately the annual energy requirement of an average person in the United States. If this amount of energy was provided by the electricity generated by the fission of uranium-235 in a nuclear reactor, instead, what mass of natural uranium would be required? About 0.7% of natural uranium is uranium-235 (almost all the rest is uranium-238, which does not fission like U-235 does). Assume that the fission of each U-235 atom provides 200 MeV of energy, and that the nuclear power plant has an efficiency of 34% in transforming the energy from the fission reactions into electricity. In your calculation, assume that the atomic mass of natural uranium is 238 grams per mole.

_____ kg

Answer 1

E per 20 tonne = 5*10^11 J
Now energy provided by every atom of U235 fission = 0.34*200*10^6 eV = 0.34*3.2*10^-11 J = 1.088*10^-11 J
Njmber of atoms = 5*10^11/1.088*10^-11 = 4.5955*10^22
Mass = 4.5955*10^22 * 235/6.022*10^23 = 17.933 g
this is 0.7 pc of uranium
0.007*U = 17.933 g
U = 2561.947 grams = 2.561 kg