Answer to Question #160181 in Electrical Engineering for Keleko

Question #160181
An all-electric home uses approximately 2000KWh of electric energy per month. How much uranium-235 would be required to provide this house with its energy needs for one year? Assume 100% conversion efficiency and 208 MeV released per fission.
1
Expert's answer
2021-02-25T04:07:45-0500

Convert 2000 kW·h to J:


"E=2000\u00b71000\u00b73600=7.2\u00b710^9\\text{ J}."

Find how many fissions (uranium atoms) can provide this amount of energy:


"n=E\/E_U=\\frac{7.2\u00b710^9}{208\u00b71.6\u00b710^{-13}}=2.16\u00b710^{20}\\text{ fissions}."

One uranium-235 weighs 235·1.66·10-27 kg. So, the number of fissions we found above corresponds to the following mass of uranium:


"M=nm=2.16\u00b710^{20}\u00b7235\u00b71.66\u00b710^{-27}=8.43\u00b710^{-5}\\text{ kg}."


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