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InterviewSolution
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A nuclear reactor using `.^(235)U` generates `250 MW` of electric power. The efficiency of the reactor (i.e., efficiency of conversion of thermal energy into electrical energy) is `25%`. What is the amount of `.^(235)U` used in the reactor per year? The thermal energy released per fission of `.^(235)U` is `200 MeV`. |
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Answer» Rate of electrical energy generration is `250 MW =250 xx 10^(6) W(or J s^(-1))`. So, electrical energy generation is `250 MW =250 xx 10^(6) W(or J s^(-1))`. Therefore, electrical energy generated in `1 year` is `(250 xx 10^(6) J s^(-1)) xx (365 xx 24 xx 60 xx 60 s)=7.884 xx 10^(15) J` Thermal eneregy from fission of one `.^(235)U` nuvleus is `200 MeV =200 xx 1.6 xx 10^(-13) = 3.2 xx 10^(-11J` . Since the efficiency is `25%`, the electrical energy obtained from the fission of one `.^(235)U` nucleus is `E_(1)=3.2 xx 10^(-11) xx(25)/(100) =8.0 xx 10^(-12) J` Therefore, the number of fission of `.^(235)U` required in one year will be `N=(7.884 xx10^(15))/(8.0 xx 10^(-12) =9.855 xx 10^(26))` Number of moles of `.^(235)U` required per year is `n=(9.855 xx 10^(26))/(6.023 xx 10^(23)) =1.636 xx10^(3)` Therefore, mass of `.^(235)U` required per year is `m=1.636 xx 10^(3) xx 235` `=3.844 xx 10^(5) g=384.4 kg`. |
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