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A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ""_(92)^(235)U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ""_(92)^(235)U and that this nuclide is consumed only by the fission process. |
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Answer» Solution :Energy generated PER GRAM of `""_(92)^(235)U = (6 xx 10^(23) xx 200 xx 1.6 xx 10^(-13))/(235) J g^(-1)` The amount of `""_(92)^(235)U` CONSUMED in 5y with 80% on - time. `=(5 xx 0.8 xx 3.154 xx 10^(16) xx 235)/(1.2 xx 1.6 xx 10^(13)) g = 1544 g` The initial amount of `""_(92)^(235)U = 3088 KG. ` |
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