A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 92235 U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 92235U and that this nuclide is consumed only by the fission process.
The half life of the fuel is 5 years we know that the energy produced in fission of Uranium is 200 MeV. So, energy produced by 1kg of Uranium: E=2356.022×1023×1000×200 ≈8.17×1013 J Since the reactor operate 80% of time, so time of operation is 4 years. The energy produced is: Er=1000×1066×4×365×24×60×60 The mass required for producing energy is: m=1000×1066×4×365×24×60×608.17×1013 J =1544 kg Since this is the amount consumed and initial amount should have been double of this. So, the initial amount of Uranium would have been 3088 kg
