1269_Physics ProblemsTechnical Physics

1269_Physics ProblemsTechnical Physics - 610 P45.46...

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610 Applications of Nuclear Physics P45.46 When mass m of 235 U undergoes complete fission, releasing 200 MeV per fission event, the total energy released is: Q m N = F H G I K J 235 200 g mol MeV A a f where N A is Avogadro’s number. If all this energy could be utilized to convert a mass m w of liquid water at T c into steam at T h , then, Q m c T L c T w w c v s h = ° + + ° 100 100 C C b g b g where c w is the specific heat of liquid water, L v is the latent heat of vaporization, and c s is the specific heat of steam. Solving for the mass of water converted gives m Q c T L c T mN c T L c T w w c v s h w c v s h = ° + + ° = ° + + ° 100 100 200 235 100 100 C C MeV g mol C C A b g b g a f b g b g b g . P45.47 (a) The number of molecules in 1.00 liter of water (mass = 1 000 g) is N = × F H G I K J × = × 1 00 10 6 02 10 3 34 10 3 23 25 . . . g 18.0 g mol molecules mol molecules e j . The number of deuterium nuclei contained in these molecules is ′ = × F H G I K J = × N 3 34 10 1 1 01 10 25 22 . . molecules deuteron 3300 molecules deuterons e j . Since 2 deuterons are consumed per fusion event, the number of events possible is = × N 2 5 07 10 21 . reactions, and the energy released is
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