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In the process of nuclear fission of 1 gram uranium, the mass lost is 0.92 milligram. The efficiency of power house run by it is 10%. To obtain 400 megawatt power from the power house, how much uranium will be required per hour? (c=3xx10^8 "ms"^(-1)) |
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Answer» Solution :Power to be obtained from power house = 400 mega watt `therefore` Energy obtained per HOUR = 400 mega watt x 1hour= (`400xx10^6` watt) x 3600 second = `1.44xx10^10` joule Here only 10% of input is utilized. In order to obtain `144 XX 10^10`joule of useful energy, the output energy from the power house `(10E)/100=144xx10^10` J, E=`1.44xx10^11`joule Let, this energy is obtained from a mass-loss of `Deltam` kg. Then `(Deltam)c^2=144xx10^11` joule `Deltam=(144xx10^11)/(3xx10^8)^2=16xx10^(-5)` kg = 0.16 gm Since 0.92 milli gram (= `0.92xx10^(-3)` gm)mass is LOST in 1 gm URANIUM , hence for a mass loss of 0.16 gm, the uranium required is `=(1xx0.16)/(0.92xx10^(-3))` = 174 gm Thus to run the power house, 174 gm uranium is required per hour. |
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