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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 can by it is 10 %,Toobtain 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 `:.` Energy obtained per HOUR = 400 mega watt `xx` 1hour = `(400xx10^(6) "watt") xx 3600 ` second `=144xx10^(10)` joule Here only 10% of input is utilized . In order to OBTAIN `144xx10^10` joule of useful energy, the output energy from the power house `(10E)/(100)=144xx10^(10)J` `E = 144 xx10^(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^(18))^2 = 16xx10^(-5)kg ` = 0.16 gm Since 0.92 milli gram `(=0.92 xx10^(-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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