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In the process of nuclear fission of 1 g...

In the process of nuclear fission of `1 g` uranium, the mass lost is `0.92 mg`. The efficiency of power house run by the fission reactor is `10%`.To obtain `400` megawatt power from the power house, how much uranium will be required per hour? `(c=3xx10^(8) m s^(-1))`.

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Power to be obtained from power house `= 400` megawatt In this case, energy obtained per hour
`=400` megawatt` xx 1 hour`
`(400 xx 10^(6) W) xx 3600 s`
`=144 xx 10^(10) J`
Here, only `10%` of output is utilized. In order to obtain `144 xx 10^(10)` joule of useful energy, the output energy from the power house is given by
`E =((144 xx 10^(10) xx100))/10 =144 xx 10^(11) J`
Let this energy is obtained from a mass loss of `Deltamkg`. Then,
`(Delta m)c^(2)=144 xx 10^(11) J`
or `Delta m=144 xx 10^(11)/(3 xx10^(8))^(2) =16 xx 15^(-15) kg =0.16g `
Since `0.92 mg` (=0.92 xx 10^(-3) g) ,mass is lost in `1 g` uranium, hence, for a mass loss of `0.16 g`, the uranium required is given by
`Delta m'=1 xx (0.16)/(0.92) xx 10^(-3) =174 g`
Thus, to run the power house, 174 g uranium is required per hour.
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