What is the power output of a `._(92) U^(235)` reactor if it is takes `30` days to use up `2 kg` of fuel, and if each fission gives `185 MeV` of useable energy ?.

What is the power output of a `._(92) U^(235)` reactor if it is takes

1.	What is the power output of a `._(92) U^(235)` reactor if it is takes `30` days to use up `2 kg` of fuel, and if each fission gives `185 MeV` of useable energy ?.
Answer» `235` am u of uranium gives `185 MeV `energy. Therefore, the energy given by `1` am u of `._(92)U^(235)` is `(185)/(2.35) MeV=1(85)/(235) xx1.6 xx10^(-13) J` But `1` am u `=1.66 xx 10^(-27) kg`. Therefore, energy released by `1.66 xx 10^(-27)kg` of`._(92)U^(235)` is `(185 xx 1.6 xx 10^(-13)/(235)` Hence, energy released by `2 kg` of `._(92)U^(235)` , `W=(185 xx 1.6 xx10^(-13) xx2)/(235 xx1.66 xx10^(-27)) =1.517 xx 10^(14) J` Therefore , Therefore , pwer output of reactor `=(W)/(t) =(1.517 xx 10^(14))/((30 days))` `=(1.517 xx1`0^(14))/(30 xx 24 xx 50 xx 60)` `=5. 785 xx 10^(7) W`.

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What is the power output of a `._(92) U^(235)` reactor if it is takes `30` days to use up `2 kg` of fuel, and if each fission gives `185 MeV` of useable energy ?.

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