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How much .^235U is consumed in a day in an atomic power house operating at 400 MW, provided the whole of mass .^235U is converted into energy? |
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Answer» Solution :power = 400 MW= `400xx10^6` W , time = 1 day = 86,400 s. Energy produced , E= power x time `=400xx10^6xx86,400=3.456xx10^13` J As the whole of mass is CONVERTED into energy , by Einstein.s mass-energy `E=Mc^2` `M=E/c^2=(3.456xx10^13)/(3xx10^8)^2=3.84xx10^(-4)` KG =0.384 g. |
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