How much U235 is consumed in a day in an atomic power house operating at 400 MW, provided the whole of the mass of U235 is converted into energy?

Question

How much U235 is consumed in a day in an atomic power house operating at 400 MW, provided the whole of the mass of U235 is converted into energy? (A) 0.384g (B) 0.386g (C) 0.382g (D) 0.381g

Tardigrade Admin · Accepted Answer

Here, rate of production of energy at the atomic power house,

P = 400 M W = 400 \times 1 0^{6} J s^{- 1}

Therefore, total energy produced in a day i.e.,

24 \times 60 \times 60 s

,

E = P \times 24 \times 60 \times 60 = 400 \times 1 0^{6} \times 24 \times 60 \times 60

= 3.456 \times 1 0^{13} J

If mass of

U^{235}

consumed per day in m (in kg) so as to produce the required amount of energy, then

E = m c^{2}

Or

3.456 \times 1 0^{13} = m c^{2}

Or

m = \frac{3.456 \times 1 0 ^{13}}{c ^{2}} = \frac{3.456 \times 1 0 ^{13}}{( 3 \times 1 0 ^{8} ) ^{2}}

= 0.384 \times 1 0^{- 3} k g

= 0.384 g

Q. How much U235 is consumed in a day in an atomic power house operating at 400MW, provided the whole of the mass of U235 is converted into energy?

0.384g

0.386g

0.382g

0.381g

Q. How much $U^{235}$ is consumed in a day in an atomic power house operating at $400 M W,$ provided the whole of the mass of $U^{235}$ is converted into energy?