Answer:
The amount of nuclear fuel required is 1.24 kg.
Step-by-step explanation:
From the principle of mass energy equivalence we know that energy generated by mass 'm' in an nuclear plant is
[tex]E=m\cdot c^2[/tex]
where
'c' is the speed of light in free space
Since the power plant operates at 1200 MW thus the total energy produced in 1 year equals
[tex]E=1200\times 10^6\times 3600\times 24\times 365=3.8\times 10^{16}Joules[/tex]
Thus using the energy produced in the energy equivalence we get
[tex]3.8\times 10^{16}=mass\times (3\times 10^{8})^2\\\\\therefore mass=\frac{3.8\times 10^{16}}{9\times 10^{16}}=0.422kg[/tex]
Now since the efficiency of conversion is 34% thus the fuel required equals
[tex]mass_{required}=\frac{0.422}{0.34}=1.24kg[/tex]
