Question

A nuclear power plant operates at 40.0% efficiency with a continuous production of 1202 MW of usable power in 1.00 year and consumes 1.23×106 g of uranium-235 in this time period. What is the energy in joules released by the fission of a single uranium-235 atom?

Answer 1

Usable power produced by nuclear power plant per year = 1202 MW

The power plant operates at an efficiency of 40 %.

Let the actual power produced by the plant per year be X MW.

Then 40 % of X = 1202 MW

(40 / 100) * X = 1202 MW

X= 1202* (40 / 100) MW = 3,005 MW

Therefore, actual power produced by the plant per year = 3,005 MW = 3,005 * 10⁶ W

Time = 1 year

= 1 yr(365 days/yr) (24 hours/day) (60 min/hour)(60 s/ min)

= 31,536,000 s

We know, Energy = Power * time

Energy produced per year by the power plant =( 3,005 * 10⁶ W) * (31,536,000 s )

=9.4766 * 10¹⁶ J

This is the energy produced by 1.23*10⁶ g of U-235.

We need to find the energy released by fission of 1 atom of U-235.

Molar mass of U-235 = 235 g /mol

Therefore,

235 g of U-235 contains =6.022 *10²³ atoms

No. of atoms in 1.23*10⁶ g of U-235. = [(6.022*10²³ atoms / mol) / 235 g/mol ] * 1.23*10⁶ g

= 3.15194 *10²⁷

Energy produced by 3.15194 *10²⁷ atoms of U-235 = 9.4766 * 10¹⁶ J

Energy produced by 1 atom of U-235 =(9.4766 * 10¹⁶ J) ( 3.15194 *10²⁷ atoms )

= 3.00658 *10^-11 J / atom of U-235