Question

Calculate the number of joules that can be obtained from the fissioning of 1 kg of Uranium-235 (U-235), assuming 198 MeV average energy release per fission.

a) How much energy can be obtained from the U-235 in 1 kg of natural uranium?

b) Calculate how much energy, in joules, can be obtained from burning 1 kg of coal.

c) What is the total energy contained in the US resources of coal and U-235?

Answer 1

Ans : (Atomic mass) / (particles) = 235 u

Particles Mass = (235 u * 1.66 * 10^-27 kg/u) = 390.1 * 10^-27 kg

Total particles = N = 1kg/(particles mass) =1kg/(390.1 * 10^-27 )kg = 2.56 * 10²⁴

Total energy = 198 meV * N = (198 * 2.56 * 10²⁴ * 1.602 * 10^-13) J = 8.12 * 10¹³ J

Ans a: U-235= 235.0439299 g/mol

Natural Uranium contains only 0.72% of U-235 and rest of U-238 isotope.

Therefore, 0.72% * 100% = 7.2 g 0f U-235

Number of moles of U-235 in 1g = (7.2g) / (235.04 g/mol) = 0.030633 mol of U-235

Number of atoms= 0.030633 * 6.022 * 10²³ = 1.844768 * 10²² atoms of U-235

If every atoms undergoes fission then, 1.844768 *198 MeV = 3.6526 * 10²⁴ MeV.

1 MeV = 1.602 * 10^-13 J

Hence, 3.6526 * 10²⁴ * 1.602 * 10^-13 J = 5.85217 * 10¹¹ J

Ans b : The energy density i,e its heating value is roughly 24 megajoules per kilogram. Burning of 1kg of coal produces 24 * 10⁶ J energy.

Ans C: It varies from year to year.