Of all the hydrogen in the oceans 00300 of the mass is deute

Of all the hydrogen in the oceans, 0.0300% of the mass is deuterium. The oceans have a volume of 311 million mi^3. (Enter your answer using one of the following formats: 1.2e-3 for 0.0012 and 1.20e+2 for 120..) If nuclear fusion were controlled and all the deuterium in the oceans were fused to^4_2He, how many joules of energy would be released? World power consumption is about 7.00 Times 10^12 W. If consumption were 80 times greater, how many years would the energy calculated in part (a) last?

Solution

When deuterium fuses to He-4

²H₁ + ²H₁ -> ⁴He₂ + energy

mass of Deterium = 2.014102 u

mass of He-4 = 4.002603 u

mass difference =2.014102*2 - 4.002603 = 0.025601 u

                             = 23.847 Mev

volume of oceen = 311e+6 ml³ , it is not cler if it is ml or cu.m , if cu.m multiply by fctor of e+6, here it is ssumed s ml³

H-2 in oceen wter = 311e+6 *2/18 gms

deuterium = 311e+6 *2/18 * 0.03/100 = 1.037e+6 gms

no of deuterium toms = 1.037e+6 * 6.02e+23/2 = 3.12137e+29

no. H-2 fusions = 1.561e+29

energy relesed = 1.561e+29 *23.847 Mev

                         = 1.561e+29 *23.847 * 1.609e-13 J

                         = 59.89e+16 J

b) power consumption 7.0 e+12 w

      the enrgy will lost for 59.89e+16 /7.0 e+12 hrs

                                    = 8.56e+4 hrs

                                   = 9.77 yrs

if the consumtion ws 80 times it would lost

    9.77/80 = 0.122 yrs

note if the volume given is in m³ then it lost

0.122e+6 yrs