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 310 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 90 times greater, how many years would the energy calculated in part (a) last?

Solution

\"... Fusion of deuterium with tritium creating helium-4, freeing a neutron, and releasing 17.59 MeV of energy, as an appropriate amount of mass changing forms to appear as the kinetic energy of the products, in agreement with kinetic E = mc2, where m is the change in rest mass of particles....\"

So we will use 17.59 MeV as the energy released for the fusion of each deuteron.

(2) I used Master Converter version 2.8.1, evaluation mode, to find conversion factors for the following calculations:

convert the volume of the oceans to metric units

(310 million mi3.) = (3.10 x 10^8 mile^3)*(1609.344 meter/mile)^3 = (3.10 x 10^8 mile^3)*(4.1655 x 10^9 m^3/mile^3) = 1.29 x 10^18 m^3

Convert this further to liters:

(1.29 x 10^18 m^3)*(1000 liter/m^3) = 1.29 x 10^21 L

(3) \"density of seawater\"

Average density at the surface is 1.025 g/ml [sic: g/mL]...\"

(4) convert the volume of water to mass

(1.29 x 10^21 L)*(1025 g/L) = 1.322 x 10^21 g

(5) calculate the amount of D in grams in this mass of water. Note: by \"... , 0.030 0% of the mass is deuterium ..\" it is assumed that the percentage is % (w/w).

(1.322 x 10^21 g)*(0.030 0%/100%) = 3.966 x 10^17 g deuterium

(6) .. Isotope mass 2.01410178 u .

(7) calculate the number of moles of deuterium

(3.966 x 10^17 g deuterium)/(2.0141 g/mole) = 1.969 x 10^17 moles of D

(8) calculate the number of deuterium atoms in that number of moles



(1.969 x 10^17 moles of D)*(6.022 × 10^23 mol^-1) = 1.185 x 10^41 D atoms

(9) calculate the energy released by fusion of that number of deuterium atoms

(1.185 x 10^41 D atoms)*(17.59 MeV/D atom) = 2.085 x 10^42 MeV

(10) convert this energy in MeV to Joules

(2.085 x 10^42 MeV)*(1.6021x 10^-13 joule/MeV) = 3.34 x 10^29 J

That\'s the <first answer> (!).

But we have a problem with this part of the problem:

\"(b) World power consumption is approximately 7* 10^12 W. If consumption were 90 times greater, how many years would the energy calculated in part (a) last?\"

We can\'t do a comparison here without a time base (1 joule = 2.778 x 10^-10 megawatt-hour).

I will assume that \"World power consumption\" was supposed to be \"Annual World power consumption\" and work it out using that.

(11) calculate the 90x power value

(7 x 10^12 W-yr)*(90) = 63 x 10^13 W-yr

(12) convert W-yr to W-hr:

(63 x 10^13 W-yr)*(8760 hour/yr) = 5.5 x 10^17 W-hr

(13) convert that to J

(5.5 x 10^17 W-hr)*(3600 Joule/W-hr) = 1.98 x 10^21 J

(14) divide the energy from the fusion of the deuterons by this number to obtain the number of years

(3.34 x 10^29 J )/(1.98 x 10^21 J) = 1.686 x 10^8 years