Question

We know already that a typical household uses about 1000 kWh of energy in one month. Let's investigate the advantages of nuclear energy by considering the following problem: if you were given a gallon of water, how long could it supply fusion energy for your home? In this case, you are considering a fusion reaction between two deuterium nuclei which releases 4.0 MeV. We will follow this problem through to the end in stages. Part I: How many deuterium nuclei are in the gallon of water? 1) How many grams of water are in one gallon? 2) What is the mass of a single H;O (water) molecule? 3) How many individual molecules of H2O (water) are in the gallon? 4) If the fraction of deuterium is only 0.015 % of hydrogen in water, how many individual molecules of D,O (heavy water) are in the gallon? Part II: How much total energy can you get out of the gallon? s) How much energy (in MeV) is released when two deuterium nuclei make a fusion reaction? 6) How much energy (in MeV) is released when all of the deuterium nuclei in the gallon of water undergo fusion? Be careful of factors of 2 when you do this! 7) Convert this total energy from MeV into Joules, and then from Joules into kWh.

Answer 1

To calculate total energy obtainable from fusion of all deuterium nuclei in 1 gallon of water

Part I gallon a 3.78591 lt z 3.78541 x 103 m3 Mass of water in I gallon = (3.7854 1 x 10 / 1000 ( density 23.78541 kg of water zloookg/m² z 3785.41gm .23 Mass of a single H₂O molecule = 18:02 am u 1 - 24 2 18002 1.66 054 x 10 gm (lamu z 166054x10gm 2.2.99229 X 10^gm = 299* 299 X 10 gm (Answers) 3) No of individual molecules of H₂O in I gallon 2 3785.41_ 2.99 22 9X10- 23 2126505 X 1026. a 1.26 *1026 Answer ) 4) No of individual molecules of D2o in I gallon = 1.26505 x 1025 x 0.015 100 Z 1.89757 X 1022 ñ 1990 x1022 (Answer) Part II 3 Energy released in fusion reaction of 2 denterium . mudei = 4:0 MeV (Answer) .... 6) Energy szeleased when all of the denteri um nuclei in the gallon of water undergo fusion a number of pairs of denterium nuclei x 4.0 Mev 2 no of D₂0 moleculer x 4.0 Mev . 21.8.97.57. X 102X4027.59026 X 10 MeVx7.5 9X10 MeV. 22 Answer )

7. Total Total entry energy = 715902 8 8 10' 2(7:59028 x 1022)x (160218 x1017) ClMev. 21.60218X10 T 21216099 x 10' J 1:21 X 10' J Anawer) kWh 1 Total eneron 21.216099 X 10 orgy ²1 21 3.6 x 106 23.378 X109 kWh 10 3380 kWh (Answer) kWh: 36% J) E l