Consider the following nuclear fusion reaction: Each fusion event releases appro
ID: 1309644 • Letter: C
Question
Consider the following nuclear fusion reaction: Each fusion event releases approximately 4.03 MeV of energy. How much total energy, in joules, would be released if all the deuterium atoms in a typical 0.290-kg glass of water were to undergo this fusion reaction? Assume that approximately 0.0115% of all the hydrogen atoms in the water are deuterium. Number 6.159 x 10 11 J A typical human body metabolizes energy from food at a rate of about 104.5 W, on average. How long, in days, would it take a human to metabolize the amount of energy calculated above? Number daysExplanation / Answer
In the above reaction,
2 atoms of deuterium fuse together to form 1 tritium and 1 hydrogen producing 4.03Mev energy
Given data,
IN 0.290 kg glass of water ,
0.015% is deuterium
ie; 0.015/100 * 290 = 0.0435g of deuterium.
In the above reaction, 2 atoms of deuterium fused for the reaction.
mass of an atom of deuterium is 3.34*10-24 gm
mass available is 0.0435g
energy liberated is 0.0435/3.34*10-24 *4.03*1.6*10-19J = 8397.844J
2.Power consumption on a daily basis = 104.5W
energy available = 8397.844J
Time = 8397.844/104.5 = 80.36 Days