Consider the following nuclear fusion reaction: Each fusion event releases appro
ID: 1309643 • 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. 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?Explanation / Answer
I think your answer is wrong. in 0.290 kg of water there would be 16.11(290/18) moles of water.
Each mole of water contains 6.023x10^23 molecules of water.
So a glass of water will contain= 16.11x6.023x10^23 molecules of water
subsequently 1 molecule of water contains 2 atoms of Hydrogen.
So, a glass of water will contain 2x16.11x6.023x10^23 atoms of hydrogen.
Since it is given that 0.0115% of hydrogen atoms are deutrium.
therefore total no of deutrium atoms would be: 0.0115x2x16.11x6.023x10^21
2 atoms of deutrium produce 4.03 MeV of energy.
So a glass of water would produce: 4.03x0.0115x16.11x6.023x10^21 MeV of energy.
1 MeV = 1.6x10^(-13) J
therefore a glaas of water would produce: 1.6x4.03x0.0115x16.11x6.023x10^8 J of energy
= 7.2x10^8 J
2. 1W = 1J/s
therefore time required : (7.2*10^8)/104.5 s
= 79.69 days
Approximately 80 days