Consider the following nuclear fusion reaction: Each fusion event releases appro
ID: 2305518 • 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 0.230-kg glass of water were to undergo this fusion reaction? Assume that approximately 0.0135% of all the hydrogen atoms in the water are deuterium. Number A typical human body metabolizes energy from food at a rate of about 108.5 W, on average. How long, in days, would it take a human to metabolize the amount of energy calculated above? Number daysExplanation / Answer
number of water molecules in 0.230 kg water,
n = (0.230) / (18 x 1.66 x 10^-27)
n = 7.697 x 10^24 molecules
number of reaction that can happen,
= (7.697 x 10^24)(0.0135/100)
= 1.0392 x 10^21 reaction
total energy released = (1.0392 x 10^21)(4.03 MeV)
= 4.1878 x 10^21 MeV
= 6.7 x 10^8 J .......Ans
Q= P t
t = (6.7 x 10^8 J) / 108.5J/s
t = 6.18x 10^6 sec
t = 71.5 days .....Ans