Consider the following nuclear fusion reaction: Each fusion event releases appro
ID: 2040858 • 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 0.0105% of all the hydrogen atoms in the water are deuterium. Number A typical human body metabolizes energy from food at a rate of about 100.5 W, on average. How long, in days, would it take a human to metabolize the amount of energy calculated above? Number daysExplanation / Answer
we kow,
1 MeV = 1*10^6*1.6*10^-19 J
mass of deuterium present in 0.23 kg of water,
m = (0.0105/100)*0.23
= 2.415*10^-5 kg
no of deuterium molecules present in the samplle of water,
N = m/mass of deuterium necleus
= 2.415*10^-5/(3.3435*10^-27)
= 7.22*10^21
no of ractions take place = 7.22*10^21/2
= 3.61*10^21
energy released, E = 3.61*10^21*4.03*10^6*1.6*10^-19 J
= 2.33*10^9 J <<<<<<<<---------------------Answer
Power = Energy/time
time = Energy/power
= 2.33*10^9/100.5
= 23184079 s
= 23184079/(24*60*60)
= 268 days <<<<<<<<---------------------Answer