Consider the following nuclear fusion reaction: Each fusion event releases appro

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 days

Explanation / Answer

1.
0.290 kg water =290/18 =16.11 mole water that contains 16.11*2= 32.22 mole H2.
Now Deuterium in that is 32.22*0.0115/100=3.7*10-3 mole
So, Total energy released =3.7*10-3 /2*6.023*1023*(4.03*106*1.6*10-19)= 7.19*108 J

2. It will take  7.19*108/ 104.5 =6885650 s = 6885650 /24/60/60=79.69 =80 days approx.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---