Question

Consider the following nuclear fussion reaction:

2/1 H + 2/1 H -----> 3/1 H + 1/1 H

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 0.250-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 96.5 W, on average. How long, in days, would it take a human to metabolize the amount of energy calculated above?

Answer 1

a )

here given

0.25 kg of water ,

then we can get ( 250 / 18 ) = 13.88

the glass of water contains = 13.88 X 6.023 X 10²³ molecules of water

1 molecule H₂O having 2 atoms of H

so 2 X 13.88 X 6.023 X 10²³ molecules of water

the % given is 0.0115 %

so

0.0115 X 2 X 13.88 X 6.023 X 10²³

and 2 atoms have 4.03 MeV of energy

then 4.03 X 0.0115 X 2 X 13.88 X 6.023 X 10²¹ MeV of energy

but we have 1 MeV = 1.6 X 10^-13 J

then the glass of water will Produce,

1.6 X 10^-13 X 4.03 X 0.0115 X 2 X 13.88 X 6.023 X 10²¹ J of energy

= 12.3981 X 10⁸ J

b )

we have Watt = Joule /sec

so

the time is = 12.3981 X 10⁸ / 96.5

= 12847772.02 sec

12847772.02 sec = 148.701 days.