Consider the nuclear fusion reaction Each fusion event releases approximately 4.

Question

Consider the 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 0.250 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. energy released: 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 released? time to metabolize the amount of energy released: days

Explanation / Answer

(a)

Number of moles of H2O molecules = 250 / 18= 13.88 moles

number of molecules = 13.88*6.02*10^23 = 83.6*10^23

number of H atoms = 2*83.6*10^23 = 167.22*10^23

Total number of deuterium atoms in reaction,

N = (0.0135 / 100) *  167.22*10^23

N = 2.25*10^21

Energy realesed,

E = (2.25*10^21 / 2) * 4.03*10^6*1.6*10^(-19)

E = 7.27*10^8 J

(b)

Power P = E / t

so, time t = E / P

t = 7.27*10^8 / 96.5 = 7.53*10^6 s

t =  7.53*10^6 / 3600*24

t = 87 days

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