Consider the following nuclear fusion reaction Each fusion event releases approx
ID: 1493957 • 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 0.250-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 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
no. of moles in 0.25kg (250g) of water (molar mass = 18) = (250/18) moles
No. of hydrogen atom in 250/18 moles of water = 2*(250/18)*avagadro number = (500/18)*6.022*1023
= 1.673*1025 H atoms
No. of deuterium = (0.0105/100)*1.673*1025 = 1.75665*1021
Two deuterim atoms give one reaction, hence no. of reactions = No. of deuterium/2 = 0.878325*1021
Energy released = 0.878325*1021 * 4.03 MeV = (0.878325*1021)*(4.03*106)*(1.6*10-19) J = 5.66*108 J
b)
Power = 108.5 W = 108.5 J/s
Time = (5.66*108 J) / (108.5 J/s) = 5216580 seconds = 60.4 days