releases 3.27 MeV of energy. Assume that a fusion reactor operates strictly on t
ID: 1379785 • Letter: R
Question
releases 3.27 MeV of energy. Assume that a fusion reactor operates strictly on the basis of this reaction and answer the following.
J
(a) How much energy could it produce by completely reacting 8 kg of deuterium?
J
(b) At seven cents a kilowatt-hour, how much would the produced energy be worth?
$
(c) Heavy water (D2O) costs about $299 per kilogram. Neglecting the cost of separating the deuterium from the oxygen via electrolysis, how much does 8 kg of deuterium cost, if derived from D2O?
$
(d) Would it be cost-effective to use deuterium as a source of energy? Discuss, assuming the cost of energy production is nine-tenths the value of energy produced.
Explanation / Answer
A.
By reaction of 4*1.67*10^-27kg deutrium 3.27Mev energy is releasedso energy released by reaction of 8kg deutrium is
E = (3.27*10^6*1.6*10^-19)*8/ (4*1.67*10^-27) J
E = 6.27*10^14 J
B
7cents for 1Kilowatt hour = 1000J
total money = 7*(6.27*10^14)/1000 = 43.89*10^11cents
C.
mass of deutrium per kilogram D2O is = (4/16)Kg = 025
water required for 8kg deutrium = 8/0.25 =32kg
cost = 32*299=9568dollars
D.
It is cost effective
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.