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Problem 44 Hard Difficulty

The fusion reaction $\frac{2}{1} \mathrm{D}+\frac{2}{1} \mathrm{D} \rightarrow_{2}^{3} \mathrm{He}+_{0}^{1} \mathrm{n}$ releases 3.27 MeV of energy. If a fusion reactor operates strictly on the basis of this reaction, (a) how much energy could it produce by completely reacting 1.00 $\mathrm{kg}$ of deuterium? (b) At eight cents a kilowatt-hour, how much would the produced energy be worth? (c) Heavy water $\left(\mathrm{D}_{2} \mathrm{O}\right)$ costs about $\$ 300$ . per kilogram. Neglecting the cost of separating the deuterium from the oxygen via electrolysis, how much does 1.00 $\mathrm{kg}$ of deuterium cost, if derived from $\mathrm{D}_{2} \mathrm{O}$ ? (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.

Answer

a. the number of deuterium atoms is $3.0 \times 10^{26}$ atoms.
b. The cost of energy per $\mathrm{kWh}$ is $\$ 0.08$ and hence the cost of total energy produced is
$\left(2.18 \times 10^{7} \mathrm{kWh}\right)\left(\frac{\mathrm{S} 0.08}{1 \mathrm{kWh}}\right)=\$ 1.74 \times 10^{6} .$
c. Therefore, the cost of one $\mathrm{kg}$ of deuterium is $\$ 1500$ .
d. It depends on the net energy produced and how much it is worth to fuse deuterium.

Discussion

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Meghan M.

McMaster University

Video Transcript

this question. You given that the fusion off duty in Liza's and energy off 3.27 MTV per reaction, you won't find what is the amount of energy that we can produce with one kilogram off deuterium Think so? First off in, you'll find how many duty rub they are in one kilogram. So in one kg, how many deuterium number off duty room you take one kg, you fight by d myself east, you tear that is to points to one atomic mass units cold weather into kilograms by multiplying by 1.66 times 10. Paul Minister is seven kilograms put you we get about tree time stamp artery six new clients. So this is the number of deuterium. Remember that the fusion off to tear and takes up actually to two new clips off deuterium. So the number of reactions that can occur it goes toe half the number of deuterium. And so the total energy released from one kilogram B number off reactions multiplied by 3.27 MTV. They're gonna come put this into actually juice to the richest E S I units, right for energy. By multiplying by 1.6. President, Home Minister. Teen juice per and TV. All right from this should get 67.85 times 10 power 13 jokes. Next to find what is the value off this amount off energy were given that the rate is zero point, you're eight dollars per kilowatt hour. Now to find out West Devalue, we'll first need to convert our energy e Toto into kiddo hours by multiplying by one of our over 0.6, the state power six juice. All right, so why kill? Ours is 3.6 time 10 power, six juice. So this the commission factor. And then from this we multiplied. I devalue off each kilowatt hour to get what he's devalue. Every street gets 1.74 Step six. So this is devalue off our energy to find out how much it costs to buy this. Ah, amount off deuterium from heavy water. So heavy water is simply the hydrogen off water replaced by two teary Um, in this case, the mess off duty. My defection off mess that is deuterium. It goes to fall over the entire, which is over 20 right? It's four is from each deuterium is to atomic mass units. So two deuterium gives us four atomic mass units over the total mess, which is four plus 16 16 ste. Mess off the oxygen. You, sir, spoke one or five. And so, in order to get one kg off deuterium, we will need five times the mess off heavy water. So the mess off ET all required will be five kilograms to get one kilogram off duty. So it cost off buying there will be five times 300. His argument is 300 per kilogram. And so that is 1500. All right, so would be cost effective if you were to just buy this amount off deuterium to gain this amount or failure off energy. Well, it depends, right? We still have in factor in the cost off extracting, extracting the deuterium from, oh, heavy water and the cost in itself could be simply the energy needed to supply in. Or it could be the cost off buying certain materials so than catalyst, you know, and so their students to be other factors accounting for it. Although the value is way higher than the cost due by the heavy water. So there is a lot off leeway to allow for this to be possible to be cost effective

National University of Singapore
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Meghan M.

McMaster University