# Two Cans & String

ForumTechnical Corner ► Physics Questions

Also, a quick explanation for how you got your answer(s) would be greatly appreciated.

Yes, this is high school physics. Don't hate me please. So you need us to do your homework for you.

One problem he's not sure how to start is pretty different from "here can you guys do this entire packet for me?" Not to mention the fact that the answers and explanations for all previous AP exams are available online (which Arthur should look at if he hasn't yet). I think this is a reasonable request. If he just wanted an answer, he could google it. He wants a different explanation so he understands it, though, which is admirable. I don't remember enough about this kind of physics to help, though :/

Oh.

My apologies.

A) Mass defect is the difference in mass on the two sides of the equations. Plug in numbers, multiply, divide.

B) Energy released must be enough to balance the mass defect. 1u is ~931 MeV/c^2, so you can use this to find the energy in MeV where c is the speed of light in vacuum. The convert to Joules from MeV.

C) Once you do B, you will just multiply and divide.

D) Again, just multiply and divide with the information from B and C. You would do well to see what the actual atomic mass of oxygen is.

If you would like to see the actual process, I am happy to go through it later.

The final answer seems way too high, but I'll show how I would have done this. (It's been, like, 8 years since physics class.) You probably also could reduce the number of steps in part D.

a) The mass defect is the amount of mass lost in the equation. Subtract the mass of the products from the mass of the reagents.
3 * 2.0141u - ( 4.0026u + 1.0078u + 1.0087u ) = 0.0232u

b) E = mc2
0.0232u * (299,792,458 m/s)2 = 3.5 * 10-12J

c) Divide energy required by energy-per-reaction, multiply by number of deuterium per reaction.
1020 / 3.5 * 10-12 * 3 = 9 * 1031

d) We are apparently assuming that seawater is 100% H20, and the Hydrogen is only either regular-type or deuterium.

The average molecule of H20 weighs:
(2.0141u * 0.00015 + 1.0078u * .99985) * 2 + 16u = 18.0159u
The contribution from deuterium is negligible, so you could probably leave it out.

The number of H20 molecules in a kilogram is then 1kg / 18.0159u, or 3.3427 * 1025, and the number of deuterium is that * 0.015%, or 5.014 * 1021.

The number of kilograms needed is (deuterium needed / deuterium per kg).
9 * 1031 / 5.014 * 1021 = 1.795×1010kg

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Almost 10 years later and I'm soon set to defend my doctoral dissertation in physics. Oh how time flies. I miss high school.

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