00:01
All right, question 26 states when one kilogram of coal is burned, approximately 3 .0 times 10 to the 7 joules of energy is released.
00:10
If the energy released during each fusion of uranium 235 is 2 .0 times 10 to the 2, mev, again, per fission, so i'm calling that per n, or per number of decays, whichever you choose.
00:23
The question does ask, how many kilograms of coal must be burned to produce the same energy as one kilogram of uranium 235.
00:33
Okay, so we're told a couple of things here.
00:37
If we have a mass of one kilogram of uranium, we know that the energy released per mass of coal.
00:45
So i guess maybe i'll clarify here.
00:46
This is for coal.
00:49
If the energy released during a uranium fission, so this is this q of renan's energy per uranium, we need to know how many kilograms of coal.
00:58
So when you're looking for mass, i'll call this of coal is what we're trying to solve for.
01:05
But we're assuming that we want to know how much energy, well, how many kilograms of coal must be burned to produce the same energy as this expression here.
01:19
Right.
01:19
So we need to know a couple of things about what we're dealing with.
01:23
The first of which, we want to find out how much energy is burned when we burn one kilogram of coal.
01:34
So if the energy burned for this, oh sorry, for uranium, uranium, that does deal with uranium, right? how much energy do we burn when we burn one kilogram of uranium? well, we can simply find that to be the charge per, not the charge, this is not charge anymore.
01:51
Excuse me, is the energy per fission multiplied by the number of fission that will occur with one kilogram of uranium.
02:01
So if we know the rate per fission 2 .0 times 10 to the 2 mab per fission reaction, well, how many fissions occur? i'm going to do this on the side here.
02:17
Well, we know we're dealing with one kilogram of uranium.
02:24
So the number of fission that would occur, we think back to our chemistry and do with these elements.
02:30
So if we're given a specific mass, the number of atoms, presence or fissions would be the mass over the, which i'm calling capital m, the molar mass, multiplied by avogadra's number, should give us the number of atoms that we're dealing with.
02:48
So i'm going to substitute this in.
02:49
We'll check our units as we go.
02:52
So we have 1 ,000 grams, right, for one kilogram.
02:55
That's our m term here.
02:57
So just, just from obvious, this term is being substituted in here.
03:06
So we have 1 ,000 grams.
03:09
Our molar mass will take to be what we have.
03:12
For uranium 235 .0439...