# Homework Help: Nuclear Energy Question

1. Oct 29, 2009

### wilson_chem90

1. The problem statement, all variables and given/known data
Nuclear energy changes are significantly greater than chemical changes. The detonation of 1.00 g of the explosive trinitrotoluene (TNT) releases 2.760 kJ. How many grams of TNT would be needed to match the energy released be one gram of U-235? (HINT: 1.00 g of U-235 contains 2.56 x 10^21 nuclei).

2. Relevant equations
E = mc^2

3. The attempt at a solution

This is all i can think of...

First, convert 1.00 g to 0.001 kg

then:

E for U-235 = mc^2
= 0.001 kg (2.998 x 10^8 m/s)^2
= 8.99 x 10^13 J
then: 1 kJ = 1000 J

Therefore: (8.99 x 10^13 J) /1000 J = 8.99 x 10^10 kJ

Then divide 8.99 x 10^10 kJ by 2.760 kJ to find the answer

3.26 x 10^10 g

Therefore, you would need 3.26 x 10^13 g to match the energy released by 1.00 g of U-235.

I'm almost positive this is incorrect...

2. Oct 29, 2009

### Pengwuino

You need to determine the reaction involved in U-235. The entire atom isn't converted into energy.

3. Oct 29, 2009

### wilson_chem90

okay, well there was a question before this one that was linked to this one, and it involved a U-235 reaction and the answer for the energy released was 4.79 x 10^-10 J. would that be it?

4. Oct 29, 2009

### wilson_chem90

so what do i do with the nuclei? do i divide them by the amount of energy during the reaction?

5. Oct 29, 2009

### Pengwuino

You need to be more specific on what you're doing. Is the energy release from I suppose the previous problem the same reaction you use for this problem? In that case, you multiply the energy/reaction by the number or reactions (or atoms) and you'll have some amount of energy. Then you can determine howmany grams of TNT would result in that same amount of energy release.

6. Oct 29, 2009

### wilson_chem90

well for example, the question number is 70, and the nuclear reaction is part a, and this question is part b. So i'm guessing they're linked together. And once i find the other energy amount, i just divide it by the TNT energy level to find the amount of grams in TNT

7. Oct 29, 2009

### Pengwuino

Yup, sounds like you got it.

8. Oct 29, 2009

### wilson_chem90

nice! thank you for your help