Calculating Energy Released: TNT vs. U-235 | Nuclear Energy Homework

[wilson_chem90]

Homework Statement

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).

Homework Equations

E = mc^2

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...

 

[Pengwuino]

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

 

[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?

 

[wilson_chem90]

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

 

[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.

 

[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

 

[Pengwuino]

Yup, sounds like you got it.

 

[wilson_chem90]

nice! thank you for your help