Solve Nuclear Physics Homework: How Many Grams of TNT Needed?

[Mitchtwitchita]

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 by one gram of U-235?

Homework Equations

Delta E = delta m * c^2

The Attempt at a Solution

Mass U = 234.993 u
Mass e- = 92(0.000549 u)
Mass nucleus = 234.993 u - 92(0.000549 u)
=234.942492 u

92 protons: 92(1.007276 u) = 92.669392 u
143 neutrons: 143(1.008665 u) = 144.239095 u
Total = 236.908487 u

Delta m = 236.908487 - 234.942492
=1.965995 u

Energy released:

Delta m = (1.965995 u)(1.6605 x 10^-27 kg/u)
=3.2645347 x 10^-27 kg

Delta E = delta m * c^2

Therefore, (3.2645347 x 10^-27 kg)(2.998 x 10^8 m/s)^2
=2.934 x 10^-10 J

Therefore, the difference in mass between 1 U-235 nucleus and its constituent nucleons is converted into 2.934 x 10^-10 J of energy.

Since 1.00 g of U-235 = 2.56 x 10^21 nuclei,

(2.934 x 10^-10 J)(2.56 x 10^21 nuclei)
=7.51 x 10^11 J
=7.51 x 10^8 kJ

Therefore, 7.51 x 10^8 kJ/2.760 kJ
=2.72 x 10^8

Therefore, 2.72 x 10^8 g of TNT would be needed to match the energy released by one gram of U-235.

I HAVE NO IDEA IF THIS IS RIGHT OR IF I'M ON THE RIGHT TRACK, COULD SOMEBODY PLEASE LET ME KNOW!

 

[denverdoc]

Not sure, it depends on how one interprets the question. What would help are the actual rxns being compared. I think a fairer comparison would be to take a fission reaction for Ur and compare the weights of the products vs the parents and use that in computing the harvestable energy. There is no way to completely pluck apart an atom so complex into its constituent parts.

 

[Mitchtwitchita]

From a previous question I attained a number of 2.82 x 10^-11 J of energy released from the fission of a U-235 nucleus.

Since 1.00 g of U-235 = 2.56 x 10^21 nuclei,

(2.82 x 10^-11 J)(2.56 x 10^21 nuclei)
=7.22 x 10^10 J
=7.22 x 10^7 kJ

Therefore, 7.22 x 10^7 kJ/2760 kJ = 2.6 x 10^7

Therefore, 2.6 x 10^7 g of TNT would be needed to match the energy released by one gram of U-235.

Does this seem like the route you were talking about?

 

[denverdoc]

Exactly.


(PS: I'm not sure if this was the latter part of a multipart question, but please all viewers, we need the entire problem to be of greatest assistance, including the pieces you have solved.)

 

[Mitchtwitchita]

Thanks immensely denverdoc. Usually they have part a) and part b) to the question. However, in this case, they were two entirely different questions. That's what made it so confusing to me as well. Thanks again!