Question

2013/2014 NEW YEARS CHALLENGE
By Frostbite
I write my solution the 1/2-2014

Answer 1

A New Years rocket is fuelled by TNT (they are not but let’s assume so) [TNT ís 2,4,6-trinitrotoluene with the formula C7H5(NO2)3] with a enthalpy of decomposition (explosion), ΔH = -1112 kJ/mol. Assume all energy is used to propel the rocket skywards. Further assume that there is 1 g TNT in the rocket and that the total rocket weight is 500g.

a) How far does it rise (you may assume no change in rocket weight during flight).
b) What is the velocity of the rocket when it hits ground again? Will it hurt, if hits you?

Happy New Year and good luck with all your studies!

Answer 2

nice question :) don't u think its also involved physics in it :)

Answer 3

It involves general physical chemistry :)

Answer 4

what challenge?

Answer 5

To solve the problem, I've set up.

Answer 6

Using the molecular formula, we've got 227 g/mol TNT and we have 1 gram of it and know that it's -1112 kJ/mol so we can pretty easily see that overall the amount of energy will be about 5 kJ/mol. I'm not really going to care about significant digits since there are about a bazillion approximations throughout the whole problem anyways.

Now we can see since kinetic energy is 1/2mv^2 we can use our kinetic energy from the TNT and the mass of the rocket to find the initial velocity. Then using gravity we can just use the regular old kinematic equations for translational motion to find the time it takes to reach the ground again and find the velocity at that time as well. Then depending on who you are, you can judge if you think it'll hurt... I'm sort of tired so I am not going to calculate the math out, maybe later.

Answer 7

Okay here is my solution and it is just as Kainui suggest:

The totally contribution of energy goes to the propel the rocket skywards, we therefor need to calculate the total amount of energy we get from the decomposition of 1 gram TNT.
The molar mass for TNT is\[\large M(\sf TNT)=227.13 ~ \frac{ g }{ mol }\]
We thereby get the amount of substance to
\[\large n(TNT)=\frac{ m(TNT) }{ M(TNT) }=\frac{ 1 ~ g }{ 227.13 \frac{ g }{ mol }} =0.0044 ~ \sf mol\]
The total energy from the decomposition of 1 gram TNT is then:
\[\large E=\Delta H \times n(TNT)=1112 \frac{ kJ }{ mol } \times 0.0044 ~ mol= 4896 ~ \sf J\]
We choose to assume the maximum in hight most be when the potential energy is at it's maximum, in other words when the kinetic energy contribution 0, mathematically we get
\[\Large E_{tot}=E_{pot}+E _{kin}=E _{pot}+0=m \times g \times h\]
Where \(m\) is the mass of the object of interest, \(g\) is the standard acceleration due to free fall and \(h\) is the height. To answer a) we get \(h\) to:
\[\large h=\frac{ E }{ mg }=\frac{ 4896 ~ J }{ 0.500 ~ kg \times 9.82 \frac{ m }{ s^{2} } }=997 ~ \sf m\]

So the answer to a) is about 1 km.

Now when it is about to go down, we want to find it's maximum velocity, which is when the potential energy is equal to 0.
\[\Large E_{tot}=E_{pot}+E _{kin}=0+E_{kin}=\frac{1}{2} \times m \times v^{2}\]
The velocity is then:
\[\Large v=\sqrt{\frac{ 2E }{ m }}=\sqrt{\frac{ 2 \times 4896 ~ J }{ 0.5 ~ kg }}=140 \frac{ m }{ s } \sim 500 \frac{ km }{ h }\]
And this is the answer to b)

Now does it hurt to get rocket in the head? One of the perhaps biggest approximations done is that we do not take account for the wind resistance which will slow the rocket down, on both it's way up and down.