# Thermodynamics, bullet melting ice problem

Tags:
1. Jul 3, 2017

### Vitalius6189

1. The problem statement, all variables and given/known data
An ice cube at the melting temperature that has a mass of 20 g, is struck by a bullet with
a mass of 9 g, flying at a certain speed. Determine the speed of the bullet, if it is known that
one third of his energy was consumed to break the ice, and the remainder to melt it.
Latent ice melting heat λt= 335 kJ / kg.

2. Relevant equations
q=1/2mv^2

3. The attempt at a solution
335 kJ/kg means 335 J/g

Heat = 335 J/g * 20 g = 6700 J

This was 2/3 of the starting kinetic energy of the bullet, the 1/3 is spent on chunks flying everywhere.
6700 J = 0.5(0.009 kg)(v^2) * 2/3
v = 1494 m/s

The problem is that the answer is supossed to be aproximately 1060 m/s but i can't get it.
I feel like i'm missing something but don't know what.

Any help will be appreciated.

2. Jul 3, 2017

### Staff: Mentor

It is a bit odd to have so much energy to break the ice, as melting is a similar process just for all remaining atoms. Unless your pieces of ice fly away at high speed, the end result is the same, 20 grams of water.
Anyway, I don't see how you could get 1060 m/s. That would correspond to a kinetic energy of 5056 J, not even enough to melt the ice.

5056 J * 4/3 = 6741 J is about the energy needed to melt the ice, but that calculation doesn't make sense.

3. Jul 3, 2017

### Staff: Mentor

Your reasoning and results look okay to me. Perhaps the problem's given values were modified at some point (to make it a "new" question) but the answer key was not updated to reflect the changes.

4. Jul 3, 2017

### Vitalius6189

Thank you for help.

5. Jul 3, 2017

### haruspex

Or they forgot the ½ in ½mv2.
And I agree with mfb that the energy that goes into breaking the ice doesn't then disappear; most of it would end up as heat.