Thermodynamics, bullet melting ice problem

AI Thread Summary
The discussion revolves around a thermodynamics problem involving a bullet striking an ice cube, where the bullet's kinetic energy is partially used to break and melt the ice. The key calculations indicate that 6700 J is required to melt the 20 g ice cube, which corresponds to two-thirds of the bullet's initial kinetic energy. The calculated bullet speed is 1494 m/s, but the expected answer is approximately 1060 m/s, leading to confusion about the energy distribution. Participants suggest that the problem's values may have been altered without updating the answer key, and they emphasize that energy used to break the ice converts to heat rather than disappearing. Overall, the calculations and reasoning presented seem sound, but discrepancies in expected outcomes raise questions about the problem's integrity.
Vitalius6189
Messages
9
Reaction score
1

Homework Statement


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.

Homework Equations


q=1/2mv^2

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.
 
Physics news on Phys.org
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.
 
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.
 
Thank you for help.
gneill said:
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.
 
gneill said:
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.
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.
 
  • Like
Likes gneill
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top