Thermodynamics, bullet melting ice problem

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Homework Help Overview

The problem involves an ice cube at melting temperature being struck by a bullet, with the objective of determining the bullet's speed based on energy transfer during the melting process. The subject area is thermodynamics, particularly focusing on energy conservation and phase changes.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the energy required to break and melt the ice, questioning the relationship between kinetic energy and the energy needed for phase change. There are considerations about the calculations leading to different bullet speed results and the implications of energy distribution.

Discussion Status

The discussion is ongoing, with participants offering insights into the calculations and questioning the problem's parameters. Some suggest that the problem's values may have been altered without updating the answer key, while others express uncertainty about the energy calculations related to breaking and melting the ice.

Contextual Notes

Participants note potential discrepancies in the problem's given values and calculations, including the possibility of missing factors in the kinetic energy formula. There is also mention of the energy distribution between breaking the ice and melting it, indicating a need for clarity on these assumptions.

Vitalius6189
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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.
 
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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.
 
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