How much ice melts when a projectile is fired at it?

  • Thread starter Thread starter Walley1
  • Start date Start date
  • Tags Tags
    Ice Projectile
Click For Summary
SUMMARY

The discussion centers on calculating the mass of ice melted when a projectile of mass 0.025 kg, traveling at 240 m/s, is fired into a block of ice at 0 degrees C. The kinetic energy (KE) of the projectile is calculated using the formula e=(mv^2)/2, resulting in 720 J. This energy is then divided by the heat of fusion of ice, 3.35 x 10^5 J/kg, yielding a mass of approximately 0.002 kg of ice melted. The calculations are confirmed as correct, with emphasis on the omission of rotational kinetic energy.

PREREQUISITES
  • Understanding of kinetic energy calculations
  • Knowledge of the heat of fusion of water
  • Familiarity with basic thermodynamics principles
  • Ability to perform unit conversions in physics
NEXT STEPS
  • Research the concept of kinetic energy in projectile motion
  • Study the heat of fusion and its implications in phase changes
  • Explore the effects of temperature on the melting process
  • Learn about energy conservation in thermodynamic systems
USEFUL FOR

Students in physics, particularly those studying thermodynamics and energy transfer, as well as educators seeking to clarify concepts related to kinetic energy and phase changes in materials.

Walley1
Messages
10
Reaction score
0

Homework Statement


Newt needs to melt some ice, so he fires a test projectile (0.025 kg) at 30.0 degrees C at a speed of 240 m/s into a large block of ice at 0 degrees C, in which it becomes embedded (comes to rest). Approximately what mass of ice melts? (i.e. ignore the heat lost by the projectile in going from 30-0 degrees C).


Homework Equations


e=(mv^2)/2


The Attempt at a Solution


The heat of fusion of water is 3.35x10^5J.
e=(0.025 * 240^2)/2=720J
720J/3.35x10^5J=0.002, so 0.002kg of ice melts.
I get the feeling that's incorrect, especially since I didn't do anything with the temperature (although it mentions that I should ignore the heat lost by the projectile).

Thanks for any help you can provide.
 
Physics news on Phys.org
Energy release of the projectile is its kinetic energy. So assuming the ice is at 0 C, the amount that melts equals the projectile KE divided by the latent heat of the ice. The rotational kinetic energy of the projectile is omitted. This is what you have done and it is correct. I assume the heat of fusion of ice is 3.35X10^5 J/kg. You left off the kg.
 
LawrenceC said:
Energy release of the projectile is its kinetic energy. So assuming the ice is at 0 C, the amount that melts equals the projectile KE divided by the latent heat of the ice. The rotational kinetic energy of the projectile is omitted. This is what you have done and it is correct. I assume the heat of fusion of ice is 3.35X10^5 J/kg. You left off the kg.

Thanks a million!
 

Similar threads

Replies
11
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K