Given a change in internal energy, how can I find the temperature

Click For Summary
SUMMARY

The discussion focuses on calculating the final temperature and phase of two lead bullets after a head-on collision, where the change in kinetic energy is converted entirely into internal energy. The bullets, with masses of 15.0 g and 7.65 g, collide at speeds of 270 m/s and 390 m/s, respectively, resulting in a total change in internal energy of 1098 J. The specific heat of lead is 128 J/(kg K), and the melting point is 327.3°C. The initial temperature of both bullets is 30.0°C, and the calculations involve determining how much energy is used to raise the temperature and subsequently to melt the bullets.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Familiarity with specific heat capacity calculations
  • Knowledge of phase changes and latent heat concepts
  • Ability to apply the equations Q=mcΔT and Q=LΔm
NEXT STEPS
  • Calculate the energy required to raise the temperature of lead to its melting point using Q=mcΔT
  • Determine the energy needed for the phase change from solid to liquid using Q=LΔm
  • Explore the implications of energy conservation in inelastic collisions
  • Investigate the thermal properties of lead and other materials for comparative analysis
USEFUL FOR

Students studying physics, particularly those focused on thermodynamics and energy conservation, as well as educators looking for practical examples of inelastic collisions and phase changes.

mp0295
Messages
1
Reaction score
0

Homework Statement


Two speeding lead bullets, one of mass 15.0 g moving to the right at 270 m/s and one of mass 7.65 g moving to the left at 390 m/s, collide head-on, and all the material sticks together. Both bullets are originally at temperature 30.0°C. Assume the change in kinetic energy of the system appears entirely as increased internal energy. We would like to determine the temperature and phase of the bullets after the collision. (Lead has a specific heat of 128 J/(kg K), a melting point of 327.3°C, and a latent heat of fusion of 2.45 104 J/kg.)

(e) What is the temperature of the combined bullets after the collision?

(f) What is the phase of the combined bullets after the collision?
mbullet,solid= ____ g
mbullet,liquid= ____ g


Homework Equations



I have already correctly found the change in internal energy to be 1098J.


Q=mc\DeltaT
Q=L\Deltam

The Attempt at a Solution



I really don't know how to do this.

I tried finding the heat required to raise the temperature from 30C to the melting point using Q=mc\DeltaT => Q=.0226kg*128J/(kgK)*297.3K and found this to be 860J but I don't know where to go from there.
 
Physics news on Phys.org
Imagine this was ice instead - you start with ice at -30degC and you heat it ... the ice warms up to 0degC right? Then what?

Same with the bullets - so 860J out of the 1098J goes to warming the bullets until they are ready to melt.
How much energy goes into melting the bullets?
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
2K