Can Kinetic Energy be Substituted for Heat in Latent Heat Calculations?

Click For Summary
SUMMARY

The discussion centers on the substitution of kinetic energy for heat in latent heat calculations, specifically in the context of a copper block sliding on ice. The problem involves a 1.60 kg copper mass moving at 2.50 m/s, and the goal is to determine the mass of ice melted due to the copper's kinetic energy. Key assumptions include the mass of the ice being large enough that not all melts, no sublimation occurs, and all kinetic energy is converted to heat for melting the ice. These assumptions are crucial for accurately applying the Q=mc(delta)T equation in this scenario.

PREREQUISITES
  • Understanding of the Q=mc(delta)T equation in thermodynamics
  • Basic principles of kinetic energy and its conversion to heat
  • Knowledge of latent heat and phase changes in materials
  • Familiarity with assumptions in physics problem-solving
NEXT STEPS
  • Explore the concept of latent heat and its calculations in detail
  • Study the principles of energy conservation in thermodynamic systems
  • Learn about the effects of pressure on phase changes, particularly sublimation
  • Investigate real-world applications of kinetic energy conversion in thermal systems
USEFUL FOR

Students and educators in physics, particularly those focusing on thermodynamics and energy transfer, as well as anyone interested in practical applications of kinetic energy in heat calculations.

mateomy
Messages
305
Reaction score
0
I think I might have a hard time adequately explaining my issue, but here we go…


So everyone knows the Q=mc(delta)T equation, I have this problem that I am working on and it was driving me CRAZY! So I checked out the solutions manual after some time.

Basically the stated a mass of copper was sliding across a slab of ice (Ice, air, and copper were all at 0 degrees Celsius), it said the block eventually comes to rest due to friction. It only gave the mass of the copper (1.60kgs) and the speed (2.50 m/s) and wanted to know the mass of the ice that melted due to the sliding copper. It mentioned looking at the internal energy of the system; both ice and copper, and determining for each what the changes were.

Here's the solution to the first part of the problem…

ScreenShot2011-07-28at20715AM.png


The thing that I am not sure about is substituting the Kinetic value as the Q value as shown in the latent heat calculation. Can you do that? Can you just substitute any energy as the "Q" variable?

Does that make sense? I am on about 15 hours of studying and I think I might be brain-dead.

Thanks.
 
Physics news on Phys.org
You can, but only under certain assumptions.

It would seem necessary to assume the following in order to solve the problem as they have done:

- The mass of the ice slab is sufficiently large, such that not all of the ice melts.

- Sublimation (ice -> vapor) does not occur.

- All of the kinetic energy is transformed into heat that is directly or over time indirectly "given" to the ice to cause it to melt (ice -> liquid).

Almost always in physics you have to come up with the right assumptions, and you need to specify why they are good assumptions. It's hard to think of any problem in physics where you don't assume something. This is often the trickiest part of doing physics. Realizing what assumptions you need to make, and why you make them. In this example (your problem), it is wise and perhaps even adequate to assume that all of the kinetic energy is transferred to the surrounding ice and results in the melting of said ice. Although it would be possible to look further into the problem at hand, the results shouldn't deviate that much.

The above assumptions are somewhat good enough to get a rough idea of what happens, and they are valid assumptions as the problem doesn't state specifically the pressure of the surrounding air and other important factors (that would govern the process of sublimation).

I hope this helped! :)
 
Definitely helped, thank you.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K