What is the final temperature of the contents of the container?

Click For Summary
SUMMARY

The final temperature of the system, consisting of a 50-gram piece of ice at -20°C and 250 grams of water at 25°C, is calculated to be 5.81°C. The calculations involve the specific heat capacities of ice and water, as well as the latent heat of fusion for ice. The correct approach requires considering the sensible heat gain of the ice as it warms to 0°C, the latent heat required for melting, and the heat lost by the water as it cools to the final temperature. Miscalculations often arise from incorrect assumptions about temperature ranges and specific heat values.

PREREQUISITES
  • Understanding of specific heat capacity and latent heat concepts
  • Familiarity with heat transfer calculations in thermodynamics
  • Knowledge of temperature scales and unit conversions
  • Ability to set up and solve equations involving heat exchange
NEXT STEPS
  • Review the principles of heat transfer in thermodynamics
  • Learn about the calculation of final equilibrium temperatures in mixed systems
  • Study the specific heat capacities of various substances
  • Practice problems involving phase changes and latent heat calculations
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone preparing for exams involving heat transfer and phase changes in materials.

pkossak
Messages
52
Reaction score
0
A 50 gram piece of ice at -20 C is dropped into an insulated container holding 250 grams of water at 25 C. What is the final temperature of the contents of the container?

I can't seem to get this problem. I was doing the following:

Lf water = 333 J/g
c water = 4186 J/kg*C
c ice at -10 C = 2220 J/kg C

.05 kg*2220 J/kg*C * (-10 + 20) (should I use this step?) + .05 kg*4186 J/kg*C * Tf + 10) + 50 g*333 J/g = - .25 kg*4186 J/kg C * (Tf - 25)

And I keep getting it wrong. Any suggestions? Thanks a ton
 
Physics news on Phys.org
Anyone? thanks
 
Units of temperature?
 
To start with, I'm going to assume that the ice will melt.

First, how much heat is required to bring the ice up to 0 degrees C? Giving up that much heat will reduce the temperature of the water to what? Now, how much heat does it take to melt the ice? Giving up that much heat will reduce the temperature of the water to what? (If 0 or below you're done- there will be a mixture of ice and water at 0 degrees.) Finally, determine the temperature that will give you an equilibrium for both the orginal water and the water from the ice.

IF giving up enough heat to bring the ice up to 0 degrees would reduce the temperature of the water below 0, the ice will not all melt. You would need to calculate how much heat will be given up when the water freezes and then find the equilibrium temperature (below zero) for the ice. Again, if the heat given up when the water freezes is more than enough to reduce the temperature of the ice to 0, you will have a mixture of water and ice at 0 degrees.
 
Thank you very much. I think I followed what you said pretty well, but for some reason my answer is about 1 or 2 C off. What did I do wrong? Thanks a lot.

(2220 J/kgC*0.05 kg*(0 + 20 C)) = -(4186 J/kgC*0.25 kg*(Tf - 25 C))
Tf = 24.03 C

(4186 J/kgC*0.05*(Tf - 0)) + (333 J/g*50 g) = -(4186 J/kgC*0.25 kg*(Tf - 24.03 C))
Tf = 6.767 C

The answer is actually 5.81 C, I can't tell what I've done wrong! Thanks a lot for any help...
 
I don't mean to keep begging, but if anyone could take a look and tell me if they see anything I'm missing i am really appreciative. My final is tomorrow.
 
The LHS of the equation should include sensible heat gain by ice from -20C to 0C, (Don't get misguided by the specific heat at -10C. This is to give you an average specific heat over a range of -20 to 0C.) latent heat gain from 0C ice to 0C water and finally sensible heat gain from 0C water to a final temperature Tf. The RHS is the heat lost by water from 25C to Tf. Apply this and you will get your answer.
 
Thank you! I don't see why the way I was doing it doesn't produce the same answer, but it's definitely working right now. Thanks a ton.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
Replies
12
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
5K