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

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Homework Help Overview

The problem involves a thermal exchange scenario where a piece of ice at a low temperature is introduced into a warmer body of water within an insulated container. The objective is to determine the final temperature of the system, which includes both the melted ice and the water.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessary heat calculations for both the ice and the water, questioning the assumptions about the melting of ice and the temperature changes involved. Some participants explore the heat transfer equations and the specific heat capacities of ice and water.

Discussion Status

There are multiple interpretations of the heat transfer process, with some participants suggesting different approaches to calculating the final temperature. Guidance has been offered regarding the steps to consider, particularly in relation to the phases of the ice and water and the heat exchanges involved.

Contextual Notes

Participants note potential discrepancies in temperature calculations and the importance of using the correct specific heat values. There is also mention of the need to account for the latent heat of fusion when considering the melting of ice.

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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
 
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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.
 

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