What Determines the Final Temperature of Ice and Water Mix?

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Homework Help Overview

The discussion revolves around a problem involving the thermal interaction between an ice cube and water, specifically determining the final temperature of a mixture consisting of 81 g of ice at 0°C and 878 g of water at 25°C. Participants are exploring the principles of heat transfer and phase changes, particularly focusing on the latent heat of fusion for ice.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants initially attempt to apply the heat transfer equation without considering the latent heat of fusion. Questions arise regarding the melting of ice and the energy required for this phase change. There is a discussion on whether the ice will completely melt and how this affects the final temperature calculation.

Discussion Status

The discussion is ongoing, with various attempts to incorporate the latent heat of fusion into the calculations. Some participants suggest adjustments to the equations based on the state of the ice and the water, while others are verifying the correctness of their calculations. There is no explicit consensus yet on the final temperature.

Contextual Notes

Participants are working under the assumption that no heat is lost to the surroundings, and they are questioning the initial setup and definitions used in their calculations. The complexity of the problem is acknowledged, particularly in relation to the melting of ice and the subsequent heat exchange.

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Homework Statement


A 81-g ice cube at 0°C is placed in 878 g of water at 25°C. What is the final temperature of the mixture?


Homework Equations


No heat is loss therefore m*Cp*deltaT+m*Cp*deltaT=0
Cp ice=2.09
Cp water=4.19


The Attempt at a Solution


878*4.19*(Tf-25)+81*2.09*(Tf-0)=0
3678.82Tf+169.29Tf=91970.5
Tf=23.9C

The answer is wrong...don't know why...
 
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What about the latent heat of fusion for ice?
 
So you think you'll have ice at 23 degrees?
The ice will melt (at least partially). This will take energy (heat) from the water so the final temperature will be lower.
You have to take into account the melting heat Q=m_ice*l_melting
where m_ice is the mass of ice that melts (may be equal to the initial mass) and l_melting is the latent heat (heat of fusion) for ice.

In order to see if the ice melts completely (or not) you can compare the heat necessary to melt all th e ice with the heat released by the water when it cools to zero degrees.
 
Qmelt=81g*333j/g
=26973
So...
878*4.19*(Tf-25)+81*2.09*(Tf-0)+26973=0
3848.11Tf=64997.5
Tf=16.89

Which is still wrong...what am I doing wrong now?
 
You don't have ice heating up. It's already at 0 degrees. There is no term like 81*2.09*(Tf-25)
It's the water from melting that will heat up.
 
So then it should be...

878*4.19*(Tf-25)+26973=0
Tf=17.668

That's not correct either...
 
After melting completely, the ice water takes heat from water to reach the final temperature.
 
Ok...if that is so...then...
81g*4.190j/kg*Tf + 878g*4.19j/kg*Tf -91970.5 + 81*333 = 0
so Tf = 16.1757
 
Appears to be correct.
 

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