Understanding Heat Transfer in a Ice-Water System

AI Thread Summary
The discussion focuses on calculating the final temperature and the amount of ice melted when an 18g ice cube at -4.0ºC is placed in 75g of water at 10ºC. The relevant equations for heat transfer, including Q=MCΔT and the latent heat of fusion for ice, are provided. The user attempts to raise the ice temperature to 0ºC and calculates the energy required, concluding that the water may not have enough heat to melt all the ice. They derive a final temperature of approximately 0.483ºC for the water after accounting for the heat exchange. The calculations suggest that the initial assumptions about energy transfer and phase change need further verification.
KylieB
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Homework Statement



An 18g ice cube at -4.0ºC is placed into 75g of water at 10 ºC in an
insulated container.
a. What is the final temperature of the system?
b. How much of the ice melts?

I have a sheet full of these & ice to water to steam I really need somone to explain how to find the results so I can get my brain around these

Homework Equations


Q=MCΔT
C ice =2.09j/g°c
Cwater = 4.186j/g°c
M=Mass in g
T = temp in Degrees C

Q=ML
M=Mass in g
Lice =334j/g

The Attempt at a Solution


So here's what i think is right so far
Raise the temp of ice to 0
18 x 2.09 x 4 = 150 joules
Remove this energy from the water
150.48=75 x 4.186 x(10-Tf)is this part right?
150.48= 3139.5-313.95(Tf)
I am not sure how to get to the resultant temp of the water propery before I consider a phase change in the ice but I have a feeling there won't be enough heat energy in the water to melt all of the ice.
 
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So here's what i think is right so far
Raise the temp of ice to 0
18 x 2.09 x 4 = 150 joules
Remove this energy from the water
150.48=75 x 4.186 x(10-Tf)is this part right?
150.48= 3139.5-313.95(Tf)
I am not sure how to get to the resultant temp of the water propery before I consider a phase change in the ice but I have a feeling there won't be enough heat energy in the water to melt all of the ice
 
or to remove the energy is it simply 150.48 = 75 x 4.186 x Tf
150.48 = 315.95 (T?)
T? = 150.48/313.95
T? = 0.483degrees C
then remove this from the initial temp of water
10 - 0.483 = 9.517 Deg C is this right so far?
 
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