# Homework Help: Mixing Ice & Water help me Obe Wan

1. Mar 3, 2013

### KylieB

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. The attempt at a solution
So heres 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 wont be enough heat energy in the water to melt all of the ice.

Last edited: Mar 3, 2013
2. Mar 3, 2013

### KylieB

So heres 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 wont be enough heat energy in the water to melt all of the ice

3. Mar 3, 2013

### KylieB

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?