# A 8.0×10^-2 kg ice cube at 0.0 degrees C is dropped into a Styrofoam cup

1. Feb 7, 2010

### ihearyourecho

1. The problem statement, all variables and given/known data
A 8.0×10^-2 kg ice cube at 0.0 degrees C is dropped into a Styrofoam cup holding 0.35 kg of water at 12 degrees C.

A) Find the final temperature of the system. Assume the cup and the surroundings can be ignored.

B) Find the amount of ice (if any) remaining.

C) Find the initial temperature of the water that would be enough to just barely melt all of the ice.

2. Relevant equations
Q=m*C*deltaT
Q=m*Lfusion

3. The attempt at a solution

Qwater=m*C*deltaT = .35kg * 4186J/kg*k *12k = 17581.2J
Qiceto0celcius=0 because it's already at 0 celcius
Q1=Mice*Lfusion = (8*10^-2kg)*(33.5*10^4 J/kg) = 26,800J

17581.2 J - 26800J = -9218.8 J

Q2=(Mwater+Mice)*C*deltaT
-9218.8 J = (.35kg + .08kg)*4186 J/kg*k * deltaT
deltaT=-5 degrees celcius

This answer is wrong, but I'm not sure where I went wrong. Any help would be great :)

2. Feb 8, 2010

### ihearyourecho

By the way, those are the steps my professor did with a very similar problem, so I don't know how he got the right answer and I didn't...

3. Feb 8, 2010

### mgb_phys

Almost
Remember you don't know the final temperature yet so you need to just work through the equations calling it T

eg eqn 1
Qwater=m*C*deltaT = .35kg * 4186J/kg*k * (T-12)k

And similarly the melted ice heats from 0 to T

Last edited: Feb 8, 2010
4. Feb 8, 2010

### willem2

You started to compute the amount of heat that was released when you cooled the water to 0C , then you computed the amount of heat needed to melt the ice. The second amount was bigger.
does this mean that
a. the water now cools to -5 to release more heat to melt all the ice
b. some of the ice remains frozen.

5. Feb 8, 2010

### ihearyourecho

I worked it through, thanks!