Ice cube melting, solving for Initial Temp of Ice.

[Skeetss]

Homework Statement

.019 Kg icecube @ unknown temperature (In deg C)
Placed in 200 mL (.2 KG) water at 35 C.
After melting, water is at 22.4 C.
Find initial temperature of ice using conservation of energy.

Homework Equations

I've done the math a ton of times and keep getting -60 C which is completely unreasonable seeing as this was an incube taken out of a standard freezer.

I'm using the equation:

$-mC\Delta T_{water}=mC\Delta T_{ice}+mH_{F_{ice}}+mC\Delta T_{water(ice)}$

HF= Heat of fusion of water/ice

Plugging in for ΔT, I get:

$-mC(T_{F}-T_{I})=mC(T_{F}-T_{I})+mH_{F_{ice}}+mC(T_{F}-T_{I})$

$-.2(4180)(22.4-35)=.019(2080)(0-T_{I})+.019(3.34*10^{5}+.019(4180)(22.4-0)$

$T_{I}=-60°C$

That can't be correct.

Any help would be greatly appreciated. I'm sure it's my math somewhere.

 

[gneill]

I can't spot any errors in what you've done.

If this is data from an actual lab experiment can you identify other sources or sinks of heat energy that might have interfered? For example, what about the container? What was it made of? How was it insulated? How long did the mixture have to sit before taking the final temperature?

 

[technician]

Agree with the answer. I agree that it sounds extreme but it is not impossible !
It is a 'tiny' piece of ice compared to the mass of water and the temp change is 'quite large'

 

[Skeetss]

It was a styrofoam cup so it was insulated pretty well. My guess is my labmates misread or interfered with the final temp by accident. We were looking to find the initial temp of the ice and the lab question was to find the temperature of the physics department's freezer so who knows what went wrong. Thanks for the help.

 

[asteriatic]

Your approach is correct, but it seems like there may be an error in your calculations. The initial temperature of the ice can be found by equating the heat lost by the water to the heat gained by the ice and the heat of fusion of the ice. Using the equation you provided, we can set up the following equation:

-mCΔTwater = mCΔTice + mHfice

Where:
m = mass
C = specific heat capacity
ΔT = change in temperature
Hf = heat of fusion

Plugging in the known values:

-(0.2 kg)(4180 J/kg°C)(35°C - 22.4°C) = (0.019 kg)(2080 J/kg°C)(Tice - 0°C) + (0.019 kg)(3.34 x 10^5 J/kg)

Solving for Tice, we get:

Tice = 0.38°C

This means that the initial temperature of the ice was approximately 0.38°C. It is possible that there was a mistake in your calculations or in the values used. Double check your work and make sure to use the correct units for each variable.