What Determines the Final Temperature of Ice and Water Mix?

[ymehuuh]

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...

 

[LowlyPion]

What about the latent heat of fusion for ice?

 

[nasu]

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 teh 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.

 

[ymehuuh]

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?

 

[nasu]

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.

 

[ymehuuh]

So then it should be...

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

That's not correct either...

 

[rl.bhat]

After melting completely, the ice water takes heat from water to reach the final temperature.

 

[ymehuuh]

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

 

[rl.bhat]

Appears to be correct.