Homework Help: Final temperature of system

1. Nov 28, 2007

Nghi

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

A large punch bowl holds 3.95 kg of lemonade (which is essentially water) at 20 C. A 1.74 kg ice cube at -10.2 C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat change with the bowl of surroundings.

2. Relevant equations

Q = mCT
Q = mLf (Lf = Latent heat of fusion)
Cwater = 4186 J/(kg x K)
Cice = 2090 J/(kg x K)
Lf = 33.5e4 J/kg

3. The attempt at a solution

To find the final temperature, I figured out the amount of heat it would take to bring water from 20.0 C to 0.00 C. It's basically plug and chug.

Qwater = mCT
= (3.95)(4186)(20)
= 330,694 J

Then I found the amount of heat it would take to bring ice from -10.2 C to 0 C. I also calculated the amount of heat it wold take to melt the ice (using latent heat).

Qice = mCT
= (1.74)(2090)(10.2)
= 37,093.32 J

Qice = mLf
= (1.74)(33.5e4)
= 582,900 J

Qice total = 619,993.32 J

Then I subtracted Qice from Qwater, and here was what I got:

Q = Qw - Qi
= 330,694 - 619,993.32
= -289,299.32 J

Since the leftover heat is negative, it means that ice is left over. So that would mean that the final temperature of water would be 0.00 C. I don't know if the thinking process is correct (Can someone help me verify this?), but I do know that the final answer is correct.

The problem I'm having is the amount of ice remaining. Since delta Q is negative, it would mean that ice would INCREASE, right? But I don't know how to solve it. :/

2. Nov 28, 2007

Galileo's Ghost

Remember the Q lost by the cooling water must be equal to the Q gained by the warming ice + the Q that melts the ice.

So the difference between the mcT(water) and mcT (ice) would give you the amount of heat still available to melt the 0 degree ice. Use that to find the mass of the ice that melted which will lead to the amount that still remains.

3. Nov 28, 2007

Nghi

Thank you so much! :)