Phase change Melting ice with hot tea.

[lwaiyipl]

Homework Statement

175 cm3 of hot tea at 87°C are poured into a very thin paper cup with 80 g of crushed ice at 0°C. Calculate the final temperature of the "ice tea". (Hint: think about two processes: melting the ice into liquid and, maybe, warming the liquid.)

Homework Equations

Q=mc(deltaT)
Q=mL

The Attempt at a Solution

First, I calculated the heat loss by the tea:
Q1=(.0175)(4186)(87)=63731.9J

Then the heat needed to melt the ice:
Q2=(0.08)(33.5e4)=26800J

Heat gain by the melted ice-water
Q3=(.08)(4186)(87)=29134.6J

Then Q(final)= Q1-Q2-Q3
=7797.34J

Set it equal to the entire system
7797.34J=(0.255)(4186)(T)

My answer is 7.3 deg Celsius, which is wrong. I can't seem to find what I did wrong!
Please help! Thanks.

 

[rl.bhat]

Hi lwaiyipl, welcome to PF.
Since Q1> Q2, temperature of the tea does not reach zero degree. So rewrite the first and third equation taking θ as the final temperature of the mixture.

 

[lwaiyipl]

Hey rl.bhat, thanks for the reply. But I still don't quite get it. If I make the final temperature as a variable, θ, then I will get θ=102.437 deg Celsius, which is hotter than the original 87 deg.
ARRGH, I am pulling my hairs out for this problem!

 

[rl.bhat]

In Q1 temperature will be (87- θ)
In Q3 temperature will be θ and mass of the mixture will be ...?

 

[lwaiyipl]

I see that the work I showed at the first post is incorrect, it's 0.175 instead of 0.0175.
So
Q1=(0.175)(4186)(87-θ)

Q2=(0.08)(33.5e4)

Q3=(0.175+0.08)(4186)(θ-0)

Am I setting it up correctly?
I have a question about Q3, why do we add the tea with the melted ice-water? Aren't we calculating for the amount of heat needed to raise the melted ice-water from 0 deg to θ?

 

[rl.bhat]

Yes. You are right.

 

[lwaiyipl]

Thanks rl.bhat. Sorry about late reply. I got the answer now!