Finding the final temperature of a final solution. putting ice in soda/water

[apbuiii]

Homework Statement

Suppose you have a 24 ounce mug of soda (treat as water) at room temperature, which is 22 C. Warm soda tastes bad, so you add 125 g of ice at -25 C to the soda. What is the temperature (in C) of the soda once all the ice melts and the solution reaches a uniform temperature? The heat of fusion is 6.01 kJ/mol and the heat of vaporization is 40.7 kJ/mol. The specific heat capacity of ice and water are 2.09 and 4.18 J/g-K, respectively. Assume no heat is transferred to the soda from the surroundings.

Homework Equations

q= (mass)(Specific heat)(change in temp); q=(moles)(delta-enthalpy)

The Attempt at a Solution

So I did q(soda lost)= q(ice gained). I found the q of the ice by (125g)(0--25)(2.09) + (125g)(1mol/18g)(6.01 KJ/mol)(1000J/KJ)= 48087.35 J. So now I set it equal to the q(soda). I had 24 oz of soda which is 681 grams: 48087.35= (681g)(4.18)(TempFinal-22). But then I get 39 C which is wrong I'd assume! Please help me by showing all the steps. Thanks!

 

[Borek]

In general you are on the right track. Pay attention to signs, my bet is that that's where you got lost.

 

[morrobay]

You should convert all units to calories and grams here;
specific heat of water = 1 cal/gram deg C
specific heat ice = .5 cal/gram deg C
latent heat of fusion ice = 80 cal/gram
Now take the 681 grams of soda from 22 deg C to 0 deg C and obtain 14982
calories
The 125 grams of ice will absorb 1562 calories going from -25 C to
0 C and will absorb 10000 cal melting
So now you have 806 grams of water at 0 deg C and 3420 calories.
Therefore 3420 calories = (Temp) ( 806 grams)

 

[Borek]

You should convert all units to calories and grams here

There is no need for that. cal is a nice unit when working with water, but it loses its simplicity when you have a system in which water changes state of matter.

 

[apbuiii]

Yeah, thanks guys. I ended up figuring it out :)