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

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

2. Relevant equations
q= (mass)(Specific heat)(change in temp); q=(moles)(delta-enthalpy)

3. 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!
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Admin In general you are on the right track. Pay attention to signs, my bet is that that's where you got lost.
 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)