Solving Temperature Problem w/ 30g Ice & 60g Water

[daisyi]

I have a question sort of similar to the question that a different person asked earlier, but a little bit different.

okay, here it is: Find the final temperature if 30 gr. of ice at -20 deg C is mixed with 60 gr. of water at 60 deg C.


this is what I have done, but I'm not sure if I'm on the right track or where to go from here.

I have used the formula Q = c*m*change in temperature.

I calculated the amount of energy lost from the 60 grams of water to be 2400 through the above formula, and the amount of energy that the 30 grams of ice gains is 48015.

As far as what to do from there, I'm not quite sure.

 

[wisky40]

the amount of energy lost by the water is the same amount of energy gained by ice plus the energy use in the phase change of ice into water. Therefore, Q(of water lost)+Q(of ice gained)+Q(of fusion)=0
=> C(of water)*M(of water)*( Tf-(Ti of water))+C(of ice)*M(ice) *(Tf-(Ti of ice))+ M(ice)*L(which is the heat of fusion)=0. I know that you have all the data except for Tf, which is the quantity you're looking for. The only problem now is that you have to deal with the numbers and the units of this eqation.

 

[daisyi]

thanks, i see now where I wasn't understanding. Thanks so much!