What is the final temperature and physical state of water when 250 grams of water at 85 degrees C is added to 80.0 grams of ice at -15 degree C?

Question

Is the physical state of water liquid?

The explanation from bobpursly was this

The sum of heats gained =0
80(cice)(0+15)+80Hf+80cw(Tf-0)+250cw(Tf-85)=0

Answer 1

The 80 x specific heat ice x 15 = heat absorbed in changing the T of ice at -15 to zero C.

Then you melt the ice and convert solid ice to liquid water with 80 x heat fusion ice.

Then you combine the moving T to zero with melting and see if the 250 g water at 85 is enough to give you a positive temperature. I quickly looked at the figures and I think so. I appears Tfinal is about 55 C or so.

You can look at it another way.
heat gained by ice in moving from -15 to zero C is the first part.
Second part is melting the ice at zero to liquid water at zero.
Add those together.
Then calculate how much heat it is possible to lose with the 150 g H2O going from 85 to zero C. If you have more heat to lose than you gain by ice moving from -15 to zero then melting, you must have enough heat in the water to do that. If the heat in the 250 g H2O isn't enough then there isn't enough to melt all of the ice. Hope this helps.