Question

When ice at 0C melts to liquid water at 0C, it absorbs 0.334 kj of heat per gram. Suppose the heat needed to melt 31.5 g of ice is absorbed from the water contained in a glass. If this water has a mass of .210 kg and a temperature of 21.0C, what is the final temperature of the water? (Note that you will also have 31.5 g of water at 0C from the ice.)

My Homework
q=ms∆T
q=.210kg x 0.334 kJ/g x (final temp-21C)

This is where I get stuck I feel like I am forgetting a step (or maybe I am completely doing this wrong)but cannot find anything in my notes or text that helps. Can anyone help?

Answer 1

The sum of the heats gained is zero.
Assuming the ice starts at zero (Ice is usually colder).
31.3*334+210*4.18*(Tf-21)=0
solve for Tf

Answer 2

that did not help at all, in order for that expression to equal zero, the deta T or "(Tf-Ti)" term has to equal zero. Im stuck where you are Capacino, idk what im doing wrong, this was a test question on my last exam and had me stumped.