Asked by Capacino
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?
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?
Answers
Answered by
bobpursley
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
Assuming the ice starts at zero (Ice is usually colder).
31.3*334+210*4.18*(Tf-21)=0
solve for Tf
Answered by
Reid
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.