Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 17 - Think and Solve - Page 333-334: 40


The final temperature of the water will be the same as that of the very large block of ice, $0^{\circ}C$.

Work Step by Step

We start by finding the heat released when 50 g of $80^{\circ}C$ water cools to 50 g of $0^{\circ}C$ water. This is discussed in the Check Point on page 331, item 2, using 50 grams instead of 1 gram. 50 g of $80^{\circ}C$ water cooling to $0^{\circ}C$ gives up 4000 calories. The 4000 calories, is used to melt ice. We know that 1 g of $0^{\circ}C$ ice requires 80 calories to melt to 1 g of $0^{\circ}C$ water, so we can melt (4000/80) = 50 grams of ice, the same mass of $80^{\circ}C$ water that we started with.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.