Conceptual Physics (12th Edition)

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

Chapter 16 - Plug and Chug - Page 317: 35

Answer

$$Q = cm \Delta T$$ $$=\frac{1 cal}{g \cdot ^{\circ}C}(20 g)(90^{\circ}C - 30^{\circ}C ) = 1200 cal$$
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.