Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 14 - Heat - General Problems - Page 410: 48

Answer

$4\times10^{15}J$

Work Step by Step

Use equation 14–2. $$Q=mc\Delta T=\rho Vc\Delta T $$ $$Q=(1000kg/m^3)(1000m)^3 (\frac{4186J}{kg\cdot C^{\circ}})(1C^{\circ})$$ $$=4\times10^{15}J$$
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.