Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 16 - Exercises and Problems - Page 300: 53


1.8 kilograms

Work Step by Step

We first must find the amount of energy supplied by the burner over this time. Recalling that a Watt is a Joule per second, it follows: $ 2000 \frac{J}{s} \times 5.4 min \times \frac{60s}{min} = 648,000\ J$ Now, we set this equal to the equation for $\Delta Q$: $ 648,000\ J = 1.2 (447)(80) + (4184)(80)m \\ m = 1.8 \ kg$
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