Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 5 - Section 5.2 - Introduction to Thermodynamics - Checkpoint - Page 195: 5.2.2


c) w= $-5.4\times10^{2}$ kJ, done by the system.

Work Step by Step

q = -928 kJ. (Heat is released, so q is negative.) $\Delta U = -1.47\times10^{3} kJ.$ $\Delta U$ = q+w, Or w=$\Delta U-q = -1.47\times10^{2} kJ -(-928 kJ)$ = $-5.4\times10^{2}$ kJ. Negative sign indicates that the work is done by the system.
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