Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 17 - Exercises and Problems: 56

Answer

Please see the work below.

Work Step by Step

We know that $Q=Pt$ $Q=500\times 1200=600KJ$ We also know that $Q=mc\Delta T+m^{\prime}L_v$ We plug in the known values to obtain: $600KJ=(0.3Kg)(4.184\frac{KJ}{KgK})(80K)+m^{\prime}(2257\frac{KJ}{Kg})$ This simplifies to: $m^{\prime}=\frac{500KJ}{2257\frac{KJ}{Kg}}=0.22Kg$ Now $mass\space of\space water \space left=0.3Kg-0.22Kg=0.08Kg$
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