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: 47

Answer

$10.27^{\circ}C$

Work Step by Step

We first find the value of Q for changing the ice to 0 degrees Celsius liquid: $Q = m_{ice}c_{ice}\Delta T_{ice}+m_{ice}L_f$ $Q=(.05)(2.05)(10)+(.05)(334)=17.7\ kJ$ We find the energy that would allow the warm water to be cooled to 0 degrees Celsius: $Q = m_{water}c_{water}\Delta T_{water}=(1)(4.184)(15)=62.8kJ$ Thus, the ice will all melt, and it will heat to: $T = \frac{62.8-17.7}{4.184\times 1.05 } = 10.27^{\circ}C$
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