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: 51

Answer

.176 kg, or 176 grams

Work Step by Step

We first find the value of Q for changing the ice to 0 degrees Celsius liquid: $Q = m_{ice}c_{ice}\Delta T_{ice}+m_{ice}L_f$ $Q=(.2)(2.05)(10)+(.2)(334)=4.1+66.8=70.9\ kJ$ We find the energy that would allow the warm water to be cooled to 0 degrees Celsius: $Q = m_{water}c_{water}\Delta T_{water}=(1)(4.184)(15)=62.8kJ$ We find that $62.8-4.1=58.7 \ kJ$ goes to getting the ice to 0 degrees Celsius. This means that: $m_{melted}=\frac{58.7}{334}=.176\ kg$ .176 kg, or 176 grams
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