Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 17 - Work, Heat, and the First Law of Thermodynamics - Exercises and Problems - Page 498: 31

Answer

$ 16.3\;\rm kJ$

Work Step by Step

We have here 5 stages of heating: 1) Heating the ice from -20$^\circ$C to 0$^\circ$C, and the heat needed for that is given by $$Q_1=mc_{ice}\Delta T_{ice}$$ 2) Melting the ice at 0$^\circ$C until all of it became water, and the heat needed for that is given by $$Q_2=mL_{{\rm f},ice} $$ 3) Heating the water from 0$^\circ$C to 100$^\circ$C, and the heat needed for that is given by $$Q_3=mc_{water}\Delta T_{water}$$ 4) Evaporating the water at 100$^\circ$C until all the water became steam, and the heat needed for that is given by $$Q_4=mL_{{\rm v},water} $$ 5) Heating the steam from 100$^\circ$C to 200$^\circ$C, and the heat needed for that is given by $$Q_5=mc_{steam}\Delta T_{steam}$$ Thus, the total heat needed is given by $$Q=Q_1+Q_2+Q_3+Q_4+Q_5$$ $$Q=mc_{ice}\Delta T_{ice}+mL_{{\rm f},ice} +mc_{water}\Delta \\T_{water}+mL_{{\rm v},water}+mc_{steam}\Delta T_{steam}$$ $$Q=m\left[c_{ice}\Delta T_{ice}+ L_{{\rm f},ice} + c_{water}\Delta T_{water}+ L_{{\rm v},water}+ c_{steam}\Delta T_{steam}\right]$$ Plugging the known; $$Q=(5\times 10^{-3})\left[(2090)(20)+ (3.33\times 10^5) + (4190)(100)+ (22.6\times 10^5)+ (2017)(100)\right]$$ $$Q=\color{red}{\bf 1.63\times 10^4}\;\rm J$$
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