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: 40

Answer

$0.973\;\rm kg$

Work Step by Step

$\bullet\bullet$We have here 5 stages of heating for the ice: 1) Heating the ice from -10$^\circ$C to 0$^\circ$C, and the heat needed for that is given by $$Q_1=m_{H_2O} c_{ice}\Delta T_{ice}$$ 2) Melting the ice at 0$^\circ$C until all of it became water, and the heat needed for that is given by $$Q_2=m_{H_2O}L_{{\rm f},ice}$$ 3) Heating the water (the ice after melting became water) from 0$^\circ$C to 20$^\circ$C, and the heat needed for that is given by $$Q_3=m_{H_2O}c_{water}\Delta T_{water}$$ $\bullet\bullet$ We have one stage of heating for the aluminum bucket which is cooling it from 70$^\circ$C to 20$^\circ$C. This amount of heat is given by $$Q_4=m_{Al}c_{Al}\Delta T_{Al}$$ Let's assume that this system [ice+aluminum] is an isolated one. Hence, $$Q_1+Q_2+Q_3+Q_4=0$$ Plugging from above and then solving for $m_{Al}$; $$m_{H_2O} c_{ice}\Delta T_{ice}+m_{H_2O}L_{{\rm f},ice}+\\m_{H_2O}c_{water}\Delta T_{water}+m_{Al}c_{Al}\Delta T_{Al}=0$$ $$m_{Al}=\dfrac{m_{H_2O} c_{ice}\Delta T_{ice}+m_{H_2O}L_{{\rm f},ice}+ m_{H_2O}c_{water}\Delta T_{water}}{-c_{Al}\Delta T_{Al}}$$ Plugging the known; $$m_{Al}=\dfrac{(0.1) (2090)(10)+(0.1)(3.33\times 10^5)+ (0.1)(4190)(20)}{-(900)(-50)}$$ $$m_{Al}=\color{red}{\bf 0.973}\;\rm kg$$
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