Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 14 - Heat - Search and Learn - Page 411: 3

Answer

a) It raises the temperature of the ice to $-15^\circ\rm C$. b) It melts 3 g of the ice. c) It raises the temperature of the liquid water to $12.4^\circ\rm C$. d) It evaporates 0.442 g of the water. e) It raises the steam temperature to $115^\circ\rm C$.

Work Step by Step

a) Let's calculate the final temperature of the ice when we add an amount of heat of 1000 J to 100 g of ice at $-20^\circ \rm C$ to see if it is enough to melt it or not. $$Q=mc_{ice}\Delta T=mc_{ice}\left(T_f-T_i\right)$$ So, $$T_f=\dfrac{Q}{mc_{ice}}+T_i\tag 1$$ $$T_f=\dfrac{1000}{100\times 10^{-3}\cdot 2100}-20$$ $$T_f=\boxed{\color{red}{\bf-15}^\circ \rm C}$$ It raises its temperature to $-15^\circ\rm C$. b) In this case, when the ice is at $0^\circ \rm C$, adding heat will melt the ice. We need to find how much of the 100-g ice will melt. $$Q=m_{melted}L_f$$ $$m_{melted}=\dfrac{Q}{L_f}=\dfrac{1000}{333\times10^3}=3\times 10^{-3}\;\rm kg$$ $$m_{melted}=\boxed{ \color{red}{\bf3}\ \;\rm g}$$ c) We need to find the final temperature of the water. Using equation (1) but for liquid water; $$T_f=\dfrac{Q}{mc_{water}}+T_i$$ $$T_f=\dfrac{1000}{0.1\cdot 4186 }+10$$ $$T_f=\boxed{\color{red}{\bf12.4}^\circ \rm C}$$ It raises the temperature of the liquid water to $12.4^\circ\rm C$. d) Adding heat to the water at $100^\circ \rm C$ will evaporate some of it. We need to find the amount that evaporates. $$Q=m_{evaporated}L_v$$ $$m_{evaporated}=\dfrac{Q}{L_v}=\dfrac{1000}{2260\times 10^3}=4.42\times 10^{-4}\;\rm kg$$ $$m_{evaporated}=\boxed{\color{red}{\bf0.442}\;\rm g}$$ e) Adding heat to steam will raise its temperature. Using equation (1) but for steam; $$T_f=\dfrac{Q}{mc_{steam}}+T_i $$ $$T_f=\dfrac{1000}{0.1\cdot 2010}+110$$ $$T_f=\boxed{\color{red}{\bf115}^\circ \rm C}$$ It raises the steam temperature to $115^\circ\rm C$.
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