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: 1

Answer

See the graph below.

Work Step by Step

To draw this graph, we need to use the information in table 14-3 for the three substances, lead, water, and ethyl alcohol. For each substance, we need to calculate the amount of heat needed in each stage. 1- From solid to the melting point: $Q_1=mc\Delta T$ 2- Latent heat of fusion: $Q_2=mL_f$ 3- From liquid to the boiling point: $Q_3=mc\Delta T$ 4- Latent heat of vaporization: $Q_4=mL_v$ 5- Heat absorbed by vapor: $Q_5=mc\Delta T$ To draw the graph, when we find the amount of one stage, we need to add it to the amount of heat of the second stage, the net heat will be added to the third stage, and so on. We chose an initial temperature of $-150^\circ \rm C$ to make sure that all 3 substances are solid. And we chose the final temperature of $2000^\circ \rm C$ to make sure that all 3 substances are vapor. $\Rightarrow$ For Ethyl alcohol: - $Q_1=mc\Delta T=1\cdot 0.58\cdot (-114-(-150))=\color{green}{20.88}\;\rm kcal$ - $Q_2=mL_f=1\cdot 25 =\color{green}{25}\;\rm kcal$ - $Q_3=mc\Delta T=1\cdot 0.58\cdot (78-(-114))=\color{green}{20.88}\;\rm kcal$ - $Q_4=mL_v=1\cdot 204=\color{green}{204}\;\rm kcal$ - $Q_5=mL_v=mc\Delta T=1\cdot 0.58\cdot (2000-78)=\color{green}{1114.76}\;\rm kcal$ Thus, the start points of each stage are $\left(0\;\rm kcal,-150^\circ C\right)$ $\left(20.88\;\rm kcal,-114^\circ C\right)$ $\left(45.88\;\rm kcal,-114^\circ C\right)$ $\left(66.76\;\rm kcal,78^\circ C\right)$ $\left(270.76\;\rm kcal,78^\circ C\right)$ $\left(1385.52\;\rm kcal,2000^\circ C\right)$ $\Rightarrow$ For Water: - $Q_1=mc_{ice}\Delta T=1\cdot 0.5\cdot (0-(-150))=\color{blue}{75}\;\rm kcal$ - $Q_2=mL_f=1\cdot 79.7=\color{blue}{79.7}\;\rm kcal$ - $Q_3=mc_{liquid}\Delta T=1\cdot 1\cdot (100-0)=\color{blue}{100}\;\rm kcal$ - $Q_4=mL_v=1\cdot 539=\color{blue}{539}\;\rm kcal$ - $Q_5=mL_v=mc_{steam}\Delta T=1\cdot 0.48\cdot (2000-100)=\color{blue}{912}\;\rm kcal$ Thus, the start points of each stage are $\left(0\;\rm kcal,-150^\circ C\right)$ $\left(75\;\rm kcal,0^\circ C\right)$ $\left(154.7\;\rm kcal,0^\circ C\right)$ $\left(254.7\;\rm kcal,100^\circ C\right)$ $\left(793.7\;\rm kcal,100^\circ C\right)$ $\left(1705.7\;\rm kcal,2000^\circ C\right)$ $\Rightarrow$ For Lead: - $Q_1=mc\Delta T=1\cdot 0.031\cdot (327-(-150))=\color{red}{14.787}\;\rm kcal$ - $Q_2=mL_f=1\cdot 5.9=\color{red}{5.9}\;\rm kcal$ - $Q_3=mc\Delta T=1\cdot 0.031\cdot (1750-327)=\color{red}{1423}\;\rm kcal$ - $Q_4=mL_v=1\cdot 208=\color{red}{208}\;\rm kcal$ - $Q_5=mL_v=mc\Delta T=1\cdot 0.031\cdot (2000-1750)=\color{red}{7.75}\;\rm kcal$ Thus, the start points of each stage are $\left(0\;\rm kcal,-150^\circ C\right)$ $\left(14.787\;\rm kcal,327^\circ C\right)$ $\left(20.687\;\rm kcal,327^\circ C\right)$ $\left(1443.687\;\rm kcal,1750^\circ C\right)$ $\left(1651.687\;\rm kcal,1750^\circ C\right)$ $\left(1659.437\;\rm kcal,2000^\circ C\right)$ Now we can draw the graph for each substance.
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