## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 1 - Exercises - Page 41: 131

#### Answer

$\underline{0.661\text{ }\Omega }$

#### Work Step by Step

$1\text{ kg}=1000\text{ g}$ Thus, mass of copper wire is converted into grams as follows: \begin{align} & m=\left( 24.0\,\text{kg} \right)\left( \frac{1000\text{ g}}{1\,\text{kg}} \right) \\ & =24,000\text{ g} \end{align} The volume is calculated as follows: $V=\frac{m}{d}$ Here, m is mass and d is density. The density of copper is $8.96\text{ g/c}{{m}^{3}}$. Thus, volume is calculated as follows: \begin{align} & V=\frac{24,000\text{ g}}{8.96\text{ g/c}{{\text{m}}^{3}}} \\ & =2,679\text{ c}{{\text{m}}^{3}} \end{align} Now, $1\text{ cm}=10\text{ mm}$. Thus, radius of copper wire is converted into centimeters as follows: \begin{align} & r=1.63\text{ mm}\left( \frac{1\text{ cm}}{10\text{ mm}} \right) \\ & =0.163\text{ cm} \end{align} The relation between volume $\left( V \right)$, radius $\left( r \right)$, and length $\left( l \right)$ is as follows: \begin{align} & \pi {{r}^{2}}\,=\,\frac{V}{l}\, \\ & \Rightarrow l=\frac{V}{\pi {{r}^{2}}} \\ \end{align} Thus, length of copper wire is calculated as follows: \begin{align} & l=\frac{2,679\text{ c}{{\text{m}}^{3}}}{\left( 3.14 \right){{\left( 0.163\text{ cm} \right)}^{2}}} \\ & =32,112\text{ cm} \end{align} Now, $1\text{ km}=1\times \text{1}{{\text{0}}^{5}}\text{ cm}$. Thus, length of copper wire in kilometers is written as follows: \begin{align} & l=\left( 32,112\text{ cm} \right)\times \left( \frac{1\text{ km}}{{{10}^{5}}\text{ cm}} \right) \\ & =0.32112\text{ km} \end{align} The resistance of copper wire is $2.061\text{ }\Omega \text{/km}$. For length of $0.32112\text{ km}$, the resistance will be as follows: \begin{align} & R=\left( 0.32112\text{ km} \right)\left( 2.061\text{ }\Omega \text{/km} \right) \\ & =0.661\text{ }\Omega \end{align} The overall resistance of the wire is $\underline{0.661\text{ }\Omega }$.

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