Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 1 - Sections 1.1-1.8 - Exercises - Cumulative Problems - Page 41: 129

Answer

$$0.661 \space \Omega$$

Work Step by Step

1. We need to find the length of the cable, then calculat the overall resistance. - In order to find the length, we can use the density of copper, to find the total volume (since we know the mass), then we assume a cylinder shape, and use the radius and the volume to calculate it. 2. $V_{cylinder} = \pi r^2 h$ $\frac V{\pi r^2} = h$ Now, using the density equation: $density = \frac{m}{V} \longrightarrow V = \frac m {density}$ Thus: $\frac {m/density}{\pi r^2} = h$ $\frac {m}{\pi r^2 (density)} = h$ If we multiply the resistance per km by the length of the cable, we will get the overall resistance: $h \times resistance \space per \space km = overall \space resistance$ $\frac {m}{\pi r^2 (density)} \times resistance \space per \space km = overall \space resistance$ 3. Convert the given data to match their units. $m = 24.0 \space kg \times \frac{1000 \space g}{1 \space kg} = 2.40 \times 10^4 \space g$ $r = 1.63 \space mm \times \frac{1 \space cm}{10 \space mm} = 0.163 \space cm$ $2.061 \space \Omega/km \times \frac{1 \space km}{10^3 \space m} \times \frac{1 \space m}{100 \space cm} = 2.061 \times 10^{-5} \Omega/cm$ 4. Calculate it: $\frac {2.40 \times 10^4 \space g}{\pi (0.163 \space cm)^2 (8.96 \space g/cm^3)} \times 2.061 \times 10^{-5} \space \Omega/cm = 0.661 \space \Omega$
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