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12P

Expert-verifiedFound in: Page 541

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An aluminum-alloy rod has a length of 10.00 cm ****at 20.00 ^{0}C**

**(a) What is the length of the rod at the freezing point of water? **

**(b) What is the temperature if the length of the rod is 10.009 cm****?**

- The length of the rod at the freezing point of water is 9.996 cm
- The temperature value is ${68}^{\mathrm{o}}\mathrm{C}$

- Aluminum-alloy rod length is ${\mathrm{L}}_{1}=10.00\mathrm{cm}$ at ${\mathrm{T}}_{1}={20}^{\mathrm{o}}\mathrm{C}$
- Aluminum-alloy rod length is ${\mathrm{L}}_{2}=10.015\mathrm{cm}$ at ${\mathrm{T}}_{2}={100}^{\mathrm{o}}\mathrm{C}$

**Thermal expansion can occur due to an increase in temperature. It has three types’ i.e. linear expansion, surface expansion, and volume expansion. For the given problem, we have to use the formula for linear expansion to find the length of the rod and the temperature of the rod.**

Formula:

The linear expansion of a body, $\u2206\mathrm{L}=\mathrm{L\alpha}\u2206\mathrm{T}$ …(i)

Using equation (i), the coefficient of linear expansion for aluminum-alloy is given as:

$\begin{array}{rcl}\mathrm{\alpha}& =& \frac{\u2206\mathrm{L}}{\mathrm{L}\u2206\mathrm{T}}\\ & =& \frac{10.015\mathrm{cm}-10.00\mathrm{cm}}{10.0\mathrm{cm}\times {\left(100-20\right)}^{\mathrm{o}}\mathrm{C}}\\ & =& \frac{0.015\mathrm{cm}}{10.0\mathrm{cm}\times {80}^{\mathrm{o}}\mathrm{C}}\\ & =& 1.88\times {10}^{-5}/{}^{\mathrm{o}}\mathrm{C}\end{array}$

We have to calculate the change in length of the rod at freezing point of the water so that the change in temperature is given as:

$\begin{array}{rcl}\u2206\mathrm{T}& =& {0}^{\mathrm{o}}\mathrm{C}-{100}^{\mathrm{o}}\mathrm{C}\\ & =& -{100}^{\mathrm{o}}\mathrm{C}\end{array}$

Length of rod at ${100}^{\mathrm{o}}\mathrm{C}$ is ${\mathrm{L}}_{2}=10.015\mathrm{cm}$

So, the change in linear expansion of the rod is given as:

$\begin{array}{rcl}\u2206\mathrm{L}& =& 10.015\mathrm{cm}\times 1.88\times {10}^{-5}/{}^{\mathrm{o}}\mathrm{C}\times \left(-{100}^{\mathrm{o}}\mathrm{C}\right)\\ & =& -1.88\times {10}^{-2}\mathrm{cm}\end{array}$

Total length at 0^{0}C is given as the sum of original length and the expanded length, that is:

$\begin{array}{rcl}{\mathrm{L}}^{\text{'}}& =& \mathrm{L}+\u2206\mathrm{L}\\ & =& 10.015\mathrm{cm}-1.88\times {10}^{-2}\mathrm{cm}\\ & =& 10.015\mathrm{cm}-0.0188\mathrm{cm}\\ & =& 9.996\mathrm{cm}\end{array}$

Hence, the value of the length of the rod is 9.996 cm

The length of the rod is given as $\mathrm{L}=10.009\mathrm{cm}$

But we know that $\u2206\mathrm{L}=10.009\mathrm{cm}-10.00=0.009\mathrm{cm}$

From equation (i), the change in the temperature due to expansion is given as:

$\begin{array}{rcl}0.009\mathrm{cm}& =& 10.00\mathrm{cm}\times 1.88\times {10}^{-5}/{}^{\mathrm{o}}\mathrm{C}\times \u2206\mathrm{T}\\ \u2206\mathrm{T}& =& \frac{0.009\mathrm{cm}}{10.00\mathrm{cm}\times 1.88\times {10}^{-5}/{}^{\mathrm{o}}\mathrm{C}}\\ & =& 47.{87}^{\mathrm{o}}\mathrm{C}\\ & \approx & {48}^{\mathrm{o}}\mathrm{C}\\ & =& \end{array}$

But, $\u2206\mathrm{T}={\mathrm{T}}_{\mathrm{x}}-{20}^{\mathrm{o}}\mathrm{C}$,

where, T_{x} is the temperature at which the length of the rod is $\mathrm{L}=10.009\mathrm{cm}$

$\begin{array}{rcl}{\mathrm{T}}_{\mathrm{x}}-{20}^{\mathrm{o}}\mathrm{C}& =& {48}^{\mathrm{o}}\mathrm{C}\\ {\mathrm{T}}_{\mathrm{x}}& =& {48}^{\mathrm{o}}\mathrm{C}+{20}^{\mathrm{o}}\mathrm{C}\\ {\mathrm{T}}_{\mathrm{x}}& =& {68}^{\mathrm{o}}\mathrm{C}\end{array}$

Hence, the value of the required temperature is ${68}^{\mathrm{o}}\mathrm{C}$.

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