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Found in: Page 541

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

When the temperature of a metal cylinder is raised from ${\mathbf{0}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{o}}}{\mathbf{C}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{{\mathbf{100}}}^{{\mathbf{o}}}{\mathbf{C}}$, its length increases by 0.23% (a) Find the percent change in density. (b) What is the metal? Use Table.

1. The percent change in density is -0.69%
2. The coefficient of expansion is $23×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}$. The given metal is Aluminum.
See the step by step solution

Step 1: The given data

Temperature of the cylinder changes from 0oC to 100oC and its length increases by 0.23%.

Step 2: Understanding the concept of thermal expansion

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. By using the formula for density, we can find the percent change in density. Also, we can use the concept of linear expansion to find the linear expansion coefficient.

Formula:

The density of a body, $\mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}}$ …(i)

The linear expansion of a body, data-custom-editor="chemistry" $∆\mathrm{L}=\mathrm{L\alpha }∆\mathrm{T}$ …(ii)

Where data-custom-editor="chemistry" $\mathrm{\alpha }$ is the coefficient of linear expansion of body, L is length, data-custom-editor="chemistry" $∆\mathrm{T}$ is change in temperature, m is mass and V is volume

Step 3: (a) Calculation of percent change in density

By differentiating equation (i), we can get

$\begin{array}{rcl}d\rho & =& -\left(\frac{\mathrm{m}}{{\mathrm{V}}^{2}}\right)dV\\ ∆\mathrm{\rho }& =& -\left(\frac{\mathrm{m}}{{\mathrm{V}}^{2}}\right)∆\mathrm{V}\\ & =& -\frac{\mathrm{m}∆\mathrm{V}}{{\mathrm{V}}^{2}}\\ & =& -\frac{\mathrm{\rho }∆\mathrm{V}}{{\mathrm{V}}^{2}}................\left(\mathrm{a}\right)\left(\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right)\right)\\ & & \end{array}$

But we know the relation of expansion of volume and length, that is:

$\frac{∆\mathrm{V}}{\mathrm{V}}=3\frac{∆\mathrm{L}}{\mathrm{L}}$

Therefore, substituting the above value in equation (a), we get

$∆\mathrm{\rho }=-\frac{3\mathrm{\rho }∆\mathrm{L}}{\mathrm{L}}\phantom{\rule{0ex}{0ex}}\frac{∆\mathrm{\rho }}{\mathrm{\rho }}=-\frac{3∆\mathrm{L}}{\mathrm{L}}$

Where $\frac{∆\mathrm{L}}{\mathrm{L}}$ is the percent change in length which is 0.23%

So, the percent change in density is given as:

$\begin{array}{rcl}\frac{∆\mathrm{\rho }}{\mathrm{\rho }}& =& -3×0.23\\ & =& -0.69%\end{array}$

Hence, the value of percent change in density is -0.69%

Step 4: (b) identifying the metal of thermal expansion

The coefficient of thermal expansion can tell us which metal it is. So, we have to find the coefficient of thermal expansion from equation (ii) as follows:

$\begin{array}{rcl}\mathrm{\alpha }& =& \frac{∆\mathrm{L}}{\mathrm{L}∆\mathrm{T}}\\ & =& \frac{0.23×{10}^{-2}}{{\left(100-0\right)}^{\mathrm{o}}\mathrm{C}}\left(\because \frac{∆\mathrm{L}}{\mathrm{L}}=0.23%=0.23×{10}^{-2}\right)\\ & =& 23×{10}^{-6}/{}^{\mathrm{o}}\mathrm{C}\end{array}$

From this value of the coefficient of thermal expansion, we can conclude that the given metal is aluminum.