91Ó°ÊÓ

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

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,$\mathrm{ÒÏ}=\frac{\mathrm{m}}{\mathrm{V}}$ …(¾±)

The linear expansion of a body, data-custom-editor="chemistry" $∆\mathrm{L}=\mathrm{³¢Î\pm }∆\mathrm{T}$ …(¾±¾±)

Where data-custom-editor="chemistry" $\mathrm{Î\pm }$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

By differentiating equation (i), we can get

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

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

$$\begin{gathered} ∆\mathrm{ÒÏ}=-\frac{3\mathrm{ÒÏ}∆\mathrm{L}}{\mathrm{L}} \\ \frac{∆\mathrm{ÒÏ}}{\mathrm{ÒÏ}}=-\frac{3∆\mathrm{L}}{\mathrm{L}} \end{gathered}$$

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:

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

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

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:

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

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