91Ó°ÊÓ

1. Edge of the cube is,$l=5.00\text{cmor}5.00×{10}^{−2}\text{m}.$
2. Initial temperature of the cube is,$T_i=10.0\text{°°ä}.$
3. Final temperature of the cube is,$T_f=60.0\text{°°ä}.$

The propensity of matter to alter its form, area, volume, and density in reaction to a change in temperature is known as thermal expansion.We can find the volume of the cube from its edge. Also, we can find the temperature difference between its initial and final temperature. Using the value of the coefficient of linear expansion, we can find its coefficient of volume expansion. Then, using the formula for the coefficient of volume expansion and inserting the above values in it, we can get the volume increase of an aluminum cube.

Formula:

The volume expansion of a body due to thermal expansion$\mathrm{Δ}V=V\mathrm{β}\mathrm{Δ}T$,…(¾±)

where$\mathrm{β}$ is the coefficient of volume expansion.

The volume of the cube is

$\begin{array}{c}{l}^3={(5.00×{10}^{−2}\text{″¾})}^3\\ =125×{10}^{−6}{\text{m}}^{\text{3}}\end{array}$

The temperature difference of the cube is

$\begin{array}{c}\mathrm{Δ}T=60.0\text{°°ä}−10.0\text{°°ä}\\ =50.0\text{°°ä}\end{array}$

The coefficient of linear expansion of Aluminum is.$\mathrm{Î\pm }=23×{10}^{−6}/\text{°°ä}$

Thus, the coefficient of volume expansion of Aluminum is

$\begin{array}{c}\mathrm{β}=3\mathrm{Î\pm }\\ =3(23×{10}^{−6}/\text{°°ä})\\ =69×{10}^{−6}/°C\end{array}$

Hence, the increase in the volume is calculatedusing equation (i) as

$\begin{array}{c}\mathrm{Δ}V=(125×{10}^{−6}{\text{″¾}}^{\text{3}})(69×{10}^{−6}/\text{°°ä})(50\text{°°ä})\\ =4.3125×{10}^{−7}{\text{m}}^{\text{3}}\\ =0.431{\text{ c³¾}}^{\text{3}}\end{array}$

Therefore, thevolume increase of an aluminum cube is.$0.431{\text{ c³¾}}^{\text{3}}$