Question

A hollow aluminium cylinder $20.0cm$ deep has an internal capacity of $2.000L$ at ${20.0^ \circ }C$. It is completely filled with turpentine and then slowly warmed to ${80.0^ \circ }C$.
A. How much turpentine overflows?
B. If the cylinder is then cooled back to ${20.0^ \circ }C$, how far below the cylinder's rim does the turpentine's surface recede?

Answer 1

Hint: The overflow of turpentine is given by the difference in the volume of aluminium can and turpentine after expansion. Cooling back to a lower temperature reverses the effect of volume expansion and the volume of the aluminium is the same as the initial volume.
Formula Used: The formulae used in the solution are given here.
$\dfrac{{\Delta V}}{{{V_0}}} = {\alpha _V}\Delta T$ where ${V_0}$ is the original volume, ${\alpha _V}$ is the volume expansion coefficient, $\Delta T$ is the temperature difference and $\Delta V$ change in volume after expansion.
${V_f} = {V_i}\left( {1 + 3\alpha \Delta T} \right)$ where ${V_f}$ is the final volume after expansion and ${V_i}$ is the initial volume.
The ${\alpha _V} = 3\alpha $ where $\alpha $ is the linear expansion coefficient.

Complete Step by Step Solution: Thermal expansion describes the tendency of an object to change its dimension either in length, area or volume due to heat. Heating up a substance increases its kinetic energy. Depending on the type of expansion thermal expansion is of 3 types– Linear expansion, Area expansion, and Volume expansion.
Volume expansion is the change in volume due to temperature. Volume expansion formula is given as
$\dfrac{{\Delta V}}{{{V_0}}} = {\alpha _V}\Delta T$ where ${V_0}$ is the original volume, ${\alpha _V}$ is the volume expansion coefficient, $\Delta T$ is the temperature difference and $\Delta V$ change in volume after expansion.
It has been given that a hollow aluminium cylinder $20.0cm$ deep has an internal capacity of $2.000L$ at ${20.0^ \circ }C$. It is completely filled with turpentine and then slowly warmed to ${80.0^ \circ }C$.
When the entire system is heated to ${80.0^ \circ }C$. The volume of the aluminium container is,
${V_f} = {V_i}\left( {1 + 3\alpha \Delta T} \right)$ where ${V_f}$ is the final volume after expansion and ${V_i}$ is the initial volume.
The ${\alpha _V} = 3\alpha $ where $\alpha $ is the linear expansion coefficient.
We have,
$\Delta T = {\left( {80 - 20} \right)^ \circ }C$
$ \Rightarrow \Delta T = {60^ \circ }C$
The initial volume is ${V_i} = 2.000L$, and the linear expansion coefficient of aluminium is $\alpha = 24 \times {10^{ - 6}}^ \circ {C^{ - 1}}$.
Substituting these values in the formula,
${V_f} = 2\left( {1 + 3 \times 24 \times {{10}^{ - 6}} \times 60} \right)$
$ \Rightarrow {V_f} = 2.00864L$
The volume of the turpentine is
${V_f} = {V_i}\left( {1 + \beta \Delta T} \right)$
$ \Rightarrow {V_f} = 2.000\left( {1 + 9.0 \times {{10}^{ - 4}} \times 60} \right) = 2.108L$
The overflow is given by the difference of volume of turpentine and volume of hollow cylinder is,
${V_{turpentine}} - {V_{alu\min ium}} = \left( {2.108 - 2.00864} \right)$
$ \Rightarrow {V_{turpentine}} - {V_{alu\min ium}} = 0.09936L = 0.099L$
Cooling the system back to ${20.0^ \circ }C$ means the aluminium container is back to $2.000L$. The volume of turpentine is,
${V_f} = {V_i}\left( {1 + \beta \Delta T} \right)$
$ \Rightarrow {V_f} = 2.00864\left( {1 + 9.0 \times {{10}^{ - 4}} \times \left( { - 60} \right)} \right) = 1.9002L$
This is the following fraction of the volume of the container.
$\dfrac{{{V_{turpentine}}}}{{{V_{alu\min ium}}}} = \dfrac{{1.9002}}{{2.000}} = 0.95009$.
It is also the following fraction of the $20.0cm$height of the container.
$\dfrac{{{h_{turpentine}}}}{{{h_{alu\min ium}}}} = \dfrac{{{h_{turpentine}}}}{{20.0}} = 0.95009$
$ \Rightarrow {h_{turpentine}} = 19.9002cm$
The distance from the top is, $20.0 - 19.9002 = 0.0998{\text{ }}cm$.

Note: In general, objects will expand with increasing temperature. Water is the most important exception to this rule. Water expands with increasing temperature (its density decreases) when it is at temperatures greater than ${4^ \circ }C\left( {{{40}^ \circ }F} \right)$. However, it expands with decreasing temperature when it is between $ + {4^ \circ }C$ and ${0^ \circ }C$ $\left( {{{40}^ \circ }F - {{32}^ \circ }F} \right)$. Water is densest at $ + {4^ \circ }C$.