An aluminum cup of 100 cm³capacity is completely filled with glycerin at 22^oC. How much glycerin, if any, will spill out of the cup if the temperature of both the cup and the glycerin is increased to 28^oC? (The coefficient of volume expansion of glycerin is$\mathbf{5}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{/}{}^{\mathbf{o}}\mathbf{C}$)

1. Initial volume of aluminum cup${\mathrm{V}}_{\mathrm{Al},\mathrm{i}}=100{\mathrm{cm}}^3$
2. Initial volume of glycerin${\mathrm{V}}_{\mathrm{g},\mathrm{i}}=100{\mathrm{cm}}^3$
3. Initial temperature${\mathrm{T}}_{\mathrm{i}}={20}^{\mathrm{o}}\mathrm{C}$
4. Final temperature${\mathrm{T}}_{\mathrm{f}}={28}^{\mathrm{o}}\mathrm{C}$
5. Coefficient of volume expansion for glycerin${\mathrm{\beta }}_{\mathrm{g}}=5.1\times {10}^{-4}/{}^{\mathrm{o}}\mathrm{C}$
6. Coefficient of volume expansion for aluminum${\mathrm{\beta }}_{\mathrm{Al}}(\mathrm{or}3{\mathrm{\alpha }}_{\mathrm{Al}})=69\times {10}^{-6}/{}^{\mathrm{o}}\mathrm{C}$

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. Thermal expansion can occur due to the increase in temperature. For the given problem, we have to find the change in volume expansion of aluminum and glycerin. From that, we can find out how much glycerin will spill out.

Formula:

The volume change in expansion of a body, $∆\mathrm{V}=\mathrm{V\beta }∆\mathrm{T}$ …(i)

where,$\mathrm{\beta }$is the volume expansion coefficient of body,$∆\mathrm{T}$is change in temperature, V and is volume

First, the change in temperature is given as:

$\begin{array}{c}\text{ΔT}={\text{T}}_{\mathrm{f}}-{\text{T}}_{\mathrm{i}}\\ =28°\text{C}-22°\text{C}\\ =6°\text{C}\end{array}$

The change in volume of the aluminum cup after expansion using equation (i) is given as:

${\mathrm{\Delta V}}_{\mathrm{Al}}={\mathrm{V}}_{\mathrm{Al},\mathrm{i}}{\mathrm{\beta }}_{\mathrm{Al}}\mathrm{\Delta T}$

The change in volume of the glycerin after expansion using equation (i) is given as:

${\mathrm{\Delta V}}_{\mathrm{g}}={\mathrm{V}}_{\mathrm{g},\mathrm{i}}{\mathrm{\beta }}_{\mathrm{g}}\mathrm{\Delta T}$

The quantity of glycerin that spills out will be given as:

$\begin{array}{c}{\mathrm{\Delta V}}_{\mathrm{g}}-{\mathrm{\Delta V}}_{\mathrm{Al}}={\mathrm{V}}_{\mathrm{g},\mathrm{i}}{\mathrm{\beta }}_{\mathrm{g}}\mathrm{\Delta T}-{\mathrm{V}}_{\mathrm{Al},\mathrm{i}}{\mathrm{\beta }}_{\mathrm{Al}}\mathrm{\Delta T}\\ =({\mathrm{V}}_{\mathrm{g},\mathrm{i}}{\mathrm{\beta }}_{\mathrm{g}}-{\mathrm{V}}_{\mathrm{Al},\mathrm{i}}{\mathrm{\beta }}_{\mathrm{Al}})\mathrm{\Delta T}\end{array}$

But, it is given that, ${\mathrm{V}}_{\mathrm{g},\mathrm{i}}={\mathrm{V}}_{\mathrm{Al},\mathrm{i}}=100{\text{cm}}^3$

Therefore, the above equation is given as:

$\begin{array}{c}{\mathrm{\Delta V}}_{\mathrm{g}}-{\mathrm{\Delta V}}_{\mathrm{Al}}=({\mathrm{\beta }}_{\mathrm{g}}-{\mathrm{\beta }}_{\mathrm{Al}}){\mathrm{V}}_{\mathrm{Al},\mathrm{i}}\mathrm{\Delta T}\\ =(5.1\times {10}^{-4}/°\text{C}-69\times {10}^{-6}/°\text{C})\times 100{\text{\hspace{0.17em}cm}}^3\times 6°\text{C}\\ =4.41\times {10}^{-4}\times 100\times 6{\text{\hspace{0.17em}cm}}^3\\ =0.26{\text{cm}}^3\end{array}$

Hence, the quantity of glycerin that will spill out is$0.26{\text{cm}}^3$