Question

You are making pesto for your pasta and have a cylindrical measuring cup 10.0 cm high made of ordinary glass $\left({\beta }_g=2.7\times {10}^{-5}{\left(C°\right)}^{-1}\right)$that is filled with olive oil localid="1665032199021" $\left({\beta }_o=6.8\times {10}^{-4}{\left(C°\right)}^{-1}\right)$to a height of 3.00 mm below the top of the cup. Initially, the cup and oil are at room temperature 122.0 °C. You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

Short answer

The oil will spill out from cup at $51.4°\mathrm{C}$.

Step-by-step solution

Given data

Height of cup: 10.0 cm

volume expansion coefficient of glass: $2.7\times {10}^{-5}{\left(\mathrm{C}°\right)}^{-1}$

volume expansion coefficient of oil: $6.8\times {10}^{-4}{\left(\mathrm{C}°\right)}^{-1}$

height of the empty part on the top of the cup: 3 mm

room temperature: $122°\mathrm{C}$

Concept/Significance of volume expansion

Volume expansion happens when the average kinetic energy of the heated item's constituents rises. The item has a greater average separation distance due to its higher vibrational kinetic energy.

$\mathbf{∆}\mathbf{V}\mathbf{=}{\mathbf{\beta V}}_{\mathbf{o}}\mathbf{∆}\mathbf{T}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, is volume expansion of the system, $\mathbf{\beta }$is the volume expansion coefficient of system, ${\mathbf{V}}_{\mathbf{o}}$is the initial volume and $\mathbf{∆}\mathbf{T}$is thetemperature change.

Determination of temperature at which the olive oil start to spill out of the cup

The volume expansion of the system is given by,

Here, $∆{\mathrm{V}}_{\mathrm{glass}}$ is the volume expansion of glass and A is the cross-sectional area of the cup.

From volume expansion formula temperature change is given by,

$∆\mathrm{T}=\frac{3\times {10}^{-3}\mathrm{A}}{{\mathrm{V}}_{0\mathrm{oil}}{\mathrm{\beta }}_{\mathrm{oil}}-{\mathrm{V}}_{0\mathrm{glass}}{\mathrm{\beta }}_{\mathrm{glass}}}$

Here, ${\mathrm{V}}_{0\mathrm{oil}}$is the initial volume of oil, ${\mathrm{\beta }}_{\mathrm{oil}}$is the volume expansion coefficient of oil, ${\mathrm{V}}_{0\mathrm{glass}}$is the initial volume of the glass, and ${\beta }_{\mathrm{glass}}$is the volume expansion coefficient of the glass.

For the given values, the change in temperature is-

$$\begin{gathered} ∆\mathrm{T}=\frac{3\times {10}^3\mathrm{A}}{9.7\times {10}^{-2}\mathrm{m}\times \mathrm{A}\times 6.8\times {10}^{-4}{\left(\mathrm{C}°\right)}^{-1}-10\times {10}^{-2}\mathrm{m}\times \mathrm{A}\times 25\times {10}^{-5}{\left(\mathrm{C}°\right)}^{-1}} \\ =31.4°\mathrm{C} \end{gathered}$$

The temperature at which oil will spill out is-

$$\begin{gathered} T_2=T_1+∆T \\ =20°\mathrm{C}+31.4°\mathrm{C} \\ =51.4°\mathrm{C} \end{gathered}$$

Thus, the temperature at which oil spill out from cup is $51.4°\mathrm{C}$.