Question

A steel tank is completely filled with 1.90 m³of ethanol when both the tank and the ethanol are at 32.0ºC. When the tank and its contents have cooled to 18.0ºC, what additional volume of ethanol can be put into the tank?

Short answer

The additional volume of ethanol that can be added is 19.0L

Step-by-step solution

Concept of Volume Expansion in solids.

Volume expansion is the expansion of the dimensions of an object with respect to temperature changes and can be expressed as,

$\mathbf{\Delta V}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{}\mathbf{\beta \Delta T}$ ...(i)

Here, V is the expansion in volume, V₀ is the expandedvolume, β is the coefficient of volume expansion and T is the absolute temperature.

 Determination of the diameter value of the rivet following the conditions mentioned.

For steel, the coefficient of volume expansion,

${\beta }_s=3.6\times {10}^{-5}{\left(C^{°}\right)}^{-1}$

For ethanol, the coefficient of volume expansion,

${\beta }_e=75\times {10}^{-5}{\left(C^{°}\right)}^{-1}$

Calculate the volume change for the steel tank using equation (i),

$∆V_s=V_0{\beta }_s∆T$

Substitute all the values in the above equation,

$$\begin{gathered} ∆{\mathrm{V}}_{\mathrm{s}}=\left(3.6\times {10}^{-5}{\left({\mathrm{C}}^0\right)}^{-1}\right)\left(1.90{\mathrm{m}}^3\right)\left(18.0\mathrm{C}°-32\mathrm{C}°\right) \\ =-9.576\times {10}^{-4}{\mathrm{m}}^{-3} \\ =-0.9576\mathrm{L}. \end{gathered}$$

Similarly,

Calculate the volume change for the ethanol using equation (i),

$∆V_e=V_0{\beta }_e∆T$

Substitute all the values in the above equation,

$$\begin{gathered} ∆{\mathrm{V}}_{\mathrm{e}}=\left(75\times {10}^{-5}{\left(\mathrm{C}°\right)}^{-1}\right)\left(1.90{\mathrm{m}}^3\right)\left(18.0\mathrm{C}°-32\mathrm{C}°\right) \\ =-1.995\times {10}^{-2}{\mathrm{m}}^3 \\ =-19.95\mathrm{L} \end{gathered}$$

Both the volumes decreases as they both cool but the contraction in ethanol is more than the tank. Thus, the amount of ethanol can be added to fill the tank fully is,

$$\begin{gathered} ∆{\mathrm{V}}_{\mathrm{e}}=∆{\mathrm{V}}_{\mathrm{s}}=-19.95\mathrm{L}-\left(-0.9576\mathrm{L}\right) \\ =-19.0\mathrm{L} \end{gathered}$$

Thus, 19.0 L volume of ethanol can be added.