Chapter 19, Problem 19.21P

(a)

To determine

The volume of turpentine overflows.

Answer to Problem 19.21P

The volume of turpentine overflows is $99.4\text{mL}$.

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is $2.000\text{L}$, the initial temperature is $20°\text{C}$, and the final temperature in the cylinder is $80°\text{C}$.

Formula to calculate the change in volume of aluminum cylinder is,

$\Delta V_{\text{Al}}=V{\beta }_{\text{Al}}\Delta T$

Here,

${\beta }_{\text{Al}}$ is the coefficient of volume expansion of aluminum.

$\Delta T$ is the temperature difference.

Formula to calculate the change in volume of turpentine is,

$\Delta V_{\text{t}}=V{\beta }_{\text{t}}\Delta T$

Here,

${\beta }_{\text{t}}$ is the coefficient of volume expansion of turpentine.

The relation between coefficient of volume expansion of aluminum and coefficient of linear expansion of aluminum is,

${\beta }_{\text{Al}}=3{\alpha }_{\text{Al}}$ (2)

Here,

${\alpha }_{\text{Al}}$ is the coefficient of linear expansion of aluminum.

The coefficient of linear expansion of aluminum is $24\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$.

Substitute $24\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for ${\alpha }_{\text{Al}}$ in the above equation.

$\begin{array}{c}{\beta }_{\text{Al}}=3\left(24\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\right)\\ =72\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\end{array}$

Thus, the coefficient of volume expansion of aluminum is $72\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$.

Write the expression for the temperature difference,

$\Delta T=T_{\text{f}}-T_{\text{i}}$

Substitute $20°\text{C}$ for $T_{\text{i}}$ and $80°\text{C}$ for $T_{\text{f}}$ in the above equation.

$\begin{array}{c}\Delta T=80°\text{C}-20°\text{C}\\ =60°\text{C}\end{array}$

Thus, the temperature difference is $60°\text{C}$.

Write the expression for overflow volume of the turpentine,

$\Delta V^{\prime }=\Delta V_{\text{t}}-\Delta V_{\text{Al}}$

Substitute $V{\beta }_{\text{Al}}\Delta T$ for $\Delta V_{\text{Al}}$ and $V{\beta }_{\text{t}}\Delta T$ for $\Delta V$ in the above equation.

$\begin{array}{c}\Delta V^{\prime }=V{\beta }_{\text{t}}\Delta T-V{\beta }_{\text{Al}}\Delta T\\ =V\Delta T\left({\beta }_{\text{t}}-{\beta }_{\text{Al}}\right)\end{array}$ (2)

The coefficient of volume expansion of turpentine is $9\times {10}^{-4}{\left(°\text{C}\right)}^{-1}$.

Substitute $72\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for ${\beta }_{\text{Al}}$, $2.000\text{L}$ for $V$, $60°\text{C}$ for $\Delta T$ and $9\times {10}^{-4}°{\text{C}}^{-1}$ for ${\beta }_t$ in the equation (2).

$\begin{array}{c}\Delta V^{\prime }=\left(2.000\text{L}\right)\left(\frac{{10}^3{\text{cm}}^3}{1\text{L}}\right)\left(9\times {10}^{-4}\left(°{\text{C}}^{-1}\right)-72\times {10}^{-6}\left(°{\text{C}}^{-1}\right)\right)\left(60°\text{C}\right)\\ =99.4{\text{cm}}^3\times \frac{1\text{mL}}{1{\text{cm}}^{\text{3}}}\\ =99.4\text{mL}\end{array}$

Conclusion:

Therefore, the volume of turpentine overflows is $99.4\text{mL}$.

(b)

To determine

The volume of turpentine remaining in the cylinder at $80°\text{C}$.

Answer to Problem 19.21P

The volume of turpentine remaining in the cylinder at $80°\text{C}$ is $2.01\text{L}$.

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is $2.000\text{L}$, the initial temperature is $20°\text{C}$, and the final temperature in the cylinder is $80°\text{C}$.

Write the expression for the new volume of turpentine in the cylinder,

$V_{\text{new}}=V\left(1+{\beta }_{\text{t}}\Delta T\right)$

The coefficient of volume expansion of turpentine is $9\times {10}^{-4}{\left(°\text{C}\right)}^{-1}$ and from the part (a) the temperature difference is $60°\text{C}$.

Substitute $2.000\text{L}$ for $V$, $9\times {10}^{-4}{\left(°\text{C}\right)}^{-1}$ for ${\beta }_{\text{t}}$ and $60°\text{C}$ for $\Delta T$ in the above equation.

$\begin{array}{c}{V}_{\text{new}}=\left(2.000\text{L}\right)\left[1+\left(9\times {10}^{-4}{\left(°\text{C}\right)}^{-1}\right)\left(60°\text{C}\right)\right]\\ =2.01\text{L}\end{array}$

Conclusion:

Therefore, the volume of turpentine remaining in the cylinder at $80°\text{C}$ is $2.01\text{L}$.

(c)

To determine

The distance for the cylinder’s rim does the turpentine surface recede.

Answer to Problem 19.21P

The distance for the cylinder’s rim does the turpentine surface recede is $0.998\text{cm}$.

Explanation of Solution

Given info: The initial volume of aluminum cylinder and turpentine is $2.000\text{L}$, the final temperature is $20°\text{C}$, and the initial temperature in the cylinder is $80°\text{C}$.

Write the expression for the volume of the turpentine in the cylinder after it cools back to $20°\text{C}$,

$V_1=V_{\text{new}}\left(1+{\beta }_{\text{t}}\Delta T^{\prime }\right)$ (3)

Write the expression for temperature difference,

$\Delta T^{\prime }=T_2-T_1$

Here,

$T_2$ is the final temperature.

$T_1$ is the initial temperature.

Substitute $20°\text{C}$ for $T_2$ and $80°\text{C}$ for $T_1$ in the above equation.

$\begin{array}{c}\Delta T^{\prime }=20°\text{C}-80°\text{C}\\ =-60°\text{C}\end{array}$

Thus, the temperature difference is $-60°\text{C}$ and from part (a) the volume of turpentine remaining in the cylinder at $80°\text{C}$ is $2.1\text{L}$.

Substitute $-60°\text{C}$ for $\Delta T^{\prime }$, $2.1\text{L}$ for $V_{\text{new}}$ and $9\times {10}^{-4}°{\text{C}}^{-1}$ for ${\beta }_{\text{t}}$ in the equation (3).

$\begin{array}{c}{V}_1=\left(2.1\text{L}\right)\left(\frac{{10}^3{\text{cm}}^3}{1\text{L}}\right)\left[1+\left(9\times {10}^{-4}°{\text{C}}^{-1}\right)\left(-60°\text{C}\right)\right]\\ =1900.17{\text{cm}}^3\end{array}$

Write the expression for the percentage of cylinder that is empty at $20°\text{C}$,

$\%V=\frac{V-V_1}{V}$

Substitute $1900.17{\text{cm}}^3$ for $V_1$ and $2.000\text{L}$ for $V$ in the above equation.

$\begin{array}{c}\%V=\frac{\left(2.0\text{L}\right)\left(\frac{{10}^3{\text{cm}}^3}{1\text{L}}\right)-1900.17{\text{cm}}^3}{\left(2.0\text{L}\right)\left(\frac{{10}^3{\text{cm}}^3}{1\text{L}}\right)}\\ =4.99\%\end{array}$

The distance for the cylinder’s rim does the turpentine surface recede is,

$4.99\%\left(20.0\text{cm}\right)=0.998\text{cm}$

Conclusion:

Therefore, the distance for the cylinder’s rim does the turpentine surface recede is $0.998\text{cm}$.