Chapter 16, Problem 19P

(a)

To determine

The amount of turpentine overflows when the turpentine and aluminum cylinder are warmed slowly.

Answer to Problem 19P

The amount of turpentine overflows when the turpentine and aluminum cylinder are warmed slowly is $\underset{\_}{99.4{\text{cm}}^3}$.

Explanation of Solution

Write the expression for change in volume when the temperature rises by $\Delta T$.

    $\Delta V=\beta V_i\Delta T$        (I)

Here, $\Delta V$ is the increased volume, $\beta$ is the average coefficient of volume expansion, $V_i$ is the initial volume, and $\Delta T$ is the change in temperature.

From the equation (I), write the expression for changing volume of the cylinder and the turpentine.

    $\Delta V={\beta }_t{V}_t\Delta T-{\beta }_{\text{Al}}{V}_{\text{Al}}\Delta T$        (II)

Here, $V_{\text{Al}}$ is the volume of the aluminum cylinder, $V_t$ is the volume of the turpentine, ${\beta }_t$ is the average coefficient of volume expansion of turpentine, ${\beta }_{\text{Al}}$ is the average coefficient of volume expansion of Aluminum, and $\Delta T$ is the changing temperature.

Substitute, $V_i$ for $V_t\text{and}{V}_{\text{Al}}$, and $3{\alpha }_{\text{Al}}$ for ${\beta }_{\text{Al}}$ in the equation (II).

    $\begin{array}{c}\Delta V={\beta }_t{V}_i\Delta T-3{\alpha }_{\text{Al}}{V}_i\Delta T\\ =\left({\beta }_t-3{\alpha }_{\text{Al}}\right)V_i\Delta T\end{array}$        (III)

Conclusion:

Substitute, $9.00\times {10}^{-4}{\left(°\text{C}\right)}^{-1}$ for $\beta$, $24.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for $\alpha$, $2000{\text{cm}}^3$ for $V_i$, and $60.0°\text{C}$ for $\Delta T$ in the equation (III), to find $\Delta V$.

    $\begin{array}{c}\Delta V=\left[9.00\times {10}^{-4}{\left(°\text{C}\right)}^{-1}-3\left(24.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\right)\right]\left(2000{\text{cm}}^3\right)\left(60.0°\text{C}\right)\\ =99.4{\text{cm}}^3\end{array}$

Therefore, the amount of turpentine overflows when the turpentine and aluminum cylinder are warmed slowly is $\underset{\_}{99.4{\text{cm}}^3}$.

(b)

To determine

The volume of the turpentine remaining in the cylinder.

Answer to Problem 19P

The volume of the turpentine remaining in the cylinder is $\underset{\_}{2.01\text{L}}$.

Explanation of Solution

The volume of the aluminum cylinder at $80.0°\text{C}$ is equal to the volume of the turpentine remaining in the cylinder at $80.0°\text{C}$.

    $\begin{array}{c}{V}_t=V_{\text{Al}}\\ =V_{\text{Al}i}+{\beta }_{\text{Al}}{V}_{\text{Al}i}\Delta T\end{array}$        (IV)

Here, $V_{\text{Al}i}$ is the initial volume of the aluminum cylinder.

Substitute, $3{\alpha }_{\text{Al}}$ for ${\beta }_{\text{Al}}$ in the equation (IV).

    $\begin{array}{c}{V}_t=V_{\text{Al}i}+\left(3{\alpha }_{\text{Al}}\right)V_{\text{Al}i}\Delta T\\ =V_{\text{Al}i}\left[1+3{\alpha }_{\text{Al}}\Delta T\right]\end{array}$        (V)

Conclusion:

Substitute, $24.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for $\alpha$, $2000{\text{cm}}^3$ for $V_{\text{Al}i}$, and $60.0°\text{C}$ for $\Delta T$ in the equation (V), to find $V_t$.

    $\begin{array}{c}{V}_t=2000{\text{cm}}^3\left[1+3\left(24.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\right)\left(60.0°\text{C}\right)\right]\\ =2008.64{\text{cm}}^3\\ =2.01\text{L}\end{array}$

Therefore, the volume of the turpentine remaining in the cylinder is $\underset{\_}{2.01\text{L}}$.

(c)

To determine

The empty height of the cylinder above the turpentine if the combination with turpentine cooled back to $20.0°\text{C}$.

Answer to Problem 19P

The empty height of the cylinder above the turpentine if the combination with turpentine cooled back to $20.0°\text{C}$ is $\underset{\_}{0.998\text{cm}}$.

Explanation of Solution

Write the expression for volume of the turpentine in the cylinder after it cools back.

    $V_t=V_{ti}+{\beta }_t{V}_{ti}\Delta T$        (VI)

Simplify the equation (IV).

    $V_t=V_{ti}\left(1+{\beta }_t\Delta T\right)$        (VII)

Conclusion:

Substitute, $9.00\times {10}^{-4}{\left(°\text{C}\right)}^{-1}$ for $\beta$, $2008.64{\text{cm}}^3$ for $V_{ti}$, and $60.0°\text{C}$ for $\Delta T$ in the equation (VII), to find $V_t$.

    $\begin{array}{c}{V}_t=\left(2008.64{\text{cm}}^3\right)\left[1+\left(9.00\times {10}^{-4}{\left(°\text{C}\right)}^{-1}\right)\left(60.0°\text{C}\right)\right]\\ =1900.17{\text{cm}}^3\end{array}$

The percentage of cylinder that is empty at $20.0°\text{C}$ can be calculated as follows.

    $\begin{array}{c}\frac{V_{\text{Al}i}-V_{ti}}{V_{\text{Al}i}}=\frac{2000{\text{cm}}^3-1900.17{\text{cm}}^3}{2000{\text{cm}}^3}\times 100\\ =4.99\%\end{array}$

Then the empty height of the cylinder above the turpentine can calculated as follows,

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

Therefore, the empty height of the cylinder above the turpentine if the combination with turpentine cooled back to $20.0°\text{C}$ is $\underset{\_}{0.998\text{cm}}$.