Chapter 10, Problem 21P

A hollow aluminum cylinder 20.0 cm deep has an internal capacity of 2.000 L at 20.0°C. It is completely filled with turpentine at 20.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 80.0°C. (a) How much turpentine overflows? (b) What is the volume of the turpentine remaining in the cylinder at 80.0°C? (c) If the combination with this amount of turpentine is then cooled back to 20.0°C, how far below the cylinder’s rim does the turpentine’s surface recede?

To determine

The volume of turpentine that has overflowed.

Answer to Problem 21P

The volume of turpentine that has overflowed is 99 mL.

Explanation of Solution

Given Info: The length of the hollow cylinder (l) is 20.0 cm. The capacity of the cylinder ( $V_0$) is 2.000 L at $T_1={20.0}^{\text{ο}}\text{C}$. The temperature is changed to $T_2={80.0}^{\text{ο}}\text{C}$. The co-efficient of volume expansion of turpentine is ${\beta }_{\text{T}}=9.0\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. The co-efficient of volume expansion of aluminum is ${\beta }_{\text{Al}}=72\times {10}^{-6}{\text{/}}^{\text{ο}}\text{C}$.

Formula to calculate the volume of turpentine that has overflowed is,

$V_{of}=\Delta V_{\text{T}}-\Delta V_{\text{Al}}$      (I)

• $\Delta V_{\text{T}}$ is the change in volume of turpentine.
• $\Delta V_{\text{Al}}$ is the change in volume of Aluminum.

Formula to calculate the change in volume of turpentine is,

$\Delta V_{\text{T}}=V_0{\beta }_{\text{T}}\left(T_2-T_1\right)$      (II)

Formula to calculate the change in volume of aluminum is,

$\Delta V_{\text{Al}}=V_0{\beta }_{\text{Al}}\left(T_2-T_1\right)$      (III)

Substitute Equations (II) and (III) in (I).

$V_{of}=V_0\left(T_2-T_1\right)\left({\beta }_{\text{T}}-{\beta }_{\text{Al}}\right)$

Substitute 2.000 L for $V_0$, ${20.0}^{\text{ο}}\text{C}$ for $T_1$, ${80.0}^{\text{ο}}\text{C}$ for $T_2$, $9.0\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$ for ${\beta }_{\text{T}}$ and $72\times {10}^{-6}{\text{/}}^{\text{ο}}\text{C}$ for ${\beta }_{\text{Al}}$ in the above expression to get $V_{of}$.

$\begin{array}{c}{V}_{of}=\left(2.000\text{L}\right)\left[\left({80.0}^{\text{ο}}\text{C}\right)-\left({20.0}^{\text{ο}}\text{C}\right)\right]\left[\left(9.0\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}\right)-\left(72\times {10}^{-6}{\text{/}}^{\text{ο}}\text{C}\right)\right]\\ =99\text{mL}\end{array}$

Conclusion:

The volume of turpentine that has overflowed is 99 mL.

To determine

The volume of turpentine remaining.

Answer to Problem 21P

The volume of turpentine remaining is 2.009 L.

Explanation of Solution

Given Info: The length of the hollow cylinder (l) is 20.0 cm. The capacity of the cylinder ( $V_0$) is 2.000 L at $T_1={20.0}^{\text{ο}}\text{C}$. The temperature is changed to $T_2={80.0}^{\text{ο}}\text{C}$. The co-efficient of volume expansion of turpentine is ${\beta }_{\text{T}}=9.0\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. The co-efficient of volume expansion of aluminum is ${\beta }_{\text{Al}}=72\times {10}^{-6}{\text{/}}^{\text{ο}}\text{C}$.

Volume of turpentine remaining ( $V_r$) is equal to the volume of aluminum cylinder. Therefore, the formula to calculate the remaining volume is,

$V_r=V_0\left[1+{\beta }_{\text{Al}}\left(T_2-T_1\right)\right]$

Substitute 2.000 L for $V_0$, ${20.0}^{\text{ο}}\text{C}$ for $T_1$, ${80.0}^{\text{ο}}\text{C}$ for $T_2$ and $72\times {10}^{-6}{\text{/}}^{\text{ο}}\text{C}$ for ${\beta }_{\text{Al}}$ in the above expression to get $V_r$

$\begin{array}{c}{V}_r=\left(2.000\text{L}\right)\left[1+\left(72\times {10}^{-6}{\text{/}}^{\text{ο}}\text{C}\right)\left[\left({80.0}^{\text{ο}}\text{C}\right)-\left({20.0}^{\text{ο}}\text{C}\right)\right]\right]\\ =2.009\text{L}\end{array}$

Conclusion:

The volume of turpentine remaining is 2.009 L.

To determine

The distance between the rim and the surface of turpentine (h).

Answer to Problem 21P

The distance between the rim and the surface of turpentine (h) is 1 cm.

Explanation of Solution

Given Info: The length of the hollow cylinder (l) is 20.0 cm. The capacity of the cylinder ( $V_0$) is 2.000 L at $T_1={20.0}^{\text{ο}}\text{C}$. The temperature is changed to $T_2={80.0}^{\text{ο}}\text{C}$. The co-efficient of volume expansion of turpentine is ${\beta }_{\text{T}}=9.0\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$. The co-efficient of volume expansion of aluminum is ${\beta }_{\text{Al}}=72\times {10}^{-6}{\text{/}}^{\text{ο}}\text{C}$.

Formula to calculate the volume of turpentine remaining at ${20.0}^{\text{ο}}\text{C}$ is,

$V_r^{\text{'}}=V_r\left[1+{\beta }_{\text{T}}\left(T_1-T_2\right)\right]$      (IV)

The fraction of empty volume is,

$f=1-\frac{V_r^{\text{'}}}{V_0}$      (V)

The fraction of empty volume is also equal to the ratio of the distance between the rim and the surface of turpentine and the length of the cylinder.

$f=\frac{h}{l}$      (VI)

From Equations (IV), (V) and (VI),

$h=l\left\{1-\frac{V_r\left[1+{\beta }_{\text{T}}\left(T_1-T_2\right)\right]}{V_0}\right\}$

Substitute 20.0 cm for l, 2.009 L for $V_r$, 2.000 L for $V_0$, ${20.0}^{\text{ο}}\text{C}$ for $T_1$, ${80.0}^{\text{ο}}\text{C}$ for $T_2$ and $9.0\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}$ for ${\beta }_{\text{T}}$ in the above expression to get h.

$\begin{array}{c}h=\left(20.0\text{cm}\right)\left\{1-\frac{\left(2.009\text{L}\right)\left[1+\left(9.0\times {10}^{-4}{\text{/}}^{\text{ο}}\text{C}\right)\left[\left({20.0}^{\text{ο}}\text{C}\right)-\left({80.0}^{\text{ο}}\text{C}\right)\right]\right]}{2.000\text{L}}\right\}\\ =1\text{cm}\end{array}$

Conclusion:

The distance between the rim and the surface of turpentine (h) is 1 cm.