Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

Answer to Problem 21P

The amount of turpentine overflows is $\underset{\_}{99.4{\text{cm}}^{\text{3}}}$.

Explanation of Solution

Write the equation of volume that overflows.

  $\left(\Delta V\right)=\left(\beta -3\alpha \right)V_i\left(T_2-T_1\right)$                                                                              (I)

Here, $\left(\Delta V\right)$ is the amount of overflow liquid, $\beta$ is the volume expansion coefficient, $\alpha$ is the linear expansion coefficient, $V_i$ is the initial volume, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Conclusion:

Substitute, $9.00\times {10}^{-4}\text{/°C}$ for $\beta$, $24.0\times {10}^{-6}\text{/°C}$ for $\alpha$, $2.000\text{L}$ for $V_i$, $20.0\text{°C}$ for $T_1$ and $80.0\text{°C}$ for $T_2$ in equation (I) to find $\left(\Delta V\right)$.

  $\begin{array}{c}\left(\Delta V\right)=\left(\left[9.00\times {10}^{-4}\text{/°C}\right]-3\left[24.0\times {10}^{-6}\text{/°C}\right]\right)\left(2.000\text{L}\right)\left(80.0\text{°C}-20.0\text{°C}\right)\\ =\left(\left[9.00\times {10}^{-4}\text{/°C}\right]-3\left[24.0\times {10}^{-6}\text{/°C}\right]\right)\left(2.000\text{L}\right)\left(60.0\text{°C}\right)\\ =99.4\text{L}\\ \text{=99}{\text{.4cm}}^3\end{array}$

Thus, the amount of turpentine overflows is $\underset{\_}{99.4{\text{cm}}^{\text{3}}}$.

(b)

To determine

The remaining volume of the turpentine in the cylinder.

Answer to Problem 21P

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

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

  $V_r=V_i+3\alpha V_i\left(T_2-T_1\right)$                                                                                   (II)

Here, $V_r$ is the remaining volume of liquid.

Conclusion:

Substitute, $24.0\times {10}^{-6}\text{/°C}$ for $\alpha$, $2.000\text{L}$ for $V_i$, $20.0\text{°C}$ for $T_1$ and $80.0\text{°C}$ for $T_2$ in equation (I) to find $V_r$.

  $\begin{array}{c}{V}_r=\left(2.000\text{L}\right)+3\left(24.0\times {10}^{-6}\text{/°C}\right)\left(2.000\text{L}\right)\left(80.0\text{°C}-20.0\text{°C}\right)\\ =\left(2.000\text{L}\right)+3\left(24.0\times {10}^{-6}\text{/°C}\right)\left(2.000\text{L}\right)\left(60.0\text{°C}\right)\\ =2.01\text{L}\end{array}$

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

(c)

To determine

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

Answer to Problem 21P

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

Explanation of Solution

Write the equation for the distance.

  $d=ph$                                                                                      (III)

Here, $d$ is the distance, $p$ is the percentage of the empty cylinder and $h$ is the depth of the cylinder.

Write the expression for the percentage.

  $p=\left(\frac{V_i-V\text{'}}{V_i}\right)\times 100\%$                                                                              (IV)

Here, $V\text{'}$ is the volume of the liquid in the cylinder after it cools back to $20\text{°C}$.

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

  $V\text{'}=V\left(1+\beta \left(T_2-T_1\right)\right)$                                                                            (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

  $d=\left[\left(\frac{V_i-\left[V\left(1+\beta \left(T_2-T_1\right)\right)\right]}{V_i}\right)\times 100\%\right]h$                                   (VI)

Conclusion:

Substitute, $9.00\times {10}^{-4}\text{/°C}$ for $\beta$, $2.01\text{L}$ for $V$, $2.000\text{L}$ for $V_i$, $80.0\text{°C}$ for $T_1$, $20.0\text{cm}$ for $h$ and $20.0\text{°C}$ for $T_2$ in equation (I) to find $\left(\Delta V\right)$.

  $\begin{array}{c}d=\left[\left(\frac{\left(2.000\text{L}\right)-\left[\left(2.01\text{L}\right)\left(1+\left(9.00\times {10}^{-4}\text{/°C}\right)\left(20.0\text{°C}-80.0\text{°C}\right)\right)\right]}{\left(2.000\text{L}\right)}\right)\times 100\%\right]\left(20.0\text{cm}\right)\\ =\left(4.99\%\right)\left(20.0\text{cm}\right)\\ =0.998\text{cm}\end{array}$

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