Chapter 13, Problem 19P

(a)

To determine

The amount of water spill out of the glass. And compare the result with the previous one.

Answer to Problem 19P

The amount of water spill out of the glass is $\underset{\_}{1.44\text{mL}}$. Less water is spilled than before.

Explanation of Solution

Write the equation to calculate the amount of spilled water.

$\left(\Delta V\right)=\left(\Delta V_{{\text{H}}_{\text{2}}\text{o}}\right)-\left(\Delta V_{\text{glass}}\right)$ (I)

Here, $\left(\Delta V\right)$ is the amount of water spilled out, $\Delta V_{{\text{H}}_{\text{2}}\text{o}}$ change in the volume of water, and $\Delta V_{\text{glass}}$ is the change in volume of the glass.

Write the expression for the change in volume of the water.

$\Delta V_{{\text{H}}_{\text{2}}\text{o}}=V_o{\beta }_{{\text{H}}_{\text{2}}\text{o}}\left(\Delta T\right)$ (II)

Here, $V_o$ is the initial amount of water, $\left(\Delta T\right)$ is the temperature difference, and ${\beta }_{{\text{H}}_{\text{2}}\text{o}}$ is the expansion coefficient of water.

Write the expression for the change in volume of the glass.

$\left(\Delta V_{\text{glass}}\right)=\left[V_o{\beta }_{\text{glass}}\left(\Delta T\right)\right]$ (III)

Use equation (II) and (III) in (I).

$\left(\Delta V\right)=\left[V_o{\beta }_{{\text{H}}_{\text{2}}\text{o}}\left(\Delta T\right)\right]-\left[V_o{\beta }_{glass}\left(\Delta T\right)\right]$

Substitute $3{\alpha }_{\text{glass}}$ for ${\beta }_{\text{glass}}$ and rewrite the above equation.

$\left(\Delta V\right)=V_o\left(\Delta T\right)\left[{\beta }_{{\text{H}}_{\text{2}}\text{o}}-\left(3{\alpha }_{\text{glass}}\right)\right]$ (IV)

Here, ${\alpha }_{glass}$ is the expansion coefficient of glass.

Conclusion:

Substitute, $268.4\text{mL}$ for $V_o$, $\left[\left(32\text{°C}\right)-\left(2\text{°C}\right)\right]$ for $\left(\Delta T\right)$, $207\times {10}^{-6}{\text{K}}^{\text{-1}}$ for ${\beta }_{{\text{H}}_{\text{2}}\text{o}}$, $9.4\times {10}^{-6}{\text{K}}^{\text{-1}}$ for ${\alpha }_{glass}$ in equation (I)

$\begin{array}{c}\left(\Delta V\right)=\left(268.4\text{mL}\right)\left[\left(32\text{°C}\right)-\left(2\text{°C}\right)\right]\left[\left(207\times {10}^{-6}{\text{K}}^{\text{-1}}\right)-\left(3\left(9.4\times {10}^{-6}{\text{K}}^{\text{-1}}\right)\right)\right]\\ =\left(268.4\text{mL}\right)\left[\left(32+273\right)\text{K}-\left(2+273\right)\text{K}\right]\left[\left(207\times {10}^{-6}{\text{K}}^{\text{-1}}\right)-\left(3\left(9.4\times {10}^{-6}{\text{K}}^{\text{-1}}\right)\right)\right]\\ =1.44\text{mL}\end{array}$

Refer the answer of problem No 19.

The amount of spilled out water is $1.67\text{mL}$.

Thus, the amount of water spill out of the glass is $\underset{\_}{1.44\text{mL}}$ which is Less water is spilled than before.

(b)

To determine

The percentage of spilled water.

Answer to Problem 19P

The percentage of spilled water is $\underset{\_}{-14\%}$.

Explanation of Solution

Write the equation of the percentage of spilled water.

$\left(\Delta V\right)\%=\left(\frac{V_f-V_i}{V_i}\right)\times 100\%$ (V)

Here, $\left(\Delta V\right)\%$ is the percentage of spilled, $V_i,V_f$ are the initial and final volume.

Conclusion:

Substitute, $1.44\text{mL}$ for $V_f$, $1.67\text{mL}$ for $V_i$ in equation (V).

$\begin{array}{c}\left(\Delta V\right)\%=\left(\frac{1.44-1.67}{1.67}\right)\times 100\%\\ =-14\%\end{array}$

Negative sign indicates less water spilled.

Thus, the percentage of spilled water is $\underset{\_}{-14\%}$.