Chapter 2, Problem 39P

Use the coefficient of volume expansion to estimate the density of water as it is heated from $60°F$ to $130°F$ at 1 atm. Compare your result with the actual density (from the appendices).

To determine

The density of water.

Answer to Problem 39P

The density calculated is $986.78\text{kg}/{\text{m}}^{\text{3}}$.

The density calculated and the actual density from appendix is nearly same.

Explanation of Solution

Given information:

The water is heated from $60°\text{F}$ to $130°\text{F}$ at $1\text{atm}$.

Write the expression for the average temperature.

  $T_{avg}=\frac{T_1+T_2}{2}$....... (I)

Here, initial temperature is $T_1$. Final temperature is $T_2$ and the average temperature is $T_{avg}$.

Write the expression for the density of water.

  $\Delta \rho =-\rho \beta \Delta T$....... (II)

Here, the change in density is $\Delta \rho$, coefficient of volume expansion is $\beta$, the temperature difference is $\Delta T,$ and the initial density is $\rho$.

Write the expression for temperature difference.

  $\Delta T=T_2-T_1$....... (III)

Here, initial temperature is $T_1$. Final temperature is $T_2$ and the average temperature is $T_{avg}$.

Write the expression for final density.

  $\begin{array}{l}\Delta \rho ={\rho }_2-\rho \\ {\rho }_2=\Delta \rho +\rho \end{array}$....... (IV)

Write the formula for interpolation of two variables.

  ${\rho }_2{}^{\prime }=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$....... (V)

Here, the temperature is denoted by variables $x$, density is denoted by variables $y,$ and the unknown density is ${\rho }_2{}^{\prime }$.

Calculation:

Substitute $60°\text{F}$ for $T_1$ and $130°\text{F}$ for $T_2$ in Equation (I).

  $\begin{array}{c}{T}_{avg}=\frac{60°\mathrm{F}+130°\text{F}}{2}\\ =\frac{\left(\frac{60°\mathrm{F}-32}{1.8}\right)°\text{C}+\left(\frac{130°\text{F}-32}{1.8}\right)°\text{C}}{2}\\ =\frac{15.56°\text{C}+54.44°\text{C}}{2}\\ =35°\text{C}\end{array}$

Refer to Table A-3, "Properties of saturated water" to obtain the value of $\beta$ as $0.00034{\text{K}}^{-1}$ at $35°\text{C}$.

Substitute $60°\text{F}$ for $T_1$ and $130°\text{F}$ for $T_2$ in Equation (III).

  $\begin{array}{c}\Delta T=130°\text{F}-60°\text{F}\\ \text{=}\left(\frac{130°\text{F}-32}{1.8}\right)°\text{C}-\left(\frac{60°\mathrm{F}-32}{1.8}\right)°\text{C}\\ =\left(54.44°\text{C}+273\right)\text{K}-\left(\text{15}\text{.56}°\text{C}+273\right)\text{K}\\ \text{=38}\text{.88}\text{K}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.00034{\text{K}}^{-1}$ for $\beta$ and $\text{38}\text{.88}\text{K}$ for $\Delta T$ in Equation (II).

  $\begin{array}{c}\Delta \rho =-\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.00034{\text{K}}^{-1}\right)\left(38.88\text{K}\right)\\ =-\left(13.219\text{kg}/{\text{m}}^{\text{3}}\right)\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $-\left(13.219\text{kg}/{\text{m}}^{\text{3}}\right)$ for $\Delta \rho$ in Equation (IV).

  $\begin{array}{c}{\rho }_2=-13.219\text{kg}/{\text{m}}^{\text{3}}+1000\text{kg}/{\text{m}}^{\text{3}}\\ =986.78\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Refer to Table $\text{A-3}$, "Properties of saturated water" to obtain the values of $x_2$ as $54.44°\text{C}$, $x_1$ as $50°\text{C}$, $x_3$ as $55°\text{C}$.

Refer to Table $\text{A-3}$, "Properties of saturated water" to obtain the values of $y_1$ as $988.1\text{kg}/{\text{m}}^{\text{3}}$, $y_3$ as $985.2\text{kg}/{\text{m}}^{\text{3}}$.

Prepare the table for temperature and density of water.

Temperature, $°\text{F}$	Density, $\text{kg}/{\text{m}}^{\text{3}}$
$50\left(x_1\right)$	$988.1\left(y_1\right)$
$54.44\left(x_2\right)$	$?\left({\rho }_2{}^{\prime }\right)$
$55\left(x_3\right)$	$985.2\left(y_3\right)$

Substitute $50°\text{C}$ for $x_1$, $54.44°\text{C}$ for $x_2$, $55°\text{C}$ for $x_3$, $988.1\text{kg}/{\text{m}}^{\text{3}}$ for $y_1$ and $985.2\text{kg}/{\text{m}}^{\text{3}}$ for $y_3$in Equation (V).

  $\begin{array}{c}{\rho }_2{}^{\prime }=\frac{\left(985.2\text{kg}/{\text{m}}^{\text{3}}-988.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(54.44°\text{C}-50°\text{C}\right)}{\left(55°\text{C}-50°\text{C}\right)}+988.1\text{kg}/{\text{m}}^3\\ =\left(\left(-2.9\text{kg}/{\text{m}}^{\text{3}}\right)\frac{4.5°\text{C}}{5°\text{C}}\right)+988.1\text{kg}/{\text{m}}^3\\ =-2.61\text{kg}/{\text{m}}^{\text{3}}+988.1\text{kg}/{\text{m}}^3\\ =985.49\text{kg}/{\text{m}}^3\end{array}$

Conclusion:

The density calculated is $986.78\text{kg}/{\text{m}}^{\text{3}}$.

The density calculated and the actual density from appendix is nearly same.