Chapter 19, Problem 57AP

A liquid has a density ρ. (a) Show that the fractional change in density for a change in temperature ΔT is Δρ/ρ = −β ΔT. (b) What does the negative sign signify? (c) Fresh water has a maximum density of 1.000 0 g/cm³ at 4.0°C. At 10.0°C, its density is 0.999 7 g/cm³. What is β for water over this temperature interval? (d) At 0°C, the density of water is 0.999 9 g/cm³. What is the value for β over the temperature range 0°C to 4.00°C?

To determine

To show: The fraction change in density for a change in temperature is given by $\frac{\Delta \rho }{\rho }=-\beta \Delta T$.

Answer to Problem 57AP

The fraction change in density for a change in temperature is $\frac{\Delta \rho }{\rho }=-\beta \Delta T$.

Explanation of Solution

Given info: The density of the liquid is $\rho$, the change in the temperature is $\Delta T$, the maximum density of fresh water at $4.0°\text{C}$ is $1.0000\text{g}/{\text{cm}}^3$, the density of fresh water at $10.0°\text{C}$ is $0.9997\text{g}/{\text{cm}}^3$ and the density of water at $0°\text{C}$ is $0.9999\text{g}/{\text{cm}}^3$.

The expression for the coefficient of the volume expansion is,

$\beta \Delta T=\frac{\Delta V}{V}$

Here,

$\beta$ is the coefficient of the volume expansion.

The formula for the density is,

$\rho =\frac{m}{V}$

Here,

$m$ is the mass of the water.

$V$ is the volume of the water.\

Differentiate the above equation with respect to volume.

$d\rho =-\frac{m}{V^2}dV$

For very small change in volume and density the expression is,

$\begin{array}{c}\Delta \rho =-\frac{m}{V^2}\Delta V\\ =-\frac{m}{V}\frac{\Delta V}{V}\end{array}$

Substitute $\rho$ for $\frac{m}{V}$ and $\beta \Delta T$ for $\frac{\Delta V}{V}$ in above equation.

$\begin{array}{c}\Delta \rho =-\rho \beta \Delta T\\ \frac{\Delta \rho }{\rho }=-\beta \Delta T\end{array}$

Conclusion:

Therefore, the fraction change in density for a change in temperature is $\frac{\Delta \rho }{\rho }=-\beta \Delta T$.

To determine

The significance of the negative sign.

Answer to Problem 57AP

the negative sign signifies that when the temperature increases, the density decreases.

Explanation of Solution

Given info: The density of the liquid is $\rho$, the change in the temperature is $\Delta T$, the maximum density of fresh water at $4.0°\text{C}$ is $1.0000\text{g}/{\text{cm}}^3$, the density of fresh water at $10.0°\text{C}$ is $0.9997\text{g}/{\text{cm}}^3$ and the density of water at $0°\text{C}$ is $0.9999\text{g}/{\text{cm}}^3$.

The expression for the fraction of the density is,

$\frac{\Delta \rho }{\rho }=-\beta \Delta T$ (1)

The negative sign in the above expression shows that as the temperature increases, the density of the water is decreases.

Conclusion:

Therefore, the negative sign signifies that when the temperature increases, the density decreases.

To determine

The value of $\beta$ for the temperature interval $4.0°\text{C}$ to $10.0°\text{C}$.

Answer to Problem 57AP

The value of $\beta$ for the temperature interval $4.0°\text{C}$ to $10.0°\text{C}$ is $5\times {10}^{-5}{\left(°\text{C}\right)}^{-1}$.

Explanation of Solution

Given info: The density of the liquid is $\rho$, the change in the temperature is $\Delta T$, the maximum density of fresh water at $4.0°\text{C}$ is $1.0000\text{g}/{\text{cm}}^3$, the density of fresh water at $10.0°\text{C}$ is $0.9997\text{g}/{\text{cm}}^3$ and the density of water at $0°\text{C}$ is $0.9999\text{g}/{\text{cm}}^3$.

Recall the equation (1) and rearrange for $\beta$.

$\begin{array}{c}\frac{\Delta \rho }{\rho }=-\beta \Delta T\\ \beta =-\frac{\Delta \rho }{\rho \Delta T}\end{array}$

Substitute $\left(0.9997\text{g}/{\text{cm}}^3-1.0000\text{g}/{\text{cm}}^3\right)$ for $\Delta \rho$, $1.0000\text{g}/{\text{cm}}^3$ for $\rho$ and $\left(10.0°\text{C}-4.0°\text{C}\right)$ for $\Delta T$ in above equation to find $\beta$.

$\begin{array}{c}\beta =-\frac{\left(0.9997\text{g}/{\text{cm}}^3-1.0000\text{g}/{\text{cm}}^3\right)}{\left(1.0000\text{g}/{\text{cm}}^3\right)\left(\left(10.0°\text{C}-4.0°\text{C}\right)\right)}\\ =5\times {10}^{-5}°{\text{C}}^{-1}\end{array}$

Conclusion:

Therefore, the value of $\beta$ for temperature interval $4.0°\text{C}$ to $10.0°\text{C}$ is $5\times {10}^{-5}{\left(°\text{C}\right)}^{-1}$.

To determine

The value of $\beta$ for the temperature interval $0°\text{C}$ to $4.0°\text{C}$.

Answer to Problem 57AP

The value of $\beta$ for temperature interval $0°\text{C}$ to $4.0°\text{C}$ is $-2.5\times {10}^{-5}{\left(°\text{C}\right)}^{-1}$.

Explanation of Solution

Given info: The density of the liquid is $\rho$, the change in the temperature is $\Delta T$, the maximum density of fresh water at $4.0°\text{C}$ is $1.0000\text{g}/{\text{cm}}^3$, the density of fresh water at $10.0°\text{C}$ is $0.9997\text{g}/{\text{cm}}^3$ and the density of water at $0°\text{C}$ is $0.9999\text{g}/{\text{cm}}^3$.

The expression for $\beta$ is,

$\begin{array}{c}\frac{\Delta \rho }{\rho }=-\beta \Delta T\\ \beta =-\frac{\Delta \rho }{\rho \Delta T}\end{array}$

Substitute $\left(1.0000\text{g}/{\text{cm}}^3-0.9999\text{g}/{\text{cm}}^3\right)$ for $\Delta \rho$, $1.0000\text{g}/{\text{cm}}^3$ for $\rho$ and $\left(4.0°\text{C}-0°\text{C}\right)$ for $\Delta T$ in above equation to find $\beta$.

$\begin{array}{c}\beta =-\frac{\left(1.0000\text{g}/{\text{cm}}^3-0.9999\text{g}/{\text{cm}}^3\right)}{\left(1.0000\text{g}/{\text{cm}}^3\right)\left(\left(4.0°\text{C}-0.0°\text{C}\right)\right)}\\ =-2.5\times {10}^{-5}°{\text{C}}^{-1}\end{array}$

Conclusion:

Therefore, the value of $\beta$ for the temperature interval $0°\text{C}$ to $4.0°\text{C}$ is $-2.5\times {10}^{-5}{\left(°\text{C}\right)}^{-1}$.