Chapter 9, Problem 17P

(a)

To determine

Measure the density of a cooking oil and express the findings in $\text{kg}/{\text{m}}^3,\text{slugs}/{\text{ft}}^3,\text{and}\text{lbm}/{\text{ft}}^3$.

Answer to Problem 17P

The density of cooking oil in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{918{\text{kg/m}}^{\text{3}}}$.

The density of cooking oil in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.781{\text{slugs/ft}}^3}$.

The density of cooking oil in ${\text{lbm/ft}}^3$ is $\underset{\_}{57.30{\text{lbm/ft}}^3}$.

The specific gravity of the cooking oil is $\underset{\_}{0.918}$.

Explanation of Solution

Find the density of the cooking oil using the procedure as follows:

• Measure the weight of the empty test container.
• Take the sample of cooking oil in a container. Measure the weight of container with oil.
• Find the mass of the oil by subtracting the weight of oil container with weight of empty container.
• Measure the volume of the liquid using a graduated cylinder.
• Find the density of the oil using the value of mass and volume.
• $\text{Density}=\frac{\text{mass}}{\text{Volume}}$
• The density of cooking oil is ${\rho }_{oil}=918{\text{kg/m}}^{\text{3}}$.

Therefore, the density of cooking oil in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{918{\text{kg/m}}^{\text{3}}}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{slugs}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{oil}=918{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{slug}}{\text{14}\text{.594}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =1.781{\text{slugs/ft}}^3\end{array}$

Therefore, the density of cooking oil in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.781{\text{slugs/ft}}^3}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{lbm}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{oil}=918{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{lbm}}{\text{0}\text{.45359}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =57.30{\text{lbm/ft}}^3\end{array}$

Therefore, the density of cooking oil in ${\text{lbm/ft}}^3$ is $\underset{\_}{57.30{\text{lbm/ft}}^3}$.

Find the specific gravity $\left(G_{oil}\right)$ of cooking oil using the relation.

$G_{oil}=\frac{{\rho }_{oil}}{{\rho }_w}$

Substitute $918{\text{kg/m}}^{\text{3}}$ for ${\rho }_{oil}$ and $1,000{\text{kg/m}}^3$ for ${\rho }_w$.

$\begin{array}{c}{G}_{oil}=\frac{918}{1,000}\\ =0.918\end{array}$

Therefore, the specific gravity of the cooking oil is $\underset{\_}{0.918}$.

(b)

To determine

Measure the density of SAE 10W-40 engine oil and express the findings in $\text{kg}/{\text{m}}^3,\text{slugs}/{\text{ft}}^3,\text{and}\text{lbm}/{\text{ft}}^3$.

Answer to Problem 17P

The density of SAE 10W-40 engine oil in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{865{\text{kg/m}}^{\text{3}}}$.

The density of SAE 10W-40 engine oil in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.678{\text{slugs/ft}}^3}$.

The density of SAE 10W-40 engine oil in ${\text{lbm/ft}}^3$ is $\underset{\_}{53.99{\text{lbm/ft}}^3}$.

The specific gravity of the SAE 10W-40 engine oil is $\underset{\_}{0.865}$.

Explanation of Solution

Find the density of the SAE 10W-40 engine oil using the procedure as follows:

• Measure the weight of the empty test container.
• Take the sample of SAE 10W-40 engine oil in a container. Measure the weight of container with oil.
• Find the mass of the oil by subtracting the weight of oil container with weight of empty container.
• Measure the volume of the liquid using a graduated cylinder.
• Find the density of the oil using the value of mass and volume.
• $\text{Density}=\frac{\text{mass}}{\text{Volume}}$
• The density of SAE 10W-40 engine oil is ${\rho }_{\text{SAE}\text{10W-40}}=865{\text{kg/m}}^{\text{3}}$

Therefore, the density of SAE 10W-40 engine oil in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{865{\text{kg/m}}^{\text{3}}}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{slugs}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{SAE 10W-40}}=865{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{slug}}{\text{14}\text{.594}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =1.678{\text{slugs/ft}}^3\end{array}$

Therefore, the density of SAE 10W-40 engine oil in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.678{\text{slugs/ft}}^3}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{lbm}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{SAE 10W-40}}=865{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{lbm}}{\text{0}\text{.45359}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =53.99{\text{lbm/ft}}^3\end{array}$

Therefore, the density of SAE 10W-40 engine oil in ${\text{lbm/ft}}^3$ is $\underset{\_}{53.99{\text{lbm/ft}}^3}$.

Find the specific gravity $\left(G_{\text{SAE 10W-40}}\right)$ of SAE 10W-40 engine oil using the relation.

$G_{\text{SAE 10W-40}}=\frac{{\rho }_{\text{SAE 10W-40}}}{{\rho }_w}$

Substitute $865{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{SAE 10W-40}}$ and $1,000{\text{kg/m}}^3$ for ${\rho }_w$.

$\begin{array}{c}{G}_{\text{SAE 10W-40}}=\frac{865}{1,000}\\ =0.865\end{array}$

Therefore, the specific gravity of the SAE 10W-40 engine oil is $\underset{\_}{0.865}$.

(c)

To determine

Measure the density of water and express the findings in $\text{kg}/{\text{m}}^3,\text{slugs}/{\text{ft}}^3,\text{and}\text{lbm}/{\text{ft}}^3$.

Answer to Problem 17P

The density of water in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{999.94{\text{kg/m}}^{\text{3}}}$.

The density of water in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.94{\text{slugs/ft}}^3}$.

The density of water in ${\text{lbm/ft}}^3$ is $\underset{\_}{62.41{\text{lbm/ft}}^3}$.

The specific gravity of the water is $\underset{\_}{1}$.

Explanation of Solution

Find the density of the water using the procedure as follows:

• Measure the weight of the empty test container.
• Take the sample of water in a container. Measure the weight of container with water.
• Find the mass of the water by subtracting the weight of water container with weight of empty container.
• Measure the volume of the liquid using a graduated cylinder.
• Find the density of the water using the value of mass and volume.
• $\text{Density}=\frac{\text{mass}}{\text{Volume}}$
• The density of water is ${\rho }_{\text{water}}=999.94{\text{kg/m}}^{\text{3}}$.

Therefore, the density of water in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{999.94{\text{kg/m}}^{\text{3}}}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{slugs}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{water}}=999.94{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{slug}}{\text{14}\text{.594}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =1.94{\text{slugs/ft}}^3\end{array}$

Therefore, the density of water in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.94{\text{slugs/ft}}^3}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{lbm}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{water}}=999.94{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{lbm}}{\text{0}\text{.45359}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =62.41{\text{lbm/ft}}^3\end{array}$

Therefore, the density of water in ${\text{lbm/ft}}^3$ is $\underset{\_}{62.41{\text{lbm/ft}}^3}$.

Find the specific gravity $\left(G_{\text{water}}\right)$ of water using the relation.

$G_{\text{water}}=\frac{{\rho }_{\text{water}}}{{\rho }_w}$

Substitute $999.94{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{water}}$ and $1,000{\text{kg/m}}^3$ for ${\rho }_w$.

$\begin{array}{c}{G}_{\text{water}}=\frac{999.94}{1,000}\\ =1\end{array}$

Therefore, the specific gravity of the water is $\underset{\_}{1}$.

(d)

To determine

Measure the density of milk and express the findings in $\text{kg}/{\text{m}}^3,\text{slugs}/{\text{ft}}^3,\text{and}\text{lbm}/{\text{ft}}^3$.

Answer to Problem 17P

The density of milk in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{1,030{\text{kg/m}}^{\text{3}}}$.

The density of milk in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.998{\text{slugs/ft}}^3}$.

The density of milk in ${\text{lbm/ft}}^3$ is $\underset{\_}{64.29{\text{lbm/ft}}^3}$.

The specific gravity of the milk is $\underset{\_}{1.03}$.

Explanation of Solution

Find the density of the milk using the procedure as follows:

• Measure the weight of the empty test container.
• Take the sample of milk in a container. Measure the weight of container with water.
• Find the mass of the milk by subtracting the weight of milk container with weight of empty container.
• Measure the volume of the liquid using a graduated cylinder.
• Find the density of the milk using the value of mass and volume.
• $\text{Density}=\frac{\text{mass}}{\text{Volume}}$
• The density of milk is ${\rho }_{\text{milk}}=1,030{\text{kg/m}}^{\text{3}}$.

Therefore, the density of milk in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{1,030{\text{kg/m}}^{\text{3}}}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{slugs}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{milk}}=1,030{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{slug}}{\text{14}\text{.594}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =1.998{\text{slugs/ft}}^3\end{array}$

Therefore, the density of milk in ${\text{slugs/ft}}^3$ is $\underset{\_}{1.998{\text{slugs/ft}}^3}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{lbm}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{milk}}=1,030{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{lbm}}{\text{0}\text{.45359}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =64.29{\text{lbm/ft}}^3\end{array}$

Therefore, the density of milk in ${\text{lbm/ft}}^3$ is $\underset{\_}{64.29{\text{lbm/ft}}^3}$.

Find the specific gravity $\left(G_{\text{milk}}\right)$ of milk using the relation.

$G_{\text{milk}}=\frac{{\rho }_{\text{milk}}}{{\rho }_w}$

Substitute $1,030{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{milk}}$ and $1,000{\text{kg/m}}^3$ for ${\rho }_w$.

$\begin{array}{c}{G}_{\text{milk}}=\frac{1,030}{1,000}\\ =1.03\end{array}$

Therefore, the specific gravity of the milk is $\underset{\_}{1.03}$.

(e)

To determine

Measure the density of ethylene glycol (antifreeze) and express the findings in $\text{kg}/{\text{m}}^3,\text{slugs}/{\text{ft}}^3,\text{and}\text{lbm}/{\text{ft}}^3$.

Answer to Problem 17P

The density of ethylene glycol (antifreeze) in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{1,125{\text{kg/m}}^{\text{3}}}$.

The density of ethylene glycol (antifreeze) in ${\text{slugs/ft}}^3$ is $\underset{\_}{2.18{\text{slugs/ft}}^3}$.

The density of ethylene glycol (antifreeze) in ${\text{lbm/ft}}^3$ is $\underset{\_}{70.22{\text{lbm/ft}}^3}$.

The specific gravity of the ethylene glycol (antifreeze) is $\underset{\_}{1.125}$.

Explanation of Solution

Find the density of the ethylene glycol using the procedure as follows:

• Measure the weight of the empty test container.
• Take the sample of ethylene glycol in a container. Measure the weight of container with ethylene glycol.
• Find the mass of the ethylene glycol by subtracting the weight of ethylene glycol container with weight of empty container.
• Measure the volume of the liquid using a graduated cylinder.
• Find the density of the ethylene glycol using the value of mass and volume.
• $\text{Density}=\frac{\text{mass}}{\text{Volume}}$
• The density of ethylene glycol (antifreeze) is ${\rho }_{\text{ethylene}}=1,125{\text{kg/m}}^{\text{3}}$.

Therefore, the density of ethylene glycol (antifreeze) in ${\text{kg/m}}^{\text{3}}$ is $\underset{\_}{1,125{\text{kg/m}}^{\text{3}}}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{slugs}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{ethylene}}=1,125{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{slug}}{\text{14}\text{.594}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =2.18{\text{slugs/ft}}^3\end{array}$

Therefore, the density of ethylene glycol (antifreeze) in ${\text{slugs/ft}}^3$ is $\underset{\_}{2.18{\text{slugs/ft}}^3}$.

Convert the unit of density from ${\text{kg/m}}^{\text{3}}$ to $\text{lbm}/{\text{ft}}^3$.

$\begin{array}{c}{\rho }_{\text{ethylene}}=1,125{\text{kg/m}}^{\text{3}}\times \frac{\text{1}\text{lbm}}{\text{0}\text{.45359}\text{kg}}\times {\left(\frac{\text{1}\text{m}}{\text{3}\text{.281}\text{ft}}\right)}^3\\ =70.22{\text{lbm/ft}}^3\end{array}$

Therefore, the density of ethylene glycol (antifreeze) in ${\text{lbm/ft}}^3$ is $\underset{\_}{70.22{\text{lbm/ft}}^3}$.

Find the specific gravity $\left(G_{\text{ethylene}}\right)$ of ethylene glycol (antifreeze) using the relation.

$G_{\text{ethylene}}=\frac{{\rho }_{\text{ethylene}}}{{\rho }_w}$

Substitute $1,125{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{ethylene}}$ and $1,000{\text{kg/m}}^3$ for ${\rho }_w$.

$\begin{array}{c}{G}_{\text{ethylene}}=\frac{1,125}{1,000}\\ =1.125\end{array}$

Therefore, the specific gravity of the ethylene glycol (antifreeze) is $\underset{\_}{1.125}$.