Chapter 13, Problem 55SP

A piece of metal has a measured mass of 5.00 g in air, 3.00 g in water, and 3.24 g in benzene. Determine the mass density of the metal and of the benzene.

To determine

The apparent mass of a $2.0\text{ cm}$ cube of metal, whensubmerged in glycerine, if it has a mass of $47.3\text{ g}$ when submerged in water.

Answer to Problem 55SP

Solution:

$45\text{ g}$

Explanation of Solution

Given data:

The side of metal cube is $2.0\text{ cm}$.

The apparent mass of metal cube in water is $47.3\text{ g}$.

The specific gravity of glycerineis $1.26$.

Formula used:

The volume of a cube is expressed as

$V=a^3$

Here, $V$ is the volume of the cube and $a$ is the side.

The density of a body is calculated using the expression

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

Here, $\rho$ is the density of the body and $m$ is the mass.

Archimedes’ principle states that the apparent upward force experienced by the solid immersed in a fluid is equal to the weight of the fluid displaced by it. This upward buoyant force on the solid is expressed as

$F_B={\rho }_l{V}_{S}g$

Here, $F_B$ is the buoyant force, ${\rho }_l$ is the density of the liquid in which the solid is submerged, $V_S$ is the volume of solid submerged in the liquid, and $g$ is the acceleration due to gravity.

The weight of a body is expressed as

$W=mg$

Here, $W$ is the weight of the body.

The apparent weight of a body submerged in a liquid is calculated as

$W_A=W-F_B$

Here, $W_A$ is the apparent weight of the body in the liquid.

The specific gravity of a liquid is expressed as

$\text{sp gr}=\frac{{\rho }_l}{{\rho }_w}$

Here, $\text{sp gr}$ is the specific gravity of liquid and ${\rho }_w$ is the density of water.

Explanation:

Draw a figure to represent the system when the cube is submerged in water:

Here, $W$ is the force exerted due to the weight of themetal cubeand $F_{B1}$ is the buoyant force on the metal cube exerted by water.

Recall the formula for calculating the volume of the metal cube:

$V=a^3$

Substitute $2.0\text{ cm}$ for $a$

$\begin{array}{c}V={\left(2.0\text{ cm}\right)}^3\\ =8{\text{ cm}}^3\\ =8{\text{ cm}}^3\left(\frac{{10}^{-6}{\text{ m}}^3}{1{\text{ cm}}^3}\right)\\ =8\times {10}^{-6}{\text{ m}}^3\end{array}$

Recall the formula for calculating the density of a metal cube:

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

Rearrange the above expression for $m$

$m=\rho V$

Substitute $8\times {10}^{-6}{\text{ m}}^3$ for $V$

$\begin{array}{c}m=\rho \left(8\times {10}^{-6}{\text{ m}}^3\right)\\ =\left(8\times {10}^{-6}\right)\rho \end{array}$

Recall expression for the weight of themetal cube:

$W=mg$

Substitute $\left(8\times {10}^{-6}\right)\rho$ for $m$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}W=\left[\left(8\times {10}^{-6}\right)\rho \right]\left(9.81{\text{ m/s}}^2\right)\\ =\left(78.48\times {10}^{-6}\right)\rho \end{array}$

It is understood that the cube is completely submerged in water, therefore, the expression for the volume of a cube submerged in water becomes

$V_S=V$

Substitute $8\times {10}^{-6}{\text{ m}}^3$ for $V$

$V_S=8\times {10}^{-6}{\text{ m}}^3$

Recall the expression for buoyant force exerted by water on the metal cube:

$F_{B1}={\rho }_w{V}_{S}g$

Here, ${\rho }_w$ is the density of water.

Consider the standard value of $g$ as $9.81{\text{ m/s}}^2$ and of density of water as $1000{\text{ kg/m}}^3$.

Substitute $1000{\text{ kg/m}}^3$ for ${\rho }_w$, $8\times {10}^{-6}{\text{ m}}^3$ for $V_S$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{B1}=\left(1000{\text{ kg/m}}^3\right)\left(8\times {10}^{-6}{\text{ m}}^3\right)\left(9.81{\text{ m/s}}^2\right)\\ =78.48\times {10}^{-3}\text{ N}\\ =\text{0}\text{.07848 N}\end{array}$

Write the expression for the apparent weight of the cube in water in terms of apparent mass:

$W_{A1}=m_{A1}g$

Here, $W_{A1}$ is the apparent weight of the cube in water and $m_{A1}$ is the apparent mass of the cube in water.

Substitute $47.3\text{ g}$ for $m_{A1}$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{W}_{A1}=\left(47.3\text{ g}\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(47.3\text{ g}\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)\\ =0.464\text{ N}\end{array}$

Recall the expression for the apparent weight of the cube in water in terms of buoyant force:

$W_{A1}=W-F_{B1}$

Substitute $0.464\text{ N}$ for $W_{A1}$, $\text{0}\text{.07848 N}$ for $F_{B1}$, and $\left(78.48\times {10}^{-6}\right)\rho$ for $W$

$\begin{array}{c}0.464\text{ N}=\left(78.48\times {10}^{-6}\right)\rho -\text{0}\text{.07848 N}\\ \left(78.48\times {10}^{-6}\right)\rho =0.464\text{ N}+\text{0}\text{.07848 N}\\ \left(78.48\times {10}^{-6}\right)\rho =0.54248\text{ N}\\ \rho =\frac{0.54248\text{ N}}{\left(78.48\times {10}^{-6}\right)}\end{array}$

Further solve as

$\rho \simeq 6912.33{\text{ kg/m}}^3$

Consider the situation when the cube is submerged in glycerine anddraw its free body diagram:

Here, $F_{B2}$ is the buoyant force on the metal cube, exerted by glycerine.

Recall the expression for the specific gravity of glycerine:

$\text{sp gr}=\frac{{\rho }_l}{{\rho }_w}$

Substitute $1.26$ for $\text{sp gr}$ and $1000{\text{ kg/m}}^3$ for ${\rho }_w$

$\begin{array}{c}1.26=\frac{{\rho }_l}{1000{\text{ kg/m}}^3}\\ {\rho }_l=1.26\left(1000{\text{ kg/m}}^3\right)\\ =1260{\text{kg/m}}^3\end{array}$

Recall the expression for calculating the buoyant force exerted by glycerine on the metal cube:

$F_{B2}={\rho }_l{V}_{S}g$

Substitute $1260{\text{kg/m}}^3$ for ${\rho }_l$, $8\times {10}^{-6}{\text{ m}}^3$ for $V_S$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{B2}=\left(1260{\text{kg/m}}^3\right)\left(8\times {10}^{-6}{\text{ m}}^3\right)\left(9.81{\text{ m/s}}^2\right)\\ =98884.8\times {10}^{-6}\text{ N}\\ =\text{0}\text{.0989 N}\end{array}$

Write the expression for apparent weight of the cube in glycerine in terms of apparent mass:

$W_{A2}=m_{A2}g$

Here, $W_{A2}$ is the apparent weight of cube in glycerine and $m_{A2}$ is the apparent mass of the cube in glycerine.

Substitute $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{W}_{A2}=m_{A2}\left(9.81{\text{ m/s}}^2\right)\\ =9.81m_{A2}\end{array}$

Recall the expression for the apparent weight of the cube in glycerine in terms of buoyant force:

$W_{A2}=W-F_{B2}$

Substitute $9.81m_{A2}$ for $W_{A2}$, $\text{0}\text{.0989 N}$ for $F_{B2}$, and $\left(78.48\times {10}^{-6}\right)\rho$ for $W$

$9.81m_{A2}=\left(78.48\times {10}^{-6}\right)\rho -\text{0}\text{.0989 N}$

Substitute $6912.33{\text{ kg/m}}^3$ for $\rho$

$\begin{array}{c}9.81m_{A2}=\left(78.48\times {10}^{-6}\right)\left(6912.33{\text{ kg/m}}^3\right)-\text{0}\text{.0989 N}\\ 9.81m_{A2}=0.5425\text{ N}-\text{0}\text{.0989 N}\\ 9.81m_{A2}=0.4436\text{ N}\\ m_{A2}=\frac{0.4436\text{ N}}{9.81}\end{array}$

Further solve as

$\begin{array}{c}{m}_{A2}\simeq 0.04521\text{ kg}\\ \simeq 0.04521\text{ kg}\left(\frac{1000\text{ g}}{1\text{ kg}}\right)\\ \approx 45\text{ g}\end{array}$

Conclusion:

The apparent mass of the cube in glycerine is $45\text{ g}$.