Chapter 13, Problem 57SP

A 2.0-cm cube of metal is suspended by a fine thread attached to a scale. The cube appears to have a mass of 47.3 g when measured submerged in water. What will its mass appear to be when submerged in glycerin, sp gr = 1.26? [Hint: Find ρ too.]

To determine

The mass density of metal and benzene if the metal piece has a measured mass of $5.00\text{ g}$ in air, $3.00\text{ g}$ in water, and $3.24\text{ g}$ in benzene.

Answer to Problem 57SP

Solution:

$2.50\times {10}^3{\text{kg/m}}^3$ and $880{\text{ kg/m}}^3$.

Explanation of Solution

Given data:

The mass of metal piece in air is $5.00\text{ g}$.

The mass of metal piece in water is $3.00\text{ g}$.

The mass of metal piece in benzene is $3.24\text{ g}$.

Formula used:

The density of a body is calculated using the formula:

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

Here, $V$ is the volume of the body, $\rho$ is the density, 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 the 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.

Explanation:

Consider the situation when the metal piece is submerged in water and draw its free body diagram:

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

Recall the expression for the weight of the metal piece:

$W=mg$

Consider, the standard value of acceleration due to gravity as $9.81{\text{ m/s}}^2$.

Substitute $5.00\text{ g}$ for $m$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}W=\left(5.00\text{ g}\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(5.00\text{ g}\right)\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\left(9.81{\text{ m/s}}^2\right)\\ =0.04905\text{ N}\end{array}$

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

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

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

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

Substitute $1000{\text{ kg/m}}^3$ for ${\rho }_w$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{B1}=\left(1000{\text{ kg/m}}^3\right)V_S\left(9.81{\text{ m/s}}^2\right)\\ =9810V_S\end{array}$

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

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

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

Substitute $3.00\text{ g}$ for $m_{A1}$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{W}_{A1}=\left(3.00\text{ g}\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(3.00\text{ g}\right)\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\left(9.81{\text{ m/s}}^2\right)\\ =0.02943\text{ N}\end{array}$

Recall the expression for the apparent weight of the metal piece in water, in terms of buoyant force:

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

Substitute $0.02943\text{ N}$ for $W_{A1}$, $9810V_S$ for $F_{B1}$, and $0.04905\text{ N}$ for $W$

$\begin{array}{c}0.02943\text{ N}=0.04905\text{ N}-9810V_S\\ 9810V_S=0.04905\text{ N}-0.02943\text{ N}\\ 9810V_S=0.01962\text{ N}\\ V_S=\frac{0.01962\text{ N}}{9810}\end{array}$

Further solve as

$V_S=2\times {10}^{-6}{\text{m}}^3$

It is understood that the metal piece is completely submerged in water, therefore, the expression for calculation of the volume of the metal piece submerged in water becomes:

$V=V_S$

Substitute $2\times {10}^{-6}{\text{m}}^3$ for $V_S$

$V=2\times {10}^{-6}{\text{m}}^3$

Recall the expression for the density of the metal piece:

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

Substitute $2\times {10}^{-6}{\text{m}}^3$ for $V$ and $5.00\text{ g}$ for $m$

$\begin{array}{c}\rho =\frac{5.00\text{ g}}{2\times {10}^{-6}{\text{m}}^3}\\ =\frac{5.00\text{ g}\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)}{2\times {10}^{-6}{\text{m}}^3}\\ =2.50\times {10}^3{\text{kg/m}}^3\end{array}$

Thus, the density of the metal piece is $2.50\times {10}^3{\text{kg/m}}^3$.

Consider the situation when themetal piece is submerged in benzene and draw its free body diagram:

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

Recall the expression for buoyant force exerted by benzene, on the metal piece:

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

Here, ${\rho }_l$ is the density of benzene.

Substitute $2\times {10}^{-6}{\text{ m}}^3$ for $V_S$ and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{F}_{B2}={\rho }_l\left(2\times {10}^{-6}{\text{ m}}^3\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(19.62\times {10}^{-6}\right){\rho }_l\end{array}$

Write the expression for apparent weight of themetal piece in benzene in terms of apparent mass:

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

Here, $W_{A2}$ is the apparent weight of themetal piece in benzene and $m_{A2}$ is the apparent mass of themetal piece in benzene.

Substitute $9.81{\text{ m/s}}^2$ for $g$ and $3.24\text{ g}$ for $m_{A2}$

$\begin{array}{c}{W}_{A2}=\left(3.24\text{ g}\right)\left(9.81{\text{ m/s}}^2\right)\\ =\left(3.24\text{ g}\left(\frac{0.001\text{ kg}}{1\text{ g}}\right)\right)\left(9.81{\text{ m/s}}^2\right)\\ =0.03178\text{ N}\end{array}$

Recall the expression for the calculation of the apparent weight of the metal piece in benzenein terms of buoyant force:

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

Substitute $0.03178\text{ N}$ for $W_{A1}$, $\left(19.62\times {10}^{-6}\right){\rho }_l$ for $F_{B2}$, and $0.04905\text{ N}$ for $W$

$\begin{array}{c}0.03178\text{ N}=0.04905\text{ N}-\left(19.62\times {10}^{-6}\right){\rho }_l\\ \left(19.62\times {10}^{-6}\right){\rho }_l=0.04905-0.03178\\ \left(19.62\times {10}^{-6}\right){\rho }_l=0.01727\\ {\rho }_l=\frac{0.01727}{\left(19.62\times {10}^{-6}\right)}\end{array}$

Further solve as

${\rho }_l\simeq 880{\text{ kg/m}}^3$

The density of benzene is $880{\text{ kg/m}}^3$.

Conclusion:

Thus, thedensity of the metal pieceis $2.50\times {10}^3{\text{kg/m}}^3$ and the density of benzene is $880{\text{ kg/m}}^3$.