Chapter 1, Problem 61AP

(a)

To determine

The densities of substances given in the table.

Answer to Problem 61AP

Density of aluminum, copper, brass, tin, and iron are $2.75\text{g}/{\text{cm}}^{\text{3}}$,$9.36\text{g}/{\text{cm}}^{\text{3}}$,$8.91\text{g}/{\text{cm}}^{\text{3}}$, $7.68\text{g}/{\text{cm}}^{\text{3}}$, and $7.88\text{g}/{\text{cm}}^{\text{3}}$ respectively.

Explanation of Solution

Write the relation between the density, radius and length of a solid cylinder.

    $\rho =\frac{m}{\pi r^{2}h}$

Here, $\rho$ is the density, $m$ is the mass of substance, $r$ is the radius of cylinder, and $h$ is the height of cylinder.

Write the relation between $r$ and diameter of cylinder.

    $r=\frac{d}{2}$

Here, $d$ is the diameter of cylinder.

Rewrite the expression for $\rho$ by substituting $\frac{d}{2}$ for $r$.

    $\begin{array}{c}\rho =\frac{m}{\pi {\left(\frac{d}{2}\right)}^{2}h}\\ =\frac{4m}{\pi D^{2}h}\end{array}$

Conclusion:

Consider the case of Aluminum.

Substitute $51.5\text{g}$ for $m$, $2.52\text{cm}$ for $D$, and $3.75\text{cm}$ for $h$ in the above equation to find $\rho$.

    $\begin{array}{c}\rho =\frac{4\left(51.5\text{g}\right)}{\left(3.14\right){\left(2.52\text{cm}\right)}^2\left(3.75\text{cm}\right)}\\ =2.75\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Consider the case of Copper.

Substitute $\text{56}\text{.3}\text{g}$ for $m$, $1.23\text{cm}$ for $D$, and $5.06\text{cm}$ for $h$ in the above equation to find $\rho$.

    $\begin{array}{c}\rho =\frac{4\left(56.3\text{g}\right)}{\left(3.14\right){\left(1.23\text{cm}\right)}^2\left(5.06\text{cm}\right)}\\ =9.36\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Consider the case of Brass.

Substitute $94.4\text{g}$ for $m$, $1.54\text{cm}$ for $D$, and $5.69\text{cm}$ for $h$ in the above equation to find $\rho$.

    $\begin{array}{c}\rho =\frac{4\left(94.4\text{g}\right)}{\left(3.14\right){\left(1.54\text{cm}\right)}^2\left(5.69\text{cm}\right)}\\ =8.91\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Consider the case of Tin.

Substitute $69.1\text{g}$ for $m$, $1.75\text{cm}$ for $D$, and $3.74\text{cm}$ for $h$ in the above equation to find $\rho$.

    $\begin{array}{c}\rho =\frac{4\left(69.1\text{g}\right)}{\left(3.14\right){\left(1.75\text{cm}\right)}^2\left(3.74\text{cm}\right)}\\ =7.68\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Consider the case of Iron.

Substitute $216.1\text{g}$ for $m$, $1.89\text{cm}$ for $D$, and $9.77\text{cm}$ for $h$ in the above equation to find $\rho$.

    $\begin{array}{c}\rho =\frac{4\left(216.1\text{g}\right)}{\left(3.14\right){\left(1.89\text{cm}\right)}^2\left(9.77\text{cm}\right)}\\ =7.88\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Therefore, density of aluminum, copper, brass, tin, and iron are $2.75\text{g}/{\text{cm}}^{\text{3}}$,$9.36\text{g}/{\text{cm}}^{\text{3}}$,$8.91\text{g}/{\text{cm}}^{\text{3}}$, $7.68\text{g}/{\text{cm}}^{\text{3}}$, and $7.88\text{g}/{\text{cm}}^{\text{3}}$ respectively.

(b)

To determine

Compare the results obtained part (a) with the values given in table 14.1.

Answer to Problem 61AP

For aluminum, copper, brass, tin, and iron, calculated densities are $2\%,5\%,5\%,5\%,\text{and}\text{0}\text{.3\%}$ larger than the respective values given in table 14.1.

Explanation of Solution

Substance	$\rho$ calculated in part (a)	Value of $\rho$ given in table 14.1	Comparison of result in part (a) with data given table 14.1
Aluminum	$2.75\text{g}/{\text{cm}}^{\text{3}}$	$2.70\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}$	$\frac{2.75\text{g}/{\text{cm}}^{\text{3}}\left(\frac{\text{1}\text{kg}}{{\text{10}}^{\text{3}}\text{g}}\right){\left(\frac{{\text{10}}^2\text{cm}}{\text{1}\text{m}}\right)}^3}{2.70\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}}=1.02$
Copper	$9.36\text{g}/{\text{cm}}^{\text{3}}$	$8.92\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}$	$\frac{9.36\text{g}/{\text{cm}}^{\text{3}}\left(\frac{\text{1}\text{kg}}{{\text{10}}^{\text{3}}\text{g}}\right){\left(\frac{{\text{10}}^2\text{cm}}{\text{1}\text{m}}\right)}^3}{8.92\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}}=1.05$
Brass	$8.91\text{g}/{\text{cm}}^{\text{3}}$	$8.4\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}$	$\frac{8.91\text{g}/{\text{cm}}^{\text{3}}\left(\frac{\text{1}\text{kg}}{{\text{10}}^{\text{3}}\text{g}}\right){\left(\frac{{\text{10}}^2\text{cm}}{\text{1}\text{m}}\right)}^3}{8.4\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}}=1.05$
Tin	$7.68\text{g}/{\text{cm}}^{\text{3}}$	$7.30\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}$	$\frac{7.68\text{g}/{\text{cm}}^{\text{3}}\left(\frac{\text{1}\text{kg}}{{\text{10}}^{\text{3}}\text{g}}\right){\left(\frac{{\text{10}}^2\text{cm}}{\text{1}\text{m}}\right)}^3}{7.30\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}}=1.05$
Iron	$7.88\text{g}/{\text{cm}}^{\text{3}}$	$7.86\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}$	$\frac{7.88\text{g}/{\text{cm}}^{\text{3}}\left(\frac{\text{1}\text{kg}}{{\text{10}}^{\text{3}}\text{g}}\right){\left(\frac{{\text{10}}^2\text{cm}}{\text{1}\text{m}}\right)}^3}{7.86\times {10}^3\text{k}\text{g}/{\text{m}}^{\text{3}}}=1.003$

Conclusion:

Therefore, for aluminum, copper, brass, tin, and iron, calculated densities are $2\%,5\%,5\%,5\%,\text{and}\text{0}\text{.3\%}$ larger than the respective values given in table 14.1.