Chapter 1, Problem 1.61AP

The data in the following table represent measurements of the masses and dimensions of solid cylinders of aluminum, copper, brass, tin, and iron. (a) Use these data to calculate the densities of these substances. (b) State how your results compare with those given in Table 14.1.

To determine

The densities of each substance.

Answer to Problem 1.61AP

The density of aluminum solid cylinders is $2.75\text{g}/{\text{cm}}^{\text{3}}$, density of copper solid cylinders is $9.36\text{g}/{\text{cm}}^{\text{3}}$, brass solid cylinders is $8.91\text{g}/{\text{cm}}^{\text{3}}$, tin solid cylinders is $7.68\text{g}/{\text{cm}}^{\text{3}}$ and iron solid cylinders is $7.88\text{g}/{\text{cm}}^{\text{3}}$.

Explanation of Solution

Given info: The mass, diameter and length of each substance are given below,

Substance	Mass $\left(\text{g}\right)$	Diameter $\left(\text{cm}\right)$	Length $\left(\text{cm}\right)$
Aluminum	$51.5$	$2.52$	$3.75$
Copper	$56.3$	$1.23$	$5.06$
Brass	$94.4$	$1.54$	$5.69$
Tin	$69.1$	$1.75$	$3.74$
Iron	$216.1$	$1.89$	$9.77$

Formula to calculate the density of substance is,

$\rho =\frac{m}{V}$ (1)

Here,

$m$ is the mass of substance.

$V$ is the volume of substance.

Write the expression for the volume of solid cylinder,

$V=\pi {\left(\frac{d}{2}\right)}^{2}l$

Here,

$d$ is the diameter of the substance.

$l$ is length of the substance.

Substitute $\pi {\left(\frac{d}{2}\right)}^{2}l$ for $V$ in the equation (1).

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

For aluminum:

Substitute $51.5\text{g}$ for $m$, $2.52\text{cm}$ for $d$ and $3.75\text{cm}$ for $l$ in the equation (2).

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

Thus, the density of aluminum solid cylinders is $2.75\text{g}/{\text{cm}}^{\text{3}}$.

For copper:

Substitute $56.3\text{g}$ for $m$, $1.23\text{cm}$ for $d$ and $5.06\text{cm}$ for $l$ in the equation (2).

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

Thus, the density of copper solid cylinders is $9.36\text{g}/{\text{cm}}^{\text{3}}$.

For brass:

Substitute $94.4\text{g}$ for $m$, $1.54\text{cm}$ for $d$ and $5.69\text{cm}$ for $l$ in the equation (2).

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

Thus, the density of brass solid cylinders is $8.91\text{g}/{\text{cm}}^{\text{3}}$.

For tin:

Substitute $69.1\text{g}$ for $m$, $1.75\text{cm}$ for $d$ and $3.74\text{cm}$ for $l$ in the equation (2).

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

Thus, the density of tin solid cylinders is $7.68\text{g}/{\text{cm}}^{\text{3}}$.

For iron:

Substitute $216.1\text{g}$ for $m$, $1.89\text{cm}$ for $d$ and $9.77\text{cm}$ for $l$ in the equation (2).

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

Thus, the density of iron solid cylinders is $7.88\text{g}/{\text{cm}}^{\text{3}}$.

Conclusion:

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

To determine

The comparison between results of part (a) and table $14.1$.

Answer to Problem 1.61AP

The density of aluminum from table is $2\%$ less than the density of aluminum from result of part (a), the density of copper from table is $5\%$ less than the density of copper from result of part (a), the density of brass from table is $6\%$ less than the density of brass from result of part (a), the density of tin from table is $5\%$ less than the density of tin from result of part (a) and the density of iron from table is $0.3\%$ less than the density of iron from result of part (a).

Explanation of Solution

Given info:

Formula to calculate the percentage error is,

$\text{percentage}\text{error}=\left(\frac{\rho -{\rho }^{\prime }}{{\rho }^{\prime }}\right)\times 100$ (3)

Here,

$\rho$ is the density of the aluminum from the result.

${\rho }^{\prime }$ is the density of aluminum from table $14.1$.

For aluminum:

From part (a), the density of the aluminum is $2.75\text{g}/{\text{cm}}^{\text{3}}$ and from table $14.1$ the density of aluminum is $2.70\text{g}/{\text{cm}}^{\text{3}}$.

Substitute $2.75\text{g}/{\text{cm}}^{\text{3}}$ for $\rho$ and $2.70\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }^{\prime }$ in the equation (3).

$\begin{array}{c}\text{percentage}\text{error}=\left(\frac{2.75\text{g}/{\text{cm}}^{\text{3}}-2.70\text{g}/{\text{cm}}^{\text{3}}}{2.70\text{g}/{\text{cm}}^{\text{3}}}\right)\times 100\\ =1.85\%\\ \approx 2\%\end{array}$

Thus, the density of aluminum from table is $2\%$ less than the density of aluminum from result of part (a).

For copper:

From part (a), the density of the copper is $9.37\text{g}/{\text{cm}}^{\text{3}}$ and from table $14.1$ the density of copper is $8.92\text{g}/{\text{cm}}^{\text{3}}$.

Substitute $9.37\text{g}/{\text{cm}}^{\text{3}}$ for $\rho$ and $8.92\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }^{\prime }$ in the equation (3).

$\begin{array}{c}\text{percentage}\text{error}=\left(\frac{9.37\text{g}/{\text{cm}}^{\text{3}}-8.92\text{g}/{\text{cm}}^{\text{3}}}{8.92\text{g}/{\text{cm}}^{\text{3}}}\right)\times 100\\ =5\%\end{array}$

Thus, the density of copper from table is $5\%$ less than the density of copper from result of part (a).

For brass:

From part (a), the density of the brass is $8.91\text{g}/{\text{cm}}^{\text{3}}$ and from table $14.1$ the density of brass is $8.4\text{g}/{\text{cm}}^{\text{3}}$.

Substitute $8.91\text{g}/{\text{cm}}^{\text{3}}$ for $\rho$ and $8.4\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }^{\prime }$ in the equation (3).

$\begin{array}{c}\text{percentage}\text{error}=\left(\frac{8.91\text{g}/{\text{cm}}^{\text{3}}-8.4\text{g}/{\text{cm}}^{\text{3}}}{8.4\text{g}/{\text{cm}}^{\text{3}}}\right)\times 100\\ =6\%\end{array}$

Thus, the density of brass from table is $6\%$ less than the density of brass from result of part (a).

For tin:

From part (a), the density of the tin is $7.68\text{g}/{\text{cm}}^{\text{3}}$ and from table $14.1$ the density of tin is $7.30\text{g}/{\text{cm}}^{\text{3}}$.

Substitute $7.68\text{g}/{\text{cm}}^{\text{3}}$ for $\rho$ and $7.30\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }^{\prime }$ in the equation (3).

$\begin{array}{c}\text{percentage}\text{error}=\left(\frac{7.68\text{g}/{\text{cm}}^{\text{3}}-7.30\text{g}/{\text{cm}}^{\text{3}}}{7.30\text{g}/{\text{cm}}^{\text{3}}}\right)\times 100\\ =5.2\%\\ \approx 5\%\end{array}$

Thus, the density of tin from table is $5\%$ less than the density of tin from result of part (a).

For iron:

From part (a), the density of the iron is $7.88\text{g}/{\text{cm}}^{\text{3}}$ and from table $14.1$ the density of iron is $7.86\text{g}/{\text{cm}}^{\text{3}}$.

Substitute $7.88\text{g}/{\text{cm}}^{\text{3}}$ for $\rho$ and $7.86\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }^{\prime }$ in the equation (3).

$\begin{array}{c}\text{percentage}\text{error}=\left(\frac{7.88\text{g}/{\text{cm}}^{\text{3}}-7.86\text{g}/{\text{cm}}^{\text{3}}}{7.86\text{g}/{\text{cm}}^{\text{3}}}\right)\times 100\\ \approx 0.3\%\end{array}$

Thus, the density of iron from table is $3\%$ less than the density of iron from result of part (a).

Conclusion:

Therefore, the density of aluminum from table is $2\%$ less than the density of aluminum from result of part (a), the density of copper from table is $5\%$ less than the density of copper from result of part (a), the density of brass from table is $6\%$ less than the density of brass from result of part (a), the density of tin from table is $5\%$ less than the density of tin from result of part (a) and the density of iron from table is $0.3\%$ less than the density of iron from result of part (a).