Chapter 5, Problem 59A

Interpretation Introduction

Interpretation: A material which density is $\text{1}.0\text{ g}/\text{mL}$ , volume is $1.2\times {10}^4{\text{in}}^{\text{3}}.$ and mass $3.5\text{lb}$ , when it is placed in a reservoir it will be float or sink needs to be explained..

Concept Introduction: The given volume of matter contained in a material is termed as density. That is, it is mass per unit volume of an object:

  $\text{Density =}\frac{\text{mass}}{\text{volume}}$

Answer to Problem 59A

The density value of the material $0.0081\text{g}/{\text{cm}}^{\text{3}}$ is equivalent to $0.0081\text{g}/\text{mL}$ which is lower than water density $1.0\text{g}/\text{mL}$ . Hence, the material will float.

Explanation of Solution

Convert given volume value from in. to centimeter as follows:

  $\begin{array}{c}1\text{in}\text{.}=2.54\text{cm}\\ \text{1}\text{.2}\times {\text{10}}^4{\text{in}}^3\text{.=}{\left(\frac{2.54\text{cm}}{1\text{in}\text{.}}\right)}^3\times \text{1}\text{.2}\times {\text{10}}^4{\text{in}}^3.\\ =16.39\times \text{1}\text{.2}\times {\text{10}}^4{\text{cm}}^{\text{3}}\\ =196680{\text{cm}}^{\text{3}}\end{array}$

The value is rounded off to significant figures. Then, the value becomes $2.0\times {10}^5{\text{cm}}^{\text{3}}$ .

Now, the weight value is converted from lb to g.

Since, $1\text{lb}=453.6\text{g}$

  $\begin{array}{c}\text{Mass}=3.5\text{lb}\\ \text{=3}.5\text{lb}\times \frac{453.6\text{g}}{1\text{lb}}\\ =1587.6\text{g}\end{array}$

To calculate density, substitute above values in below equation.

  $\begin{array}{c}\text{Density =}\frac{\text{mass}}{\text{volume}}\\ =\frac{1587.6\text{g}}{196680{\text{cm}}^3}\\ =0.0081\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Hence, density is $0.0081\text{g}/{\text{cm}}^{\text{3}}$ .

Conclusion

The density value of the material $0.0081\text{g}/{\text{cm}}^{\text{3}}$ is equivalent to $0.0081\text{g}/\text{mL}$ which is lower than water density $1.0\text{g}/\text{mL}$ . Hence, the material will float.