Chapter 6, Problem 97A

Interpretation Introduction

Interpretation:

To calculate the density of jagged rock

Concept Introduction :

The relation between mass, volume, and density is as shown:

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

Answer to Problem 97A

The mass of jagged rock is $\text{4}\text{.53 }\text{g}/{\text{cm}}^3$ .

Explanation of Solution

First, calculate the volume of the container filled with water

  $\begin{array}{c}\text{volume with water}=\text{height}\times \text{width}\times \text{length}\\ =15.2\text{ cm}\times \text{22}\text{.9 cm}\times \text{55}\text{.0 cm}\\ \text{=19,144}{\text{.4 cm}}^{\text{3}}\end{array}$

First, calculate the volume of the container filled with water and rock

  $\begin{array}{c}\text{volume with rock}=\text{height}\times \text{width}\times \text{length}\\ =15.2\text{ cm}\times \text{22}\text{.9 cm}\times \text{58}\text{.3 cm}\\ \text{=20,293}{\text{.064 cm}}^{\text{3}}\end{array}$

Now, calculate the volume of rock

  $\begin{array}{c}\text{volume of rock=20,293}{\text{.064 cm}}^{\text{3}}-\text{19,144}{\text{.4 cm}}^{\text{3}}\\ =\text{1,148}{\text{.664 cm}}^{\text{3}}\end{array}$

Now, calculate density of rock

  $\begin{array}{c}\text{density}=\frac{\text{mass}}{\text{volume}}\\ =\frac{5.21\times {10}^3\text{ g}}{\text{1,148}{\text{.664 cm}}^{\text{3}}}\left(1\text{ kg}=1000\text{ g}\right)\\ =\frac{5210\text{ g}}{\text{1,148}{\text{.664 cm}}^{\text{3}}}\\ =\text{4}\text{.53 }\text{g}/{\text{cm}}^3\end{array}$

Thus, the mass of jagged rock is $\text{4}\text{.53 }\text{g}/{\text{cm}}^3$ .