The purpose of this question is to find the density of an aluminum engine block with a given volume and mass.

The thickness of a substance indicates the density of a substance in a particular region. To put it in another way, density is the distribution of mass over a volume. Alternatively, it is the number of kilograms that a one-meter cube of the material weighs. The more each meter cube weighs, the denser the material will become. It can also be regarded as the mass per unit volume of a substance.

Let $d$ be the density, $m$ be the mass, and $v$ be the volume of the substance. Then mathematically, the density is given by $d=m/v$. Common instances of density include the water density which is one gram per cubic centimeter, and Earth’s density is about $5.51$ grams per cubic centimeter.

More specifically, density regards the fact that two cubes of different substances having the same size will weigh differently. It is an estimation of how closely a substance is packed together. This physical property is unique in every particular substance.

## Expert Answer

Let $d$ be the density, $m$ be the mass, and $v$ be the volume of the aluminum engine block, then:

$d=\dfrac{m}{v}$

Here, $m=12.88\,kg$ and $v=4.77\,L$

So, $d=\dfrac{12.88\,kg}{4.77\,L}$

Since it is required to find the density in grams per cubic centimeter, therefore, take into account the following conversions:

$1\,kg=1000,g$ and $1\,L=1000$ cubic centimeters

So that the density will be:

$d=\left(\dfrac{12.88\,kg}{4.77\,L}\right)\left(\dfrac{1000\,g}{1\,kg}\right)\left(\dfrac{1\,L}{1000\,cm^3}\right)$

$d=2.70\,g/cm^3$

## Example 1

Find the mass of the block if it has the density $390\,g/cm^3$ and the volume $3\,cm^3$.

### Solution

Given that:

$d=390\,g/cm^3$ and $v=3\,cm^3$

To find: $m=?$

Since $d=\dfrac{m}{v}$

So that $m=dv$

$m=(390\,g/cm^3)(3\,cm^3)$

$m=1170\,g$

Hence, the mass of the block is $1170$ grams.

## Example 2

Calculate the volume in liters of the glass of water having the density $1000\,kg/m^3$ and a mass of $1.4\,kg$.

### Solution

Given that:

$d=1000\,kg/m^3$ and $m=1.4\,kg$

To find: $v=?$

Since $d=\dfrac{m}{v}$

So that $v=\dfrac{m}{d}$

$v=\dfrac{1.4\,kg}{1000\,kg/m^3}$

$v=0.0014\,m^3$

Now since the volume is required in liters, so convert $m^3$ into liters $L$ as follows:

$v=0.0014\times 1000\,L$

$v=1.4\,L$

Hence, the volume of the water is $1.4$ liters.

## Example 3

Let the volume and mass of a metal be $20\,cm^3$ and $230\,kg$ respectively. Find its density in $g/cm^3$.

### Solution

Given that:

$v=20\,cm^3$ and $m=230\,kg$

$d=\dfrac{m}{v}$

$d=\dfrac{230\,kg}{20\,cm^3}$

$d=11.5\,kg/cm^3$

Since the density is required in grams per cubic centimeter, therefore:

$d=11.5\times 1000\,g/cm^3$

$d=11500\,g/cm^3$