# An aluminum engine block has a volume of 4.77 L and a mass of 12.88 kg. What is the density of the aluminum in grams per cubic centimeter?

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.

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:

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

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