Video Transcript
In this video, we’re talking about
calculating density. What we’ll see is that, given some
object, in order to calculate its density, we’ll need to know two things about
it. Beyond that, we’ll get some
practice going through the steps of calculating the densities of various
materials.
To start off, we can recall that
density is a measure of how much material, that is, mass, fits into a given space, a
volume. This point is important because it
means that density isn’t only about mass and it’s not only about volume either. For example, say that we have these
three different-sized cubes. The length of each of the sides of
the smallest cube we can call 𝐿. The next largest cube has side
length of two 𝐿. And the biggest cube of all has
sides of length five 𝐿. And let’s say further that, going
left to right, the masses of these three cubes are 10 kilograms, 12 kilograms, and
eight kilograms, respectively.
If we were going to calculate the
relative density of each of these cubes, we might guess that the most dense cube is
the largest one, the one with the largest side length and the one with the most
mass. To calculate these three densities,
let’s rewrite this sentence as an equation. The sentence told us that density
is how much mass fits in a certain volume.
Mathematically then, we can see
that density is equal to mass divided by volume. And we can even go further, letting
symbols represent these terms in the equation. To represent the density of a
material, we’ll use the Greek letter 𝜌. Then to represent an object’s mass,
we’ll use 𝑚. And to represent volume, we’ll use
capital 𝑉.
We now have a mathematical
expression for the density of an object that will let us calculate that density, so
long as we know its mass and its volume. Knowing this, we can now calculate
the densities of these three cubes. The density of the green cube —
we’ll call it 𝜌 sub 𝑔 — is equal to the mass of that cube, 10 kilograms, divided
by its volume. Since this cube has sides of length
two 𝐿, that means the volume it takes up is two 𝐿 times two 𝐿 times two 𝐿, or
written in another way, two 𝐿 quantity cubed. And when we cube this denominator,
we find a result of eight 𝐿 cubed.
So finally, if we simplify this
fraction, we find that the density of the green cube is five kilograms divided by
four 𝐿 cubed, where 𝐿 is some distance in meters. We can then store that result off
to the side and go on to calculating the density of our orange cube, what we’ll call
𝜌 sub 𝑜. For this cube, the mass is 12
kilograms. And the volume it takes up is five
𝐿 times five 𝐿 times five 𝐿. We can also write that as five 𝐿
quantity cubed. And that expands out to 125𝐿
cubed. That’s because five times five
times five is 125. We can store this result and then
go on to calculating the density of our blue cube. We’ll call it 𝜌 sub 𝑏. This smallest cube of all three has
a mass of eight kilograms and a volume of 𝐿 times 𝐿 times 𝐿, or 𝐿 cubed.
We can now compare these results
for the three different cubes. And we see that the density of the
smallest cube, the blue one, is actually the greatest. The blue cube density is
greatest. And then the green cube density is
greater than the density of the orange cube. What we find then is that our
largest cube, the orange one, has the smallest density. And then our smallest cube in terms
of volume has the greatest density.
This helps us understand that just
because an object is larger, like the orange cube in this case, doesn’t mean it has
a larger density. And in fact, because volume appears
in the denominator of the equation for calculating density, by itself having a large
volume decreases overall density.
Beyond just considering relative
differences in material densities, it’s helpful to talk a bit about the units of
density. We see that density is calculated
by dividing mass by volume. That means that the units of
density, putting these square brackets around the density symbol, will be equal to
the units of mass, typically kilograms, divided by the units of volume, typically
meters cubed. We say “typically” because
sometimes our mass is given in units of grams or milligrams or even micrograms. And sometimes our volume is given
in units of cubic centimeters or cubic millimeters, and so on.
So to equip ourselves for any units
that might come up in a density calculation, it’s helpful to recall the conversions
between different mass and volume units. The most common mass conversion is
between units of kilograms and grams. For every one kilogram, there are
1000 grams. Sometimes though, we’re given an
object’s mass in units of milligrams. And there are 1000 milligrams in
one single gram. So we can see if there are 1000
milligrams in one gram and 1000 grams in one kilogram, that means there are 1000
times 1000 milligrams in one kilogram, in other words 1000000 milligrams per
kilogram.
Moving on to the units of volume
and converting between them, this one is a bit trickier than units of mass. That’s because when we’re
calculating a volume, it has dimensions of length cubed. We can look at it this way. Say we wanna convert between meters
and centimeters. To do that, we can recall that one
meter is equal to 100 centimeters. But remember, this one meter is a
length, as is 100 centimeters, whereas for our density calculation, we want a
volume, a length cubed.
So let’s say we go to our one meter
equals 100 centimeters equation. And then we cube both sides of
it. This way, we have a volume on
either side of the equation. Looking on this left-hand side,
when we apply the cube to both the meters as well as the number, one, we find a
result of one cubed times meters cubed. But since one cubed is just one,
then it simplifies to one meter cubed.
On the right-hand side of the
expression, when we do the same thing, when we cube the unit as well as the number,
we find a result of 100 cubed times centimeters cubed. And 100 cubed, which is 100 times
100 times 100, is 1000000. We see then that when we’re talking
about a volume, a length cubed, we need to be careful to keep track of our
conversions. They’re not as easy as when we’re
working with just a length to another length.
Perhaps the best way to take care
when calculating volumes is to start with a length conversion, say, for example,
centimeters to millimeters. And then when calculating a volume,
cube both sides of the equation and then calculate each side separately. In this instance, when we do that,
we find that one cubic centimeter is equal to 1000 millimeters cubed. We now know enough to try out a few
examples that will let us calculate density for ourselves.
Gold can be hammered into very
thin sheets. If a thin sheet of gold is 30
centimeters wide and 40 centimeters long and has a mass of 9.4 grams, find the
thickness of the sheet. Use a value of 19320 kilograms
per cubic meter for the density of gold. Give your answer in millimeters
to two significant figures.
Okay, so we have here this very
thin sheet of gold. It has a width of 30
centimeters and a length of 40 centimeters. And we’ll say that it has a
thickness that we’ll call 𝑡. That’s what we want to solve
for. In the problem statement, we’re
told what the density of gold is. But interestingly, we’re given
that density in units of kilograms per cubic meter, whereas we’re given these
lengths in units of centimeters. And to complicate it even more,
we want to give our answer in units of millimeters.
So clearly, in answering this
question, we’re going to be doing some unit conversion. There are a number of different
ways we could convert units so that, in the end, we give our answer in
millimeters to two significant figures. What we’re going to do is we’ll
leave the density of gold, what we’ll refer to as 𝜌 sub 𝑔, in its original
units of kilograms per cubic meter. Then what we’ll do is we’ll
convert the given dimensions of our gold sheet, centimeters, to units of
meters. And then we’ll convert the mass
of the sheet given in units of grams to kilograms.
So let’s start doing that
now. And as an overarching
principle, if we were to write out the volume of this gold sheet in units of
cubic meters and then write out the mass of this gold sheet in units of
kilograms, we know what the ratio of those two values will be. It will be equal to 𝜌 sub 𝑔,
the density of gold. So that’s what we’ll work
on.
And to get started, let’s
recall that one meter is equal to 100 centimeters. That’s the conversion between
these two length units. This means that the volume of
our gold sheet, what we can call 𝑉 sub 𝑠, is equal not to 30 centimeters by 40
centimeters by 𝑡, but 0.30 meters by 0.40 meters times 𝑡, where 𝑡 is now in
units of meters.
Now that we have an expression
for the volume of our gold sheet written in terms of the variable we want to
solve for, 𝑡, we can now turn our attention to the mass of the sheet. As we saw, that mass is given
in units of grams. But we’d like to express it in
units of kilograms, to match the units of our given density.
We can now recall that the
conversion between kilograms and grams is that one kilogram is equal to 1000
grams. Based on this, we can write the
mass of our gold sheet, 𝑚 sub 𝑠, this way. We can say it’s equal to 0.0094
kilograms. Or writing this in scientific
notation, that mass is 9.4 times 10 to the negative third kilograms.
We now have expressions for the
mass and the volume of our gold sheet. And we can recall at this point
that, in general, the density, 𝜌, of an object is equal to the mass of that
object divided by its volume. This means the density of our
sheet, which is equal to 𝜌 sub 𝑔 since the sheet is entirely made of gold, is
equal to the mass of the sheet divided by its volume.
When we substitute in the
values we’ve found for these two terms, we find that density is 9.4 times 10 to
the negative third kilograms divided by 0.30 meters times 0.40 meters times
𝑡. And this whole fraction is
equal to what we were given for the density of gold earlier, 19320 kilograms per
cubic meter. That’s a wonderful thing
because now we can solve for 𝑡 in the left-hand side of this expression.
To do it, we’ll multiply both
sides of the equation by 𝑡 so that that term cancels out on the left-hand
side. And then we’ll multiply both
sides of the equation by the inverse of this gold density. That is, we’ll multiply by one
meter cubed divided by 19320 kilograms. Looking at the right-hand side
of our equation, that means that 19320 kilograms cancels in the denominator and
numerator, as does the meters cubed term. This means we’re left just with
𝑡 all by itself, the variable we want to solve for.
And then on the left-hand side
of the equation, notice what units we have now. On the top in the numerator,
we’re multiplying by meters cubed. And in the bottom on the
denominator, we’re dividing by meters squared. So this means two factors of
meters cancel out from top and bottom. Along with that, both numerator
and denominator have a factor of kilograms. So that unit cancels as
well.
What we’re left with is a bunch
of numbers, 9.4 times 10 to the negative third divided by 19320 times 0.30 times
0.40, and one single unit, a unit of meters. We can rearrange this left side
this way. We can write it so that now we
have one expression, which when we calculate it will just be some number, and
then a unit, the unit of meters.
And remember, this left side is
designed to calculate the thickness, 𝑡, of our gold sheet. And we see that we’ll get that
thickness now in units of meters. But at this point, we can
recall that we don’t want that answer ultimately in meters. We want it in millimeters to
two significant figures.
So then let’s recall the
conversion from meters to millimeters. One meter of distance is equal
to 1000 millimeters of distance, all of which means if we wanna replace this
meter with some amount of millimeters, that one meter will be equal to 1000
millimeters.
And then with that calculation
done, we’re finished all of our setup. We now have an expression for
the thickness of the gold sheet in units of millimeters. All that’s left for us to do is
to calculate this expression and then round it to two significant figures. When we do that, when we
calculate as well as round, we find a result of 0.0041 millimeters. That’s the thickness of this
gold sheet in those units.
Let’s take a moment now to
summarize what we’ve learned so far about calculating density. In this lesson, we learned that the
density of an object tells us how much of its mass fits within a certain volume. Written as an equation, that’s
expressed this way, that an object’s density is equal to its mass divided by its
volume. And we saw that a shorthand way to
write this is to use the Greek letter 𝜌 to symbolize density, 𝑚 for mass, and
capital 𝑉 for volume.
Furthermore, we learned that, to
calculate an object’s density, unit conversion is often needed. This is because we may wanna solve
for object density in units of kilograms per cubic meter or grams per cubic
centimeter or milligrams per cubic millimeter, or even some other set of units. To calculate density in any units,
we’ll want to recall the conversions between kilograms and grams; grams and
milligrams; and meters, centimeters, and millimeters. Those are the primary unit
conversions involved in calculating density.
