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
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
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
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.