Video Transcript
A chunk of iron with a mass in air
of 390.0 grams is found to have an apparent mass of 350.5 grams when completely
submerged in an unknown liquid. Assume that the iron has a density
of 7.86 times 10 to the third kilograms per meter cubed. What mass of fluid does the iron
replace? What is the volume of the iron? Calculate the fluid’s density.
In the statement, we’re told the
mass of the iron in air, 390.0 grams. We’ll name that mass 𝑚. We’re told that when the iron is
submerged completely in an unknown liquid, it has an apparent mass of 350.5
grams. We’ll name that 𝑚 sub 𝑎. Knowing that the density, 𝜌, of
iron is 7.86 times 10 to the third kilograms per cubic meter, we want to solve for
three things. First, the mass of fluid that the
iron replaces. We’ll name that 𝑚 sub 𝑓. Second, we want to know the volume
of the iron. We’ll name it 𝑉. And finally, we wanna solve for
this unknown fluid’s density. We’ll name that 𝜌 sub 𝑓.
To start in on our solution, let’s
consider the difference between the apparent mass and the true mass of the iron. Say that we have our chunk of iron
and we suspend it by an effectively weightless chord from a scale. The scale reads out a mass. And we’ve called that mass in air
𝑚. Then, imagine we take the same
setup but we submerge the chunk of iron in our unknown fluid. When we do that, the scale records
the value we’ve named 𝑚 sub 𝑎, the apparent mass of the iron.
By a corollary of Archimedes’
principle, the mass of the fluid displaced by the iron must be equal to the
difference between the true mass, we can call it the mass in air of the iron, and
its apparent mass. Written as an equation, we can see
that 𝑚 sub 𝑓 is equal to the mass of the iron in air minus the apparent mass, or
the measured mass when it’s in the fluid. That’s 390.0 grams minus 350.5
grams, or 39.5 grams. That’s the mass of fluid displaced
by the iron.
Next, we wanna solve for the volume
𝑉 of this chunk of iron. We recall that the density of an
object, 𝜌, is equal to its mass divided by its volume. Rearranging that expression for our
case, we can write that the volume of the iron is equal to the mass of the sample
divided by its density. We’ll use the mass of the sample
measured in air. We can call that the true mass of
the iron. Substituting in our values for that
mass of the iron in units of kilograms and the iron’s density, when we calculate
this fraction, we find it’s 49.6 centimeters cubed. That’s the volume taken up by this
chunk of iron.
Finally, we want to solve for the
density of the unknown fluid that the iron is submerged in. To figure this out, we can again
rely on our expression for density. The density of the fluid is equal
to the fluid’s mass divided by its volume, which is the same as the volume of the
submerged iron. Plugging in for these two values,
when we calculate this fraction, we find it’s 0.796 grams per centimeter cubed. That’s the density of the fluid the
iron is submerged in.
