### Video Transcript

A glass beaker has a capacity of
397 millilitres. The beaker is filled to the brim
with water that is at a temperature of 15.0 degrees Celsius. Determine the volume of water that
will overflow the beaker when the water’s temperature increases to 25.0 degrees
Celsius. Use a value of 207 times 10 to the
negative sixth inverse degrees Celsius for the coefficient of volumetric expansion
of water.

Let’s start off our solution with a
sketch of this glass beaker. We start out with this beaker of
known volume. And we’re told that it’s then
filled to the brim with water. So the water completely fills this
beaker. And it’s just about to spill over
the edge. And initially, this volume of water
filling the beaker is at a temperature of 15.0 degrees Celsius. Then, say that we heat the water,
perhaps by a flame underneath, somehow adding heat to it to raise its
temperature. The water goes from 15.0 to 25.0
degrees Celsius. Based on this temperature change
and something called the coefficient of volumetric expansion of water, we want to
figure out how much water will overflow the beaker.

Now here is what this number, the
coefficient of volumetric expansion, means. It tells us that, as the
temperature of the water increases, so does the volume of the water. In other words, the hotter the
water, the more space it takes up. Now, if we rewind to the time when
the water temperature in the beaker was 15.0 degrees Celsius, remember we’re told
that, at that temperature, the water completely fills the beaker. It’s absolutely brimming full. So if water expands when it’s
heated, that means when the temperature of the water goes up, the volume will go up
as well. And some of this water will spill
over the edges of the beaker. It’s that volume of spilled water
that we want to solve for. And we can do it using a
mathematical equation that ties together the change in the volume of a substance
with its coefficient of volumetric expansion and its temperature change.

Here it is. It’s a somewhat simple mathematical
relationship that tells us that the ratio between the change in an object’s volume
and its original volume is equal to 𝛼 sub 𝑣, its coefficient of volumetric
expansion, multiplied by its change in temperature Δ𝑇. Since we want to solve for the
amount of water that overflows, we know that that’s Δ𝑉, the change in volume,
because the initial volume 𝑉 completely filled the beaker. Here’s what we can write then. We can say that the change in
volume of the water, that is, the amount that overflows the beaker, is equal to its
initial volume multiplied by 𝛼 sub 𝑣 multiplied by the change in temperature. Now capital 𝑉, the original
volume, is given as 397 millilitres. The coefficient of volumetric
expansion is given as 207 times 10 to the negative sixth inverse degrees
Celsius. Now what about Δ𝑇? Well, that’s equal to the final
temperature of the water minus its initial temperature, 25.0 minus 15.0 degrees
Celsius.

When we plug all this in,
recognising that 25.0 minus 15.0 degrees Celsius is 10.0 degrees Celsius, notice how
the units of degrees Celsius cancel out with inverse degrees Celsius, leaving us
with the units of volumes in millilitres. That’s a good sign, because we’re
calculating a change in volume, Δ𝑉. Multiplying these numbers together
to three significant figures, we find the result of 0.822 millilitres. So this tells us that when we heat
up our original volume of water, 397 millilitres, by 10.0 degrees Celsius, then the
volume of that water expands by a little bit less than one millilitre. That’s how much overflows the
beaker.