### Video Transcript

The table lists the specific latent
heat of fusion for various metals. A student has 200 grams of an
unknown metal. The student heats the metal to its
melting point and then measures how much energy is absorbed by the metal for all of
it to melt, and gets a value of 79.6 kilojoules. Which of the four metals listed in
the table does the student have?

Okay, taking a look at this table,
we see that it has two rows. The first row lists four different
kinds of metal, aluminum, cobalt, iron, and nickel. In the second row, the specific
latent heat of fusion of those metals is given in units of kilojoules per
kilogram. When we talk about the specific
latent heat of fusion, that refers to the amount of energy that a substance needs to
go from a liquid to a solid or a solid to a liquid. In other words, it’s an amount of
energy needed to solidify or liquify one kilogram of some material.

Now, in our scenario, we’re told
that a student has a 200-gram mass of some unknown metal. The metal begins as a solid but
then is heated up to its melting point. And the student then measures how
much energy is absorbed by the metal in order for all of it to melt. So, our 200-gram sample has now
gone from completely solid to completely liquid. The student measures the energy
required for that phase transition from solid to liquid to be 79.6 kilojoules. Based on this information, we want
to figure out whether the student is working with aluminum, cobalt, iron, or
nickel. To figure that out, let’s recall a
mathematical relationship between specific latent heat, mass, and energy.

In general, the energy, 𝐸,
required to affect a phase transition is equal to the mass of the substance going
through that transition multiplied by the specific latent heat of that
substance. Looking again at our table, we’re
given the specific latent heat, in particular of fusion, for these four different
metals. In other words, for these metals,
we know capital 𝐿. But we don’t yet know the specific
latent heat of fusion of the unknown metal that our student is working with. It’s that that we want to identify
to know which of the four metals we’re using. Now, here’s what we do know about
our unknown metal and the energy involved. First, we know the mass of our
sample. That’s given as 200 grams. And we also know how much energy it
took to completely melt this sample of metal. We can call that energy 𝐸, and
we’re told that it’s equal to 79.6 kilojoules, 79.6 thousand joules.

Now, taking a look back over at our
expression for the energy 𝐸 in terms of the mass and the specific latent heat. We can see that if we divide both
sides by the mass involved, then that term cancels from the right and we arrive at a
mathematically equivalent statement. That the energy involved in this
transition divided by the mass of the substance is equal to the specific latent heat
of that substance. And it’s that value, capital 𝐿,
that we want to solve for for our as-yet-unknown metal.

So, to do it, we’ll divide the
energy we needed to melt the metal by the mass of the sample. In other words, we’ll divide 79.6
kilojoules by 200 grams. But before we do that, notice the
units in which our specific latent heats of fusion are given. They’re kilojoules per
kilogram. And we have kilojoules per
gram. So, before we do this division,
we’ll want to convert our mass into units of kilograms. We can recall that one kilogram of
mass is equal to 1000 grams, which means that to convert 200 grams into kilograms,
we’ll shift the decimal place three spots to the left, at which point we can see
that 200 grams is equal to 0.200 kilograms.

Looking at the units in our
fraction now, we see that they’re kilojoules per kilogram. They’re a match for the units in
terms of which the specific latent heats of fusion of these metals are given. So, we’re ready to divide. When we do, we find a result of 398
kilojoules per kilogram. And looking through our table, we
see that this is a match for the specific latent heat of fusion of aluminum. These tells us that the metal the
student is working with is aluminum.