# Specific heat, heat of fusion and vaporizationÂ example

## Video transcript

A couple of videos ago, we
learned that if we started with solid water or ice at a
reasonably low temperature-- maybe this temperature
right here is minus 10 degrees Celsius. And we can deal with Celsius
when we're dealing with these phase changes, because we really
just care about the difference in temperature, and
not necessarily the absolute temperature. So one degree in Celsius is the
same thing is one degree in Kelvin so the differences
are the same whether you're dealing with Celsius
or Kelvin. So we're starting with minus
10 degree Celsius ice or solid water. And we learned that as you heat
it up, as you add heat energy to the water, the
temperature goes up. The molecules, at least while
they're in that ice lattice network, they just
start vibrating. And their average kinetic energy
goes up until we get to zero degrees. Which is the melting
point of water. And at zero degrees, we already
learned something interesting happens. The added heat in the system
does not increase the temperature of the
ice anymore. At least over that little
period right here. What's happening is that heat
energy is being used to kind of break the lattice
structure. To add potential energy
to the ice. Or essentially melt it. So for it here, right
here, we're ice. Right at this point, we're
zero degree ice. And then as we add more and
more heat we get to zero degree water. So at zero degrees, you can
either have water or ice. And if you have water, to turn
it into ice, you have to take heat out of it. And if you have ice, and you
want to turn it into water, you have to put heat into it. And then the heat is used
again to warm up the water at some rate. And then at 100 degrees, which
is the boiling point of water, right here, a similar phase
change happens. Where the increased heat is
not used to increase the temperature of the water, it's
used to put potential energy into the system. So the water molecules are
forced away from each other. The same way that if I'm forced
away from the planet Earth, I have potential
energy because I can fall back to the earth. Similarly, they have the
potential energy of falling back to each other. But this energy, right here
is the energy necessary to vaporize the water. Right here you have 100 degree
water, 100 degree liquid. And here you have 100 degree
vapor, water vapor. And then as you add more and
more heat, once again it increases the temperature. But you, Sal, I already learned
this a few videos ago, I have the intuition. But I want to deal with
real numbers. I want to know exactly how
much heat is required for these different things
to happen. And for that, we can
get these numbers. And these are specific to the
different states of water. If you looked up any other
element or molecule, you would have different values for these
numbers we're going to be dealing with right now. But this first number right here
is the heat of fusion. And this is the amount of heat
that's required to fuse 100 degree water into
100 degree ice. Or the amount of energy you have
to take out of the water. So this distance right
here, or along this axis, is 333 joules. If you're going in the the
leftward direction, you have to take that much out of the
system to turn into ice. If you're going in the rightward
direction, you have to add that much to
turn into water. So heat of fusion. It's called the heat of fusion
because when you fuse something together you
make it solid. So it could also be considered
the heat of melting. Just two different words for the
same, thing depending on what direction you going. The important thing is
the number, 333. Similarly, you have the
heat of vaporization. 2257 joules per gram. That's this distance along
this axis right here. So if you had one gram of 100
degree liquid water and you wanted to turn it into
one gram of 100 degree liquid vapor. And in all of this, we're
assuming that nothing silly is happening to the pressure,
that we're under constant pressure. You would have to put 2257
joules into the system. If you had 100 degree vapor and
you wanted to condense it, you would have to take that much
energy out of the system. OK, fine, you know how much
energy is required for the phase changes. But what about these
parts right here? How much energy is required to
warm up a gram of ice by one degree Celsius or Kelvin? And for that we look at
the specific heat. It takes 2 joules of
energy to warm up 1 gram 1 degree Kelvin. When water is in the
solid state. When it's in the liquid state,
it takes about double that. It takes about 4 joules
per gram to raise it 1 degree Kelvin. And when you're in the vapor
state, it's actually more similar to the solid state. So given what we know now, we
can actually figure out how much energy it would take to go
from minus 10 degree ice to 110 degree vapor. Let's work this out. So the first thing we're going
to be doing is, we're going to be going from minus 10 degree
ice to zero degree ice. So we're going to
go 10 degrees. We have to figure out how much
heat does it take to warm up ice by 10 degrees. So the heat is going to be
equal to the change in temperature. So actually let me write the
specific heat first. So 2.05 joules per gram Kelvin. Oh, and I should tell you, we
can't different values for the amount of ice we're warming up
to vapor, so let's say we're dealing with 200 grams. So it'll be the specific heat
times the number of grams we're warming up of ice times
the change in temperature that we're trying to get. So, times 10 degrees Kelvin. Le'ts just say it's a 10 degrees
Kelvin change, it doesn't matter if we're using
Kelvin or Celsius. I could have written
a Celsius here. Let's put Celsius right there. So what is that equal to? Get the calculator. Clear it out. 2.05 times 200 times times 10
is equal to 4,100 joules. Let me do this in a
different color. This is 4,100 joules. Fair enough. Now, so what we've done is just
this part right here. This distance right here
is 4,100 joules. Now we have to turn that
zero degree ice into zero degree water. And for that we use the
heat of fusion. We're adding that
amount of heat. It's 333.5 joules per gram. So that's equal to 335.55 joules
per gram times, we have 200 grams of ice that we're
trying to melt. So that is what? So let me get the
calculator out. So I have 335.55 times 200
is equal to 67,110. Let me do it in that color so
can sum them up at the end. 67,110 joules to
melt the water. Or take it from ice to water. Now, we have to go from, and
this is the big one. We have to go from zero degree
water to 100 degree water. Or liquid water. This is in the liquid state. So now we take the specific heat
of water which was 4.178. For some reason I'm thinking
it's 4.176 but it doesn't matter. So it's 4.178. I might off a little bit on that
number, but that digit is not that significant. Joules per gram Celsius
times 200 grams times 100 degrees Celsius. And notice, the Celsius and
the Celsius cancel out. The grams and the grams
cancel out. So we are left with joules,
which is what we want. We want to know how much heat
or how much energy we're adding to the system. Let me get the calculator out. So this stage is going to be
4.178 times 200 times 100 is equal to 83,560 joules. Does that look about right? Let's see. 4 times 200 is 800, 800 times
100; yeah, that's about right. Now, we're dealing with 100
degree water vapor, and we have to turn that 100
degree water vapor to 110 degree vapor. So we use the specific
heat of vapor. 1.89 joules per gram Kelvin. Multiplied by the amount of
vapor we're dealing with, 200 grams. That obviously
doesn't change. We're not adding or taking away
mass from the system. Times the temperature
change, times 10. So what is that? Let me get the calculator
out again. So we're dealing with
a 10 degree change. 10 degree Celsius times 200
times 1.89 is equal to 3,780. And I just realized, I made
a horrible mistake. It's not an irrevocable mistake,
otherwise I would rerecord the video. But I just figured out the
amount of energy to take it from zero degree water to 100
degree water, which is this energy right here. And now I just calculated how
much energy to go from 100 degree vapor to 110
degree vapor. Which is this right here, that
distance right here. I forgot to figure out how much
energy to turn that 100 degree water into 100
degree vapor. So that's key. So I really should have done
that up here before I calculated the vapor. But I'll do it down here. So to do 100 degree water
to 100 degree vapor. That's this step right here,
this is the phase change. I multiply the heat of
vaporization, which is 2,257 joules per gram times
200 grams. And this is equal to 451,400. I'll do it in that blue color. 451,400 joules. So this piece right here is
451,000 for our sample of 200 grams. This piece right here
was 83,000 joules. This piece right here
was 3,780 joules. So to know the total amount of
energy, the total amount of heat that we had to put in the
system to go from minus 10 degree ice all the way to 110
degree vapor, we just add up all of the energies we had to
do in all of these steps. Let's see. And I'll do them in
order this time. So to go from minus 10 degree
ice to zero degree ice. Of course we have
200 grams of it. It was 4,100. Plus the 67,000. I'm just doing this
off the screen. So plus 67,110. Plus 83,000. That's to go from zero degree
water to 100 degree water. Plus 83,560. So we're at 154,000 right
now just to get to 100 degree water. And then we need to turn that
100 degree water into 100 degree vapor. So you add the 451,000. So, plus 451,400 is
equal to 606. And then finally, we're at 100
degree vapor, and we want to convert that to 110
degree vapor. So it's another 3,700 joules. So plus 3,780 is equal
to 609,950 joules. So this whole thing when we're
dealing with 200 grams, when you're going from
minus 10 to 110. And remember, this is
for 200 grams. It took us 609,950 joules. Or 609 kilojoules to do this. So that by itself is an
interesting thing. You might think, wow, this
is a huge number. But actually, it turns out
joules isn't a lot of work. Kilojoules starts to become
a little bit interesting. You might realize, 200 grams
of ice, that's about half a pound of ice. So to take half a pound of ice
and warm it up on your stove would take 609 kilojoules
of heat to do that. So that's something that you
probably could do on your stove at home.