- [Voiceover] Let's talk about specific heat
and the heat of fusion and vaporization.
Let's say you had a container of some liquid
and you wanted to increase the temperature,
you'd probably add heat.
But how much heat should you add?
There's a formula for it.
Let's try to figure out what should go in here.
It's gonna depend on a few things.
For one it's gonna depend on
how much you want to increase the temperature.
So, what is the amount by which you want to
increase the temperature?
The more you wanna increase the temperature
the more heat you'll need to add.
It's also gonna depend on how much of the material you have.
In other words, the mass of this liquid in here.
The more you have the more heat you're gonna have to add
to change that temperature.
And it depends on one more thing,
the specific heat of that particular material.
So, different materials will be harder to
increase in temperature than other materials.
And if a material has a high specific heat
it'll take more joules of heat
in order to increase the temperature.
So let's get a little more specific.
Let's say our liquid was water
and it was at a temperature of 20 degrees Celsius.
Let's say it was a big container that had
two kilograms of water in it.
Now, the specific heat,
you can look those up, or sometimes they're given.
It turns out the specific heat of water
is 4,186 joules
per kilogram degree Celsius.
And these units give you an idea
of what the meaning of the specific heat is.
It's telling you that water takes 4,186 joules
to heat up one kilogram by one degree Celsius.
So you can already tell,
this is gonna take a lot of heat.
Water has a very high specific heat.
It can store a lot of heat energy
without raising its temperature by much.
And so let's say the question was
we wanted to get this temperature up to
50 degrees Celsius.
How much heat energy are we going to add
in order to get it to 50 degrees Celsius?
Well, we can come over to here.
The amount of heat that we're going to need will be
the mass is two kilograms,
so two kilograms times the specific heat
which is 4,186
times the change in temperature.
This really means T final minus T initial.
So what's T final gonna be?
T final is 50 degrees Celsius,
that's where we wanted to get,
minus the initial, it started at 20 Celsius.
And now we can solve this for the amount of heat.
And if you multiply all this out
you get 251,160 joules.
That's a lot of heat energy
just to get water to increase by 30 degrees Celsius.
That's why we often use water as a heat sink.
You can put a lot of heat in water
and not change the temperature by all that much.
But that example was pretty simple.
Let's look at a harder one.
Let's say instead of heating the cube
with a fire underneath
we just take a hot piece of metal.
Say we've got a 0.5 kilogram piece of copper
and we drop it into the water.
We've heated up this copper
and now we drop it in the same container of water.
And we wanna know what equilibrium
temperature will they reach?
The copper is gonna cool down, the water is gonna heat up.
Eventually they're gonna reach equilibrium
at some temperature.
What temperature will that be?
Well we're gonna need to know a few things.
I already told you the mass of the copper.
We're gonna need to know its initial temperature,
so the initial temperature of the copper.
I said we made it really hot.
Let's say it's 90 degrees Celsius.
We'll need to know the specific heat of copper
and it turns out the specific heat of copper is 387.
And let's say that water has the same
properties it had initially.
There were two kilograms of it.
The specific heat is always 4,186 for water.
And let's say it started at a temperature
of 20 degrees Celcius.
So we know the equilibrium temperature,
the temperature at which these are going to meet
is somewhere between 20 and 90.
We have to figure out exactly where it's gonna be.
And the trick we're gonna use is if you think about it
the copper is gonna lose heat,
the water is gonna gain heat,
how do those heats compare?
They've gotta be the same
assuming no heat is being lost to the environment.
We're gonna assume no heat's lost
so you want this to happen in
what's often called a calorimeter,
something insulated,
something that prevents any heat from getting out.
And if no heat gets out
then whatever heat the water gains
has to be the same as the heat lost by the copper.
Basically, if you add up the heat from the copper
plus the heat from the water you're gonna get zero
because one of these are gonna be negative
and one of these gonna be positive,
and they're gonna be the same absolute value.
How do you find these?
We had a formula.
Remember Q equals MC delta T which I like to remember
because it looks like MCAT,
so MC, this delta looks like an A to me,
so this looks like Q equals MCAT.
So I've got to use the mass of copper
there was 0.5 kilograms of copper
times the specific heat which is 387,
times the change in temperature.
I don't know the final temperature,
that's okay.
I'm going to name my ignorance and give it a variable here.
I'm gonna call it T final minus,
I don know the initial.
It started at 90 degrees Celsius,
so minus 90 degrees Celsius,
plus the heat gained by the water
which we can use the same formula for, MCAT.
So the mass is two kilograms.
The specific heat is 4,186.
T final, I still don't know T final,
but I do know the T initial,
the initial temperature is 20 degrees Celsius.
And I've got to set this all equal to zero.
Ran out of room there.
Sorry about that.
This looks a little intimidating now
you got this big equation.
You've got your unknown hidden in here.
Is this solvable?
Yeah, it's solvable.
Look, you've only got one unknown.
The unknown is T final.
These are both the same variable.
This is the temperature at which
the water and copper are gonna meet.
This whole term here, this orange term,
is gonna come out to be some negative number
because the copper is gonna lose heat energy,
and the water term is gonna come out positive
because it's gonna gain heat energy.
The two will cancel out.
They'll give you zero.
That's the condition that we're requiring.
We just have to solve for T final
which means we multiply all this out,
combine the T final terms,
and then solve for T final.
First, I'm just gonna multiply everything out.
I'm gonna combine the T final terms,
and the two terms that don't have T final.
Then I'm gonna move everything over to the other side.
I solve for T final and I get 21.58 degree Celsius.
And when you look at this you might think,
"Hey, we must have screwed up."
21.58?
The water started at 20.
It barely increased its temperature at all.
Yeah, that's what we were saying is that
this specific heat for water is so high
you can add a lot of heat
and it doesn't change its temperature much.
Note that we could've added in this container.
This container might absorb some of that heat,
so we could've had another term over here,
Q of the container,
and taken that into account,
or we could've dropped another cube of something in here
and we could've added that over here.
If you add up all the Qs from everything involved
that might gain heat or lose heat
you can set that equal to zero
because if no heat's getting out
that heat just has to be interchanged within here.
No heat is gonna be created or destroyed.
It just gets transferred amongst the
materials that are interacting.
So this is the key problem solving idea
when you're doing these specific heat problems.
You set it up with this
and then you solve for the unknown.
In this case it was T final.
Sometimes the thing you won't know
would be the mass of one of them
or the specific heat of one of them
regardless, you solve for the thing you wanna find.
Let me ask you another question.
Let's say we took the same amount of water,
two kilograms at a temperature of 20 degrees Celsius,
but this time I wanna know how much heat
do I have to add in order to boil
all of this water into steam?
Well, the first thing we need to do
is get this to the boiling temperature.
And the boiling temperature of water
is 100 degrees Celsius.
So Q, I first have to use MCAT.
MC delta T.
The mass is two.
The specific heat of water is 4,186.
The change in temperature,
well, the boiling point of water is 100,
so I have to get this water from 20 to 100.
That means T final is 100.
T initial was 20.
And I get that the heat that needs to be added
in order to get this to the boiling temperature is
669,760.
I'm being sloppy with significant figures
but this is the number you get from that calculation.
But that's not enough to boil it.
That's just the heat required
to get the water up to 100 degrees Celsius.
That's not enough.
If you get water to a 100 degrees Celsius
and let is sit there, it'll just sit there.
It won't actually boil.
You gotta keep adding heat.
How much more heat are we gonna have to add?
Once this water gets to 100,
in order to boil all of this water into steam,
for that you need to know about the heat
of fusion and vaporization.
In this case since it's boiling
the heat of vaporization because
we're turning liquid into vapor.
If we were turning liquid into a solid
it'd be the heat of fusion.
The formula for the heat of fusion and vaporization
looks like this.
Q, the amount of heat you need to add
in order to change the phase.
This is what happens when you change phase,
heat of fusion and vaporization.
Specific heat is what happens
when you change temperature.
So this calculation showed us
how much heat we needed to change
the temperature by 80 degrees Celsius.
This calculation is gonna tell us
once we're at a 100
how much heat do we need to add
to change the phase of all of this water into vapor?
And the formula for the heat of
fusion and vaporization looks like this.
Q equals ML.
M is the mass.
The more mass you have then the more heat you have to add.
L is the latent heat of fusion or vaporization.
This is a number similar to the specific heat,
but instead of telling you how much
heat you have to add in order to change the temperature,
this is telling you how much heat you have to add
in order to change the phase.
And it turns out the latent heat
of vaporization for water is huge.
2,260,000 joules per kilogram.
This means it takes 2,260,000 joules
to turn one kilogram of water,
which is at the boiling point,
into one kilogram of vapor.
So, if I'm trying to get this water from 20 degrees
turned into vapor,
first I need to MC delta T,
get it to 100 degree Celsius.
And then I need to add to that another amount of heat,
an amount of heat M times L.
The mass is two kilograms of this water.
L for water is 2,260,000.
So I get that it will take
669,760 joules to get the water
up to the boiling point,
and it will take another
4,520,000 joules
to turn that water into vapor
which in total gives
5,189,760 joules
in order to get this from 20
all the way up to two kilograms of vapor.
Now, there's one more thing that I wanna show you.
Let's get rid of this.
Let's say instead of starting with water
you started with some ice.
Let's say you started with a big piece of ice,
three kilograms worth of ice.
And it was very cold.
This wasn't just at zero.
Let's say this started at an initial temperature
of negative 40 degrees Celsius.
The question I wanna know
how much heat would we have to add
in order to turn this three kilogram block of ice
into three kilograms of water vapor?
But not just get it to the point of being vapor.
I wanna get it to hotter than a 100 degrees Celsius.
I wanna get this to a T final
of 160 degrees Celsius.
How much heat would I have to add
in order to do that?
And to track this, to visualize this a little better,
I'm gonna track what the temperature is
on this vertical axis as a function of
how much heat we've added into this system.
Let me quickly show you how not to do this.
A naive approach would say, "Okay, Q equals MC delta T."
My M is three kilograms.
My C is, well, whatever my C is.
We'll talk about that in a minute.
Delta T.
All right, delta T, my final temperature
is 160 degree Celsius.
My initial temperature is negative 40.
You can't forget the negative sign.
And I just plug in my specific heat
and I find my value, this is wrong.
You cannot do it this way.
For one what specific heat are you gonna put in?
The specific heat of water?
The specific heat of ice?
The specific heat of vapor?
They all have different specific heats.
And for two there's gonna be phase changes.
First the ice turns into water
and then at some later point the water turns into vapor.
You can't just neglect and gloss over those phase changes.
So this is not how you do it.
Here's what you should do.
We're gonna start at negative 40 degrees Celsius.
I know this is above the axis over here
but just pretend like this isn't
the zero point for this vertical axis.
And we're gonna add heat and that's gonna bring
this temperature up to zero degrees Celsius,
and we have to stop there.
We have to pause at zero degrees Celsius
because we have to pause every time
there's gonna be a phase change.
So how much heat does this correspond to right here?
We can use MC delta T.
M is three kilograms.
The specific heat of ice is about 2,090
and the final temperature is zero,
our initial temperature,
so T final minus our initial is negative 40.
Don't forget that negative sign.
And if you calculate that you get that amount of heat.
It's 250,800 joules.
But that's just getting the ice up to the melting point.
Now we need to melt it.
What's this graph gonna look like while it melts?
Well the temperature of the ice
is gonna stay constant while this thing melts.
As you're melting the ice cube
the temperature doesn't change.
All that energy is going into breaking those bonds
and turning this ice into water.
So, how much heat does this correspond to?
It's a phase change so you gotta use Q equals ML.
The M is three kilograms.
The latent heat, we can't use the latent heat
of vaporization.
This is a solid turning into a liquid.
That's latent heat of fusion that we need,
and the latent heat of fusion for water
is about 333,000 joules per kilogram
which gives you 999,000 joules of heat
in order to turn this ice at zero degree Celsius
into water at zero degrees Celsius.
Now you see how this works.
We've got to take this water at zero up to what?
Not up to 160.
Up to 100 degrees Celsius because that's
where it's gonna turn into steam.
And whenever there's a phase change you gotta pause
because your specific heat is gonna change its value.
So this Q that is required right here
we can do MC delta T,
specific heat of water is 4,186.
Delta T is, final is 100, initial is zero.
We get 1,255,800 joules.
Now we've got to turn that water into steam.
How much heat is that gonna take?
That's a phase change.
So we've gotta use ML, the mass is three kilograms.
The latent heat of vaporization is
2,260,000 which means we need
6,780,000 joules in order to turn
this water into vapor,
and we've got one more step to go.
We've got to turn this vapor at a 100
into vapor at a 160.
We've got to do one more MC delta T,
the mass of the steam is three kilograms.
The specific heat of steam is about 2,010.
The final temperature is 160.
The initial temperature was 100
which gives us 361,800 joules.
That's how much heat it would take
to get this ice cube at negative 40
into vapor at 160.