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## High school physics - NGSS

# Specific heat and latent heat of fusion and vaporization

Defining specific heat, heat of fusion, and heat of vaporization. How to calculate the amount of heat to change the temperature of water and the energy required to change for a phase change. Created by David SantoPietro.

## Want to join the conversation?

- Just to be clear, the overall Q required to turn -40c to 160 will be the addition of all five Q values at each point on the graph + Q5?(119 votes)
- Don't worry guys, I got this. Total net value of Q = 9,647,400 J(31 votes)

- Why is the heat of fusion different from the heat of vaporization?(43 votes)
- good question

Think about the spacing of the molecules during phase change.

Latent heat of fusion: Solid to liquid...not much space increase

Latent heat of vapourisation: Liquid to gas...big space increase

The larger distance between the molecules means much larger potential energy

More energy is required, therefore, to turn the liquid into gas to provide that potential energy between the molecules(106 votes)

- Is it coincidence that the specific heat of water is approximately twice that of both ice and steam?(37 votes)
- Well, ice happens to be a bit special as water is denser than it, unlike how it is for most solids. And since the liquid state is denser than both the gas and solid, that is likely (I could be wrong since I don't know how to mathematically prove this) the reason why water's specific heat is greater than both.

And since ice is denser than vapour, phase change from ice to water has slightly higher heat requirement than that for phase change of water to vapour.(2 votes)

- what is heat capacity and specific heat capacity(12 votes)
- Heat Capacity: ratio of the amount of energy absorbed to the associated temperature rise.

•Example: if it takes 10 calories to raise the temperature of a glass of water by 2 °C, then the heat capacity of the glass of water is 10 calories/2°C = 5 calories per °C.

•Specific Heat: the heat capacity of a substance per unit mass

•Example: for water, it takes 1 calorie to raise the temperature of 1 gram of water by 1°C. So the specific heat for water is 1cal/gram °C(23 votes)

- what is the meaning of latent heat in fusion and in vaporization?(7 votes)
- Think of it this way: when a pot of water is kept boiling, the temperature remains at 100C until the last drop evaporates, because all the heat being added to the liquid is absorbed as latent heat of vaporization and carried away by the escaping vapor molecules(15 votes)

- Ice at 273 K has more cooling effect than water at 273 K. Why?(7 votes)
- because when ice changes state to water at this temperature it must absorb a certain amount of heat from it's surroundings. So unlike liquid water at 273 K, that only requires 4.18J/gC to raise it one degree, the ice is going to require more energy and draw that energy from its surrounding, making the surroundings even more cool.(19 votes)

- In the very last example to turn the block of ice into steam, wouldn't you have to add all the heat values up? Because 361,800J is much smaller than the other heat values, such as the phase changes.(5 votes)
- At the very end of the video, he draws circles around all five heat values, implying that you need to add up all the five steps but he doesn't state what the total is. He could have been more explicit.(20 votes)

- 0.5kg of ice at -5degC is put into a vessel containing 2kg of water at 15deg C and mixed together, the result being a mixture of ice and water at 0degC.Calculate the final masses of ice and water, taking the water equivalent of the vessel as 0.15kg.The specific heat of ice is 2.04kJ/kg/K and the latent heat of fusion is 335kJ/kg.(4 votes)
- heat to warm ice + heat to melt ice + heat to cool water + heat to cool vessel = 0

q₁ + q₂ + q₃ + q₄ = 0

m₁c₁ΔT₁ + m₁ΔfusH + m₃c₃ΔT₃ + m₄c₃ΔT₃ = 0

q₁ =0.5 kg × 2.04 kJ·K⁻¹kg⁻¹ × (0 –(-5)) K = 5.1 kJ

q₂ = m₁ × 335 kJ·kg⁻¹

q₃ = 2 kg × 4.184 kJ·K⁻¹kg⁻¹ × (0-15) K = -124.4 kJ

q₄ = 0.15 kg × 4.184 kJ·K⁻¹mol⁻¹ × (0-15) K = -9.4 kJ

5.1 kJ + 335m₁ kJ·kg⁻¹ - 125.5 kJ - 9.4 kJ = 0

5.1 + 335m₁ kg⁻¹ - 125.5 - 9.4 = 0

335m₁ kg⁻¹ = 124.4 + 9.4 – 5.1 = 129.8

m₁ = 129.8/335 kg⁻¹ = 0.39 kg

So, 0.39 kg of ice have melted.

∴ The mass of the remaining ice is (0.5 – 0.39) kg = 0.1 kg(15 votes)

- I don't understand how the temperature can stay constant during a phase change if the phase change releases heat anyway. Wouldn't the heat released from the phase change affect the overall temperature?(7 votes)
- To add on to what Andrew M said, I like to think about phase changes as converting the incoming thermal energy (when you're heating something up) to potential/phase energy. In a certain phase, such a solid, the particles can only move so much because its IMF is holding them together. At a certain point, the incoming thermal energy can't make the particles move any faster/KE can't increase because of the phase it's in. Instead, the energy is converted to phase energy and allows the particles to break free of their IMF. When the energy is converted to phase energy, you're right in that there is a tiny drop in temperature but it's replaced almost immediately when you're continuously adding energy into a substance.(6 votes)

- Can someone please explain to me how he solved for T final at6:01(7 votes)
- Heat Capacity: ratio of the amount of energy absorbed to the associated temperature rise.

•Example: if it takes 10 calories to raise the temperature of a glass of water by 2 °C, then the heat capacity of the glass of water is 10 calories/2°C = 5 calories per °C.

•Specific Heat: the heat capacity of a substance per unit mass

•Example: for water, it takes 1 calorie to raise the temperature of 1 gram of water by 1°C. So the specific heat for water is 1cal/gram °C(1 vote)

## Video transcript

- [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.