Lesson video

Hello there, I'm Mr. Forbes and welcome to this lesson from the energy of moving particles unit.

This lesson's called, specific latent heat and it's all about changes of state and the energy required for those changes of state.

By the end of this video, you're going to be able to describe the changes of state and the energy changes required for those changes.

You're also going to be able to calculate the amount of energy required for a change of state.

Here are the keywords that will help you through the lesson.

The first of them is melting, and that's the change of state when a solid turns into a liquid, and that happens at a very specific temperature.

Then there's boiling, the change of state from a liquid to a gas.

And again, that happens at a different, but fixed temperature.

Then we have specific latent heat, and that's the amount of energy transferred when one kilogramme of a material changes its state from one state to another.

And finally, we've got specific heat capacity, and that's the change in internal energy, when the temperature of a substance changes by one degree Celsius and that's per kilogramme.

You can return to this slide at any point in the lesson.

This lesson's in two parts, and in the first part, we're going to look at a heating curve.

A graph we generate when we provide energy to a substance that cause it to heat up and/or change state.

In the second part of the lesson, we're going to look at specific latent heat in particular and how to calculate the energy required for a change in state.

So when you're ready, let's start by looking at interpreting a heating curve.

When you heat a substance, you provide it with energy and you'd expect its temperature to increase over time as you provided it with that energy.

I've got a graph here showing the temperature of a substance where I'm heating it at a constant rate over a period of time.

And as you can see, there are several key parts to this graph.

In the first part of the graph, here, we can see that the temperature is increasing with time, so it is actually getting hotter.

But then in this second part of the graph, we've got a period of time where the temperature isn't changing.

Then we have a part where the temperature changes again.

So we've got a gradual increase in temperature over this period of time.

And finally we have got another part here where the temperature isn't increasing again.

So we need to be able to explain why there's those different sections of the graph.

Let's start by looking at the first section of the graph.

And at this point, we've got a solid that we're heating, so we're gonna heat that solid providing it with energy until it reaches its melting point.

And what's happening during this is the average speed of the particles is increasing as we're providing them with more energy.

So as we heat them up with a Bunsen Iron, an electrical heater, the particles gain energy and the average speed increases, they get faster.

The temperature is actually related to the average speed of the particles.

So because the average speed's increasing, the temperature is increasing as well.

If we continue to heat a solid, it's going to increase in temperature until it reaches its melting point.

And it's melting point for a pure substance is a fixed value.

So I've got different substances with some different melting points here.

The one you should be familiar with is water.

If you heat up water until it reaches zero degrees celsius, it's going to start to melt.

So the ice is going to start to change into liquid water.

But other substances have different melting points.

Ethanol's are a much lower melting point and gold are much higher, over 1,000 degrees Celsius.

Mercury is an example of a metal that is liquid at room temperature.

It melts at minus 39 and so on.

Here's a question for you now.

An engineer finds the melting point of a north 0.

5 kilogramme sample of a metal and it's 600 degrees Celsius.

What's the melting point of a one kilogramme sample of the very same metal? Pause the video, make your selection, and then restart please.

Welcome back, hopefully you selected 600 degrees Celsius.

The melting point of a substance is the same no matter how much of it there is.

It doesn't depend on the mass of the metal sample.

If a small sample is a melting point of 600 degrees celsius, a large sample will have the same melting point.

So well done, if you selected that.

Once a substance reaches its melting point, its temperature stops rising even though you're providing it with energy.

So my graph here shows that the temperature isn't rising during this next section.

So during the melting process, we are not getting a change in temperature.

Energy is still being transferred to that substance, but the speed of the particles or the average speed of the particles isn't increasing anymore because the temperature's not increasing.

The faster moving particles are instead starting to pull away from each other.

They're separating slightly, but they're being slowed down by the electrostatic forces of attraction that keep them together.

So what we're doing instead with that energy is the particles are separating without gaining any extra speed.

The temperature doesn't increase because there's no increase in speed of the particles or average speed of the particles at all.

So we're getting no temperature change because the energy is doing something else.

It's causing the particles to separate rather than get faster.

If we continue to heat the substance once it's melted, that liquid's going to gradually increase in its temperature towards its boiling point.

So that's the next section of the graph as indicated here.

What's happening there is the speeds of the particles is increasing again, and because the speed of the particles is increasing, then the temperature is rising again.

So always linking that temperature to the average speed of the particles.

So during this section we get a gradual increase in temperature as we provide energy.

That increase in temperature will continue until we reach the boiling point of the substance.

And I'm gonna talk about the boiling point of just pure substances 'cause if you mix things together then you'll get a range of boiling points.

But a pure substance has a fixed boiling point.

So for example, the boiling point of water, I'm sure you are aware, is 100 degrees Celsius.

And I've got a boiling point of some other substances here.

Ethanol's got a lower boiling point that will boil much more easily.

You can boil gold if you actually raise its temperature to almost 3000 degrees Celsius.

So it's very difficult to do, but it can be done.

And something like nitrogen, well that's boiling point, is very, very low.

And so it's a gas at room temperature.

It's already all boiled and turned into a gas.

Another check for you here.

I've got a type of wax and it's got a melting point of 55 degrees Celsius and a thermometer is placed into a mixture of solid and molten wax.

What temperature will that thermometer show when I've got solid and molten wax together? So pause a video, make a decision and restart please Welcome back, hopefully you pick 55 degrees Celsius.

The melting point is 55 and that's the only temperature at which you'll have solid and liquid wax at the same time.

So well done, if you selected that.

When that liquid reaches its boiling point, its temperature stops rising again.

So we've got another section of the graph here where there's no increase in temperature.

What's happening there is energy is still being provided to the substance.

So it's still gaining energy 'cause we're still heating it, but the average speed of the particles isn't increasing again.

Remember, average speed is linked to temperature.

So what's going to happen is that the faster particles in that liquid are gonna start to escape from the liquid and those faster particles are taking away excess energy and that means that the particles left behind, their average energy and their average speed isn't changing.

So that remains constant and so the temperature remains constant 'cause we're getting no changes in the average speed of the particles.

That temperature cannot increase anymore until all of the particles in that liquid have gained enough energy to escape the electrostatic forces of attraction between them and the whole of the substances changed into a gas.

So we get no change in temperature at the boiling point.

Okay, another check for you now, the melting point of ice is zero degrees Celsius.

Ice is kept inside a freezer and the temperature of that freezer is set to -10 degrees celsius.

As soon as the ice is taken out the freezer, it's placed in a beaker and its temperature is measured each minute until all the ice is melted.

Which of those three statements is correct please? So pause the video, made a decision and restart.

Welcome back, hopefully you select the last of those.

The temperature of any water in the beaker during the melting process is going to be zero degrees Celsius until all the ice has melted in it.

Well done if you've got that.

Okay, now it's time for the first task of the lesson and what I'd like to do is to have a look at this graph.

It shows the cooling of a pure substance over a period of time.

So I've got this pure substance and put it somewhere very cold until it changes state and is all turned into a solid.

What I'd like you to do is to label the boiling point and the melting point of the substance on the graph and describe what happens to the temperature and the behaviour of the particles in the substance during the cooling processes X and Y there.

So there's just two phases in the graph.

So pause the video, work out your answers to those and restart please Welcome back, hopefully your answers look something like this.

I've marked the boiling point and the melting point there.

The boiling point is the higher of those temperatures where there's no change in temperatures.

So the flat line towards the top is the boiling point and the flat line lower down is the melting point.

And the descriptions of what's happening to the temperatures and the particle behaviour, during phase X, the temperature is decreasing.

You can see that the temperature goes down.

So that must mean the average speed of the particles is decreasing, some of the energy is being dissipated to the surroundings.

That's where the energy goes.

During phase Y, the energy of the particles is still decreasing, I'm still cooling it there, but the temperature's not decreasing.

What's happening there is the energy is being dissipated as the particles move closer together and the forces of attraction, the electrostatic forces between them get stronger.

Well done if you've got answers like that.

Now it's time to move on to the second part of the lesson and then it we're going to be looking at specific latent heat, which is to do with the energy requirement for a change of state and we'll be doing some calculations.

So let's do that.

You'll need different amounts of energy to change the state of different substances.

It's easier to change the state of some than it is to change the state of others.

And that's because there's different strengths of bonds between the particles in that substance.

The amount of energy required to change the state is called the latent heat.

So for a particular substance we'll have a latent heat and that's a measure of how much energy it would require to change its state.

And for a fair measure of how difficult it is to melt or boil a substance, we use the specific latent heat and that's the amount of energy per kilogramme for a particular substance.

So specific latent heat is how much energy is required to change a state of one kilogramme of a substance? As we've seen already, there are two different changes of states, we can have melting and we can have boiling or conversely freezing and condensing.

So there's two different changes of state there.

So there's two different values for the specific latent heat for each substance and those can be very different.

So we've got these changes of state.

I've got water changing from a solid to a liquid and that's melting.

And then changing from the liquid to a gas and that's evaporation.

And the reverse of those processes, which is condensation and freezing.

And the amount of energy per kilogramme for those changes in state for water shown here.

So to melt one kilogramme of water, I need 334 kilojoules for every kilogramme.

To freeze it then we'd have a change of energy that's exactly the same.

It would give out 334 kilojoules per kilogramme of water when you change state.

And that's very different from changing the state from liquid to a gas.

It requires much, much more energy to do that.

2,260 kilojoules per kilogramme to boil or evaporate one kilogramme of water.

And obviously the reverse process of that energy is dissipated back into the surroundings the exact same amount.

The table here shows you some of those specific latent heats for some different substances.

So you can see the values for water I've put in there, they are 334 kilojoules per kilogramme to melt or freeze and 2,260 to evaporate or condense.

So you can see quite a lot of energy required to change the state of water.

Ethanol, it's easier to change the state.

Again, it's more difficult to change it from a liquid to a gas than it is from a solid to a liquid, but it's easier to affect than water.

Mercury is very easy to change the state.

You can see the values the much lower.

And nitrogen are even lower still.

So nitrogen's very easy to change the state of.

We need very large amounts of energy for those changes in the state.

So that's why we're measuring them in kilojoules per kilogramme.

And changes the state from a solid to a liquid or liquid to solid involve those specific latent heats of fusion.

So the changes of state involving a solid, it's called a specific latent heat of fusion.

And the changes of state from a gas or back into from a gas is the latent heat of vaporisation.

As usual, there's an equation that links the three variables we've been discussing, energy, mass and specific latent heat.

And that equation is here.

Thermal energy for change in state is the mass times the specific latent heat.

And there're symbols that can be written out like this.

Q equals m times L where Q is the energy change in joules.

So we use the Q for energy change in this equation.

m is the mass in kilogrammes and L, capital L is the specific latent heat and that's measured in joules per kilogramme.

Let's have a look at an example of using that equation.

The specific latent heat of fusion for ice when it's melting is 334 kilojoules per kilogramme.

Calculate the energy required to melt 1.

5 kilogrammes of ice.

So I write out the equation Q equals mL and I substitute the values from the question.

The Q is what we're looking for, that's the energy.

We've got the mass as 1.

5 kilogrammes and we've got the latent heat of fusion and that's 334 kilojoules per kilogramme.

And then all I have to do is multiply those two together and that gives me an energy change of 501 kilojoules.

So it takes 501 kilojoules to melt 1.

5 kilogrammes of ice.

Now at your turn.

What I'd like you to do is to calculate the energy required to evaporate 1.

5 kilogrammes of water.

And I've given you the specific latency of vaporisation for water.

So pause the video, work out the energy required and then restart please.

Welcome back, hopefully you did this.

Write out the equation Q equals mL, substitute the values from the question and that gives a final answer of 3,390 kilojoules.

That's much higher energy value than the energy required to melt the ice.

So well done if you've got that.

We can also use the same equation to find the other variables.

So for example, we can try and find a mass of something that evaporates when we're heating it.

So I've got the specific latent heat of vaporisation for ethanol and it's 108 kilojoules per kilogramme and I'm gonna calculate the mass of boiling ethanol that will evaporate when I provide it with 50 kilojoules of energy by continually heating it.

So to do that I write out the equation, I substitute the values I can from the question and I know the energy it provided it this time, it's 50,000 joules.

And I know the latent heat of vaporisation, 108,000 joules per kilogramme.

And it's the mass which I've left as a symbol m that I'm looking for.

What I do then is rearrange the equation in terms of m and that gives me this equation and I can then calculate the mass.

And the mass that has evaporated is 0.

46 kilogrammes.

Now it's your turn.

So what I'd like you to do is calculate the mass of mercury that melts when I provide it with 550 joules of energy and the specific latent heat of fusion for mercury is 11 kilojoules per kilogramme though.

So pause the video, calculate that mass and then restart please.

Welcome back, you should have done this.

You write out the equation, you substitute the values that you do know, so you know the energy provided and you know the latent heat of fusion and then you rearrange the equation in terms of m and that gives a finance of not 0.

95 kilogrammes.

So well done if you've got that.

The specific latent heat and the specific heat capacity are two equations that contain very similar terms and so they can be confused because they're both about heat.

So let's just have a look at both of those equations together, yeah? We've been looking at specific latent heat and that's the energy required to change the state of one kilogramme of a material.

So the energy required, Q, is the mass of the material times the specific latent heat.

And that doesn't involve any change in temperature.

There's no symbol for temperature change there.

So that's the equation, that's not about changing temperature, it's just about changing state.

The other equation, specific heat capacity that is to do with temperature change and that does involve a symbol for temperature change then.

So the energy change there, the delta Q is the mass times the specific heat capacity times that delta theta, that change in temperature.

So that equation is about change in temperature, whereas the latent heat one doesn't have a change in temperature because it's just a change in state.

So be careful with those two equations please.

Okay, let's have a look at an example that involves using both of those equations together so we can see how we can use them without getting confused.

So I'm gonna calculate the total energy required to melt 4.

00 kilogrammes of aluminium starting from a temperature of 20 degrees Celsius.

And I've got some data for the aluminium there.

I've got its melting point, it's specific heat capacity and the specific latent heat of fusion that the energy required to melt one kilogramme of it.

So step one, what we need to do is to heat the aluminium to its melting point.

So it's starting off at 20 degrees Celsius and I've got to get it all the way to its melting point of 660 degrees Celsius.

So I'll write out the equation to do with changing its temperature and that's the equation involving specific heat capacity.

So the energy required there is mass times the specific heat capacity, times of temperature change.

So I can fill in those values using the data in the question and the data for the aluminium.

So there's four kilogrammes, the specific heat capacity is 900 joules per kilogramme per degree Celsius.

And the change in temperature here, well I've raised it from 20 degrees up to 660 degrees, so that's 640 degrees C.

So I can calculate the energy required just to heat it to it's melting point and that gave me a value of 9.

22 megajoules.

The next step is to find the energy required to melt the aluminium and this time we're gonna use the equation that involves specific latent heat.

So we've got the energy required is going to be the mass times the specific latent heat.

I can substitute those values in.

I've got four kilogrammes and I've got a specific latent heat of fusion 395,000 Jules per kilogramme and that gives me an energy required to melt the aluminium of 1.

58 megajoules.

And the final step then is it takes me some energy to heat the aluminium to the melting point and some energy to melt it.

All I need to do is to add those two values together to get the total amount of energy required.

So that's what I do.

The total energy needed is those two values added together and that gives me a total amount of energy required of 10.

80 megajoules.

And that's to three significant figures and I round it to 10.

8.

Okay, I'd like you to try that process for yourself now.

So I'd like you to calculate the total energy required to melt 1.

25 kilogrammes of gold from starting temperature of 20 degrees C.

And I've given you all the data you need for gold there.

So pause video, work out your solution to that and then restart when you're done please.

Welcome back, well hopefully your process looks something like this.

You've got the energy needed to heat the gold to its melting point.

We write out the equation, get the values, and we substitute all those in to give us the energy needed to get the gold to its melting point.

We also need to look at the energy needed then to melt the gold.

So we write out the equation for that involving this specific latent heat and that gives us value there.

And then finally, we need to get the total energy needed by adding those two together.

That gives an answer of 247.

1 kilojoules or 247 kilojoules to three significant figures.

Well done if you've got that.

Okay, now it's time for the final task of the lesson and what I'd like you to do is to answer these three questions.

So read through them carefully.

You're going to be using the equations for specific latent heat and specific heat capacity for the last question as well.

So pause the video, work through your solutions and restart when you're done, please.

Welcome back and here the solutions to those.

So for question one, we substitute the values into the energy equals mL equation there.

That's 221 kilojoules.

For question two, a more complicated one there.

We need to calculate the energy transferred to the water, that's 30,000 joules and then we can calculate the mass using that data and the latent heat for the water, specific latent heat of vaporisation and that's 13.

3 grammes.

So well done if you've got those two.

And here's the answer to the third part and that involves using both the latent heat equation and the specific heat capacity equation.

So the first part, we use that specific heat capacity equation to find out the energy needed to heat to go to it's melting point.

Then we use the latent heat equation to find the energy needed to melt the gold and we add those together to get the total energy needed.

And that's 39.

4 kilojoules.

Well done if you've got that.

And now we've reached the end of the lesson.

So here's a summary of everything we've covered.

During changes of state, the temperature remains constant.

So when you are heating a substance and it gets to its melting point, the temperature remains at that melting point until all of the solid is turned to a liquid.

And similarly for liquids to a gas and the boiling point.

When solids melt energy is transferred, is increasing the space in between the particles rather than the kinetic energy and the temperature.

Then we had specific latent heat.

And the specific latent heat of a substance is the energy required to change one kilogramme from one state to another.

And we've got an equation for that.

Thermal energy for a change in state is mass times specific latent heat or Q equals mL.

And Q is a change in energy in joules, m is the mass in kilogrammes and L is the specific latent heat in joules per kilogramme.

There's two values for the specific latent heat because there's two different changes of state.

So we've got specific latent heat of fusion and that's for solid to liquid and we have a specific latent heat of vaporisation and that's where liquids to a gas.

Well done for reaching the end the lesson.

I'll see you in the next one.