Heating curve for water | Thermodynamics | AP Chemistry | Khan Academy

Let's look at the heating curve for water. A heating curve has temperature on the y-axis, in this case, we have it in degrees Celsius, and heat added on the x-axis; let's say it's in kilojoules.

Let's say we have 18.0 grams of ice, and our goal is to calculate the total heat necessary to convert that 18 grams of ice at negative 25 degrees Celsius to steam at 125 degrees Celsius.

So we're starting with ice at negative 25 degrees Celsius, and first, we need to heat up the ice to zero degrees Celsius, which we know is the melting point. So on our heating curve, we're going from point A to point B. To calculate the heat necessary, we need to use the equation q is equal to m c delta t, where q is the heat added, m is the mass of the ice, c is the specific heat of ice, and delta t is the change in temperature, which is the final temperature minus the initial temperature.

So we're trying to calculate q, and we know the mass of our ice is 18.0 grams. The specific heat of ice is 2.03 joules per gram degree Celsius. For the change in temperature, the final temperature would be zero degrees Celsius, initial is negative 25, so 0 minus negative 25 gives us positive 25 degrees Celsius.

So grams will cancel out, degrees Celsius cancels out, and this gives us q is equal to 9.1 times 10 to the second joules to two significant figures, or we could also write 0.91 kilojoules.

Now that the ice is at zero degrees Celsius, we know it's going to melt. So we're going to go from point B on the heating curve to point C. To calculate how much heat is necessary to melt the ice, we need to know the heat of fusion of ice, which is equal to 6.01 kilojoules per mole.

So we need to figure out how many moles of ice we have. Well, we're starting with 18.0 grams. We divide by the molar mass of H2O, which is 18.0 grams per mole, and the grams will cancel and give us one mole. So we have 1.00 mole of ice, and we multiply that by 6.01 kilojoules per mole. The moles cancel, and it takes 6.01 kilojoules.

Now that all the ice is melted, we have liquid water. So on our heating curve, we're going to heat that liquid water from 0 degrees Celsius to 100 degrees Celsius, which is the boiling point of water. So we're going from point C to point D on the heating curve to calculate the heat added; we use the q is equal to m c delta t equation again.

So we're solving for q. The mass is still 18.0 grams, but the specific heat now, since we have liquid water, we need to use the specific heat of liquid water, which is 4.18 joules per gram degree Celsius. And for the change in temperature, the final temperature is 100, so 100 minus 0 gives us positive 100 degrees Celsius.

So grams cancel out, degrees Celsius cancels out, and we find that q is equal to 7.52 times 10 to the third joules; just correct that three there: 7.52 times 10 to the third joules, which is equal to 7.52 kilojoules.

Once we reach point D in the heating curve, we're at the boiling point of water. So the heat that we add now is going to go into turning the liquid water into gaseous water.

So going from point D to point E, we're doing a phase change. We need to know the heat of vaporization of water, and that's equal to 40.7 kilojoules per mole. We already know we have one mole of H2O, so one mole times 40.7 kilojoules per mole, the moles cancel, and it takes 40.7 kilojoules of energy to convert the liquid water into gaseous water or steam.

Next, we're going to heat the gaseous water from 100 degrees Celsius to 125 degrees Celsius. So we're going from point E to point F on the heating curve, and to figure out how much heat we need to add, we use the q is equal to m c delta t equation one more time.

So we're solving for q, and we still have 18.0 grams. This time we need to use the specific heat of steam, which is 1.84 joules per gram degree Celsius. The change in temperature would be 125 minus 100, or positive 25 degrees Celsius.

So grams cancel, units cancel out, and we get q is equal to 8.3 times 10 to the second joules to two significant figures, which is equal to 0.83 kilojoules.

Finally, we need to add everything up. So going from point A to point B on the heating curve, so think about just the x-axis this time.

All right, so going from point A to point B, we calculated that to be equal to 0.91 kilojoules. From point B to point C, we calculate that to be 6.01 kilojoules. From C to D, so this distance here was 7.52. From D to E, this was the big one here; this was equal to 40.7 kilojoules. Finally, from E to F, we calculated this was equal to 0.83 kilojoules.

When you add everything up, this is equal to fifty-six point zero kilojoules. So that's how much energy it takes to convert 18.0 grams of ice at negative 25 degrees Celsius to gaseous water at 125 degrees Celsius.

Next, let's think about the slopes of the different lines on our heating curve. So let's look at the line going from B to C and also the line going from point D to point E. Both of these lines represent phase changes. Going from point B to point C was going from solid to liquid, and going from point D to E was going from liquid to gas.

Since the slope of both of these lines is zero, that means as you add heat on the x-axis, the temperature doesn't change. So during a phase change, all the energy goes into disrupting the intermolecular forces that are present, and they don't go into increasing the temperature.

So there's no increase in temperature during the phase change. Think about going from point D to point E; this was converting our liquid water into gaseous water. As the heat is being added, all that energy goes into breaking the intermolecular forces between water molecules and pulling apart those liquid water molecules, turning them into gaseous water molecules.

It's only after all of the liquid water molecules are converted into gaseous water molecules that we see the temperature increase again. So talking about from point E to point F, everything is now in the gaseous state, and then we see the increase in temperature.

Finally, let's compare the slope of the line from A to B to the slope of the line from C to D. If we look at it, the slope of the line from A to B is a little bit steeper than the slope of the line from C to D.

The reason for the different slopes has to do with the different specific heats. From A to B, we use the specific heat for ice, which is 2.03 joules per gram degree Celsius. From C to D, in our calculation, we use the specific heat for water, which is 4.18 joules per gram degree Celsius.

The higher the value for the specific heat, the more energy it takes to raise the temperature of a substance by a certain amount. So if we think about comparing these two, let's say we try to raise the temperature of ice by 25 degrees Celsius.

So think about this distance here on the y-axis; we would have to put in only a small amount of energy to get ice to increase its temperature by 25 degrees Celsius. We think about that same temperature change on liquid water. If we tried to increase the temperature of liquid water by that same amount, 25 degrees, we would have to put in more energy.

On the x-axis, we have to put in more energy to accomplish the same change in temperature, and that's because liquid water has a higher specific heat. Since it might be a little bit hard to see on that diagram, let's think about putting some heat into a substance here.

So let me just draw a horizontal line, and then we're trying to accomplish a certain temperature change. So I'll draw a vertical line, those two give me those two give me a line with a slope. Let's say we're trying to accomplish the same change in temperature, so I'll draw this y distance the same as before, but we have a higher specific heat.

So it takes more; it takes more energy, therefore this x distance is going to increase. When we increase the x distance, we see that the slope decreases. So the greater the value for the specific heat, the lower the slope on the heating curve.