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Course: AP®︎/College Chemistry > Unit 6
Lesson 4: Energy of phase changesHeating curve for water
The heating curve for water shows how the temperature of a given quantity of water changes as heat is added at a constant rate. During a phase change, the temperature of the water remains constant, resulting in a plateau on the graph. We can use the heating curve to calculate the amount of heat required to raise the temperature of the water sample by a certain amount, such as from -25°C (when the water is present as a solid) to 125°C (when the water is present as a gas). Created by Jay.
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- Summary
the greater the value of the specific heat,
the lower the curve on the heating curve.
the higher the value of the specific heat,
the more energy it takes to rise the temperature of the substance by a certain amount.(11 votes) - Why did you not show us an experiment of the heating curve of water? don't we need to see how it works first?(0 votes)
- It's literally just water heating up. Go get a pot and boil some ice to see the experiment.(23 votes)
- Water evaporates (goes from liquid to gas) even then, when it hasn't yet reached it's boiling point, right?
So does that mean that some of the energy used to raise the temperature, let's say, from 0C° to 80C° will be also used to turn H2O to gasseous state?(2 votes)- There's a slight difference between boiling and evaporating. Evaporation means the most energetic liquid particles transition to the gas phase. Boiling means that the entire mass of liquid is transitioning to the gas phase. Even at low temperatures well below 100°C there is still a degree of evaporation of water.
Hope that helps.(3 votes)
- At2:00I'm so confused why there is a straight line from B to C. Why does adding heat not change the temperature? Q=mcdelta(T), so when q is increasing (heat is being added) why doesn't delta(T) increase? What am I missing? Does adding heat somehow not increase the average kinetic energy of molecules during a phase change?(1 vote)
- When we have a phase change on a heating curve they appear as straight lines. Solid to liquid (or liquid to solid) and liquid to gas (or gas to liquid). If we're progressing to the right on the graph by adding heat then going from point B to point C would mean we are melting solid water (ice) to make liquid water. It's a horizontal line meaning there is no temperature change and so the equation Q = MCΔT won't suffice. ΔT would be 0 making the heat added also 0 which doesn't make sense since we are still adding heat.
Instead we use a different equation for phase changes. For solid moving to the liquid we use: Q = M x L, where Q is still heat, M is mass, and L is the latent heat of fusion (also known as the enthalpy of fusion). The latent heat of fusion is the new term and is the amount of heat which must be provided to a chemical with a certain mass in order for it to change phase from solid to liquid. Just like how the specific heat capacity from the previous equation has values specific to what chemical we're dealing with, latent heat of fusion also depends on what chemical we are using. For water this value is: L = 333.55 kJ/kg.
At phase changes the heat added is longer being used to increase the average kinetic energy of the sample, and instead is used just to facilitate the phase change.
Hope that helps.(5 votes)
- Just to check, were the initial and final values of -25 and 125 arbitrary?
Could the initial one have been anything below 0, and the final one been anything above 100, to illustrate the same ideas?
Or are -25 and 125 significant?(1 vote)- Yep, they’re arbitrary. They picked those numbers so the curve had to pass through 0°C and 100 °C, but it could have also been any other number less than 0 or greater than 100.
Hope that helps.(2 votes)
- Can you give us a few questions based on heating and cooling curve?(1 vote)
- Why are they using J/g and KJ/mol. Isn't that just overcomplicating it. I know 18g is 1mol of H20 but why not keep it consistent(1 vote)
- What is defination of heating curve?(0 votes)
- Heating curve topic for this video is for which grade ?(0 votes)
- How are these flat line sections (representing different states) modeled mathematically? Does the equation q =mc*delta T cover this?
Of course, if you already had the graph, you could just use piecewise linear functions to model this. But let's assume you don't. How can you figure out the heat curve for some substance? Do you have to determine it experimentally?(0 votes)- The flatline sections represent latent heat of fusion/evaporation. It's flat because it takes a considerable amount of energy to let every molecule of a substance cross from phase to phase, and it's not increasing the kinetic energy (temperature) of any particles that've already crossed.
And yeah, you just determine stuff experimentally.
Hope that helps.(1 vote)
Video transcript
- [Instructor] 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 -25 degrees Celsius to
steam at 125 degrees Celsius. So we're starting with
ice at -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 mc 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. We know the mass of our ice is 18.0 grams. The specific heat of ice is 2.03 joules per gram degrees Celsius. And for the change in temperature,
it's final minus initial. So the final temperature
would be zero degrees Celsius, initial is -25. So zero minus -25 gives
us +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 what's going to melt. So we're gonna go from point B on the heating curve to point C. And 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. After 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 moles of ice and we multiply that by
6.01 kilojoules per mole and the moles cancel out
and give us 6.01 kilojoules. Now that all the ice is
melted, we have liquid water. And so on our heating curve, we're gonna heat that liquid water from zero degrees Celsius to 100 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
mc 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 degrees Celsius. And for the change in temperature, the final temperature is 100. So 100 minus zero gives
us +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, let me
just correct three there, 7.52 times 10 to the third joules, which is equal to 7.52 kilojoules. Once we reached a point
D in the heating curve, we're at the boiling point of water. So the heat that we add now is gonna 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. And we already know we
have one mole of H2O. So one mole times 40.7
moles, the moles cancel and it takes 40.7 kilojoules of energy to convert the liquid water
in to gaseous water or steam. Next we're gonna 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 mc
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
these specific heat of steam, which is 1.84 joules
per gram degree Celsius. The change in temperature
would be 125 minus 100 or +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 in 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. And then from point B to point C, we calculated 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. And finally from E to F we calculated this was equal to 0.83 kilojoules. And when you add everything up this is equal to 56.0 kilojoules. So that's how much energy it takes to convert 18.0 grams of
ice at -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 a solid to a liquid and going from point D to E was going from a liquid to a gas. And 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 is no increase in temperature during a phase change. Think about going from point D to point E, this was converting our liquid
water into gaseous water. So 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 and turning them into
gaseous water molecules. So it's only after all of
the liquid water molecules are converted into
gaseous water molecules, that's when 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 used the
specific heat for ice which is 2.03 joules per
gram degrees Celsius. From C to D in our calculation, we used the specific heat for water which is 4.1 joules per
gram degrees 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 lets 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. So if we tried to increase the
temperature of liquid water by that same amount, 25 degrees, we would have to put in more energy. So 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 I'm gonna 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 a line with a slope. So 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 energy. Therefore this X distance
is going to increase. And 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.