Phase Changes

So far, we have seen how heat flowing in or out of a material can change its temperature, but we have not dealt with the fact that losing or gaining heat can cause a material to undergo a phase change, such as freezing or vaporization. In this section, we will learn how to calculate the heat flow involved in phase changes like these. We will integrate what we learn about phase changes with what we already know about temperature changes to draw heating and cooling curves - tools for visualizing what happens as a material gains or loses heat.

Heat in Phase Changes

Imagine you have a 50.0 g chunk of ice at -10.0 °C and you begin adding heat energy to it. What will happen to it?

You might be tempted to say "it will melt, because ice melts when you heat it," however this is not strictly accurate. Ice melts only at its melting point, a set value of 0.0 °C. So the first thing to happen would be a warming from -10.0 °C to 0.0 °C. Using the heat capacity of ice (2.05 J/g°C) and what you learned in the previous section, you can even calculate how much heat energy would be absorbed in doing so. Try doing so now for practice ... you should get 1025 J of heat.

Even once it has been warmed to the freezing point, the ice does not melt on its own. More heat energy must be added to it in order to break some intermolecular forces in the ice and allow the water molecules to move more freely. Somewhat counterintuitively, as this heat is added and the ice melts, the temperature does not change. The solid ice begins melting at 0.0 °C, with heat being added until it is entirely converted to liquid water, still at 0.0 °C.

As you can see, heat can do multiple different things. It can raise the temperature of an object, or it can be used to accomplish a phase change. In the latter case, we can calculate the heat required using a value called the "heat of _____" where "_____" refers to whatever phase change (melting, vaporization, or sublimation) is happening).

For example, ΔH_fus for water is 334 J/g. Note: unlike heat capacity, the units of ΔH_fus and ΔH_vap do not include °C, because temperature does not change when something melts or vaporizes. For our 50.0 g chunk of ice, we can now calculate how much heat would be needed to melt it.

q = 334 J/g * 50.0 g

q = 16,700 J

This is one type of calculation we can do involving heats of fusion and vaporization. Another one would be to calculate the amount of substance that could be vaporized with a certain amount of heat. For example, if ΔH_vap for rubbing alcohol is 663 J/g, how much rubbing alcohol could be vaporized with 15,000 J of heat?

15,000 J = m * 663 J/g

m = 15,000 J/663 J/g

m = 22.6 g

Finally, if we want to experimentally determine ΔH_fus or ΔH_vap for a material, we can measure the heat involved and divide it by the mass of material that melts or vaporizes. This is what you will do as part of your lab this week: determine an experimental value of ΔH_fus for water.

Heating and Cooling Curves

Let's return to our 50.0 g chunk of ice. We started it at -10.0 °C and started adding heat. It took 1025 J to bring it up to the melting point, and another 16,700 J to melt it. What happens if we keep going from here?

Well, we now have liquid water at 0.0 °C; if we add heat to this, its temperature will again start to rise. How much? That depends on how much heat we add. We can warm it to the boiling point, 100.0 °C, by adding 20,920 J of heat.

q = m * C * ΔT

q = 50.0 g * 4.184 J/g°C * 100 °C

q = 20,920 J

If we keep putting heat in, the ice-that-has-become-water will now vaporize into a gas. The value of ΔH_vap value for water is 2260 J/g, so we can calculate the heat needed to vaporize this 50.0 g sample.

q = 2260 J/g * 50.0 g

q = 113,000 J

From there, we can continue adding heat to the water vapor basically indefinitely. Its temperature will continue to rise, interrupted only at the temperature (roughly 2500 °C) when water begins to fall apart into its free atoms.

This sequence of events can be summarized and visualized using a tool called a heating curve. The heating curve for water is shown below, and has some important features you should be familiar with.

First, note that the x-axis shows heat added to the water, while the y-axis shows the temperature. Our block of ice started at -10.0 °C, which is where this curve starts. It could be started at a lower temperature, in which case the first rising segment, in which the ice is warmed to the melting point, would be longer.

There are three rising segments, which correspond to warming of the three phases of matter. As you add heat to a solid, liquid or gas, the temperature rises, so the lines slope upward.

Of course, there are two points at which adding heat does not result in a rising temperature: the melting and boiling points. These are the flat sections at o.0 °C and 100.0 °C. Here, the temperature stays constant throughout the phase change. One subtle but important feature is that the second flat segment, corresponding to boiling the water, is much longer than the first one. This is because vaporizing water (or any other substance) takes much more energy than melting it, so more heat needs to be added. Recall that water's ΔH_vap value was 2260 J/g where as ΔH_fus was only 334 J/g. The reason for this is that transitioning from a solid to a liquid breaks only about 15% of the bonds/intermolecular forces in a substance, whereas transitioning from a liquid to a gas breaks all the remaining bonds/IMFs.

We can use a heating curve like this one to help get our arms around a question like this: "how much heat is required to vaporize 150.0 g of water that starts out at 25.0 °C?"

First we locate 25.0 °C on our curve ... it is about 1/4 the way up the middle sloping line. So the first thing we have to do in order to vaporize it is heat it to boiling. We can calculate the energy needed to do this using the heat capacity equation.

q = m * C * ΔT

q = 150.0 g * 4.184 J/g°C * 75.0 °C

q = 47,070 J

Once at 100.0 °C, our substance boils, and we can calculate the energy needed to vaporize it using the ΔH_vap value from before.

q = 150.0 g * 2260 J/g

q = 339,000 J

Thus the total heat needed to vaporize the water sample, starting from 25.0 °C, is the sum of these two values.

q = 47,070 J + 339,000 J

q = 386,070 J

A heating curve is not strictly required to do a problem like this, but you will probably find it very helpful for thinking through multi-step heat calculation problems like this.