# 3.4: Solving Energy Problems Involving Phase Changes and Temperature Changes

When a cloud drop evaporates, the energy to evaporate it must come from somewhere because energy is conserved according to the 1st Law of Thermodynamics. It can come from some external source, such as the sun, from chemical reactions, or from the air, which loses some energy and thus cools. Thus, temperature changes and phase changes are related, although we can think of phase changes as occurring at a constant temperature. The energy associated with phase changes drives much of our weather, especially our severe weather, such as hurricanes and deep convection. We can quantify the temperature changes that result from phase changes if we have a little information on the mass of the air and the mass and phases of the water.

In the previous lesson, we said that all changes of internal energy were associated with a temperature change. But the phase changes of water represent another way to change the energy of a system that contains the phase-shifter water. So often we need to consider both temperature change and phase change when we are trying to figure out what happens with heating or cooling.

For atmospheric processes, we saw that we must use the specific heat at constant pressure to figure out what the temperature change is when an air mass is heated or cooled. Thus the heating equals the temperature change times the specific heat capacity, constant pressure times the mass of the air. For dry air, we designate this specific heat constant pressure as cpd. For water vapor, we designate this specific heat constant pressure as cpv. So for example, the energy required to change temperature for a dry air parcel is

$c_{pd}m ΔT = c_{pd} ρV ΔT$

where cpd is the specific heat capacity for dry air at constant pressure. If we have moist air, then we need to know the mass of dry air and the mass of water vapor, calculate the heat capacity of each of them, and then add those heat capacities together.

For liquids or solids, specific heat, constant volume, and specific heat, constant pressure, are about the same, so we have only one for each type of material, including liquid water (cw) and ice (ci).

For phase changes, there is no temperature change. Phase changes occur at a constant temperature. So to figure out the energy that must be added or removed to cause a phase change, we only need to know what the phase change is (melting/freezing, sublimating/depositing, evaporating/condensing) and the mass of water that is changing phase. So, for example, the energy needed to melt ice is lf mice.

The following tables provide numbers and summarize all the possible processes involving dry air and water in its three forms.

Specific Heat Capacity at 0 oC (units: J kg–1 K–1)
Dry air Water vapor Liquid water Ice
cpd cpv cw ci
1005 1850 4218 2106
Latent Heat (units: J kg–1)
Vaporization @ 0 oC Vaporization @ 100 oC Fusion @ 0 oC Sublimation @ 0 oC
lv lv lf ls
2.501 x 106 2.257 x 106 0.334 x 106 2.834 x 106
Temperature Change
Dry air Water vapor Liquid water Ice
cpd md = cpd ρdV cpv mv= cpv ρvV cw mliquid ci mice

Phase Change
vapor→liquid liquid→vapor vapor→ice ice→vapor liquid→ice ice→liquid
lv mvapor lv mliquid ls mvapor ls mice lf mliquid lf mice

Note

To solve energy problems you can generally follow these steps:

1. Identify the energy source and write it on the left-hand side of the equation.
2. Identify all the changes in temperature and in phase and put them on the right-hand side.
3. You should know all of the variables in the equation except one. Rewrite the equation so that the variable of interest is on the left-hand side and all the rest are on the right-hand side.

Knowing how to perform simple energy calculations helps you to understand atmospheric processes that you are observing, and to predict future events. Why is the air chilled in the downdraft of the thunderstorm? When will the fog dissipate? When might the sun warm the surface enough to overcome a near-surface temperature inversion and lead to thunderstorms? We can see that evaporating, subliming, and melting can take up a lot of energy and that condensing, depositing, and freezing can give up a lot of energy. In fact, by playing with these numbers and equations, you will see how powerful phase changes are and what a major role they play in many processes, particularly convection.

With the elements in the tables above, you should be able to take a word problem concerning energy and construct an equation that will allow you to solve for an unknown, whether the unknown be a time or a temperature or a total mass.

In the atmosphere, these problems can be fairly complex and involve many processes. For example, when thinking about solar energy melting a frozen pond, we would need to think about not only the solar energy needed to change the pond from ice to liquid water, but we would also need to consider the warming of the land in which the pond rests and the warming of the air above the pond. Further, the land and the ice might absorb energy at different rates, so we would need to factor in the rates of energy transfer among the land and the pond and the air.

So we can make these problems quite complex, or we can greatly simplify them so that you will understand the basic concepts of energy required for temperature and phase changes. In this course, we are going to solve fairly simple problems and progress to slightly more complicated ones. Let’s look at a few examples. I will give you some examples and then you can do more for Quiz 3-3.

Example Problems

A small puddle is frozen and its temperature is 0 oC. How much solar energy is needed to melt all the ice? Assume that mice = 10.0 kg.

1. The heating source is the sun and we are trying to calculate the total solar energy. Put this on the left-hand side.
2. The change that we want is the melting of the ice. We know the mass and the latent heat. We write those on the right-hand side.
3. The equation already has the unknown variable on the left-hand side.

$$\int Q d t=l_{f} m_{i c e}=\left(0.334 \times 10^{6} \mathrm{J} \mathrm{kg}^{-1}\right)(10.0 \mathrm{kg})=3.34 \times 10^{6} \mathrm{J}$$

To put this amount of energy into perspective, this energy is equivalent to a normal person walking at about 4 mph for 2 hours (assuming the person burns 400 calories per hour, which is really 400 kilocalories per hour in scientific units).

Now let’s assume that the ice is originally at –20.0 oC. Now we have to both raise the temperature and melt the ice. If we don’t warm the ice, some of it will simply refreeze. Our equation now becomes:

$$\int Q d t=l_{f} m_{i c e}+c_{i} m_{i c e} \Delta T$$

$$=\left(0.334 \times 10^{6} \mathrm{J} \mathrm{kg}^{-1}\right)(10.0 \mathrm{kg})+\left(2106 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}\right)(10.0 \mathrm{kg})(0.0--20.0) \mathrm{K}$$

$$=3.76 \times 10^{6} \mathrm{J}$$

We see that the amount of energy required increased by about 25%. Most of the energy is still required to melt the ice, not change the temperature.

Now let’s assume that the solar heating rate is constant at 191 W m–2and that the area of the puddle is 2.09 m2. How long does it take the sun to raise the temperature of the ice and then melt it?

$$\int Q d t=\frac{\overline{\Delta Q}}{\Delta A} A \Delta t=l_{f} m_{i c e}+c_{i} m_{i c e} \Delta T$$

\Delta t=\frac{l_{f} m_{i c e}+c_{i} m_{i c e} \Delta T}{\frac{\overline{\Delta Q}}{\Delta A} A}=\frac{3.76 \times 10^{6} \mathrm{J}}{\left(191 \mathrm{Jm}^{-2} \mathrm{s}^{-2}\right)\left(2.09 \mathrm{m}^{2}\right)}=9.42 \times 10^{3} \mathrm{s}=2.6 \mathrm{h}

We could now assume that the source of heating is not the sun but instead is warm air passing over the puddle. If the temperature of the air is 20.0 oC and we assume that its temperature drops to 0.0 oC after contacting the ice, what is the mass of air that is required to warm the ice and then melt it?

$$\int Q d t=c_{p d} m_{a i i} T_{a i r}=l_{f} m_{i c e}+c_{i} m_{i c e} \Delta T_{i c e}$$

$$m_{a i r}=\frac{l_{f} m_{i c e}+c_{i} m_{i c e} \Delta T_{i c e}}{c_{p d} \Delta T_{a i r}}=\frac{3.76 \times 10^{6} \mathrm{J}}{\left(1005 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}\right)(20.0 \mathrm{K})}=187 \mathrm{kg}$$

See this video (2:28) for further explanation:

Melting Ice