1.1 What is Chemistry?

17.3. Enthalpy and Phase Transitions http://www.ck12.org

FIGURE 17.6

Phase transitions

(melting, vaporization, and sublimation) require an input of energy; they are endothermic. Processes in which
attractive interactions are formed (freezing, condensation, and deposition) release energy, so they are exothermic.

The amount of energy necessary to melt one mole of a substance is known as themolar heat of fusion. Similarly,
the amount of energy necessary to vaporize one mole of a substance is referred to as themolar heat of vaporization.
The values of these parameters are given for water in the following table, along with the specific heats exhibited by
water in its solid, liquid, and gaseous forms.

TABLE17.2:

Parameter Phase Transition Enthalpy Value

molar heat of fusion (melting) solid→liquid +6.0 kJ/mol

molar heat of vaporization liquid→gas +40.7 kJ/mol

specific heat of ice 2.09 J/g•°C

specific heat of water 4.18 J/g•°C

specific heat of steam 1.84 J/g•°C

Notice that the process of freezing is the exact reverse of the process of melting, so the enthalpy change has the same
magnitude but opposite sign. The same is true for the relationship between condensation and vaporization. Also
note that the specific heat of water is dependent on the state of the material. Ice, liquid water, and steam all have
different specific heat values. Remember that specific heat refers to the the amount of energy needed to raise the
temperature of one gram of liquid water by 1°C.

Energy and Changes of State

We have already learned how to use specific heat to calculate the energy needed to change a material from one
temperature to another within a given state. Now let’s look at the energy changes required for changes of state.

Example 17.3

How much energy is needed to convert 180 grams of ice at 0°C to liquid water at the same temperature?

Answer:

The heat of fusion for water is 6.0 kJ/mol. First, we need to convert the mass of water to moles:

180 g H 2 O× 181 mol H. 02 g H^2 O
2 O

= 10 .mol H 2 O

We now can calculate the energy needed for the conversion:

∆Hf us= 6 .0 kJ/mol

∆H=10 mol H 2 O×

6 .0 kJ

1 mol H 2 O

= 60 .kJ

A more complicated situation might involve temperature increases in addition to the phase change. When this is the
case, we will need to use specific heat data along with the molar heats of fusion or vaporization.

Example 17.4

How much energy is needed to convert 180 grams of ice at -12°C to liquid water at 25°C?

Answer:

There are three components to this problem:

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