Suppose you start with a system in its solid phase and heat it at a steady rate. FIGURE 19.13, which you saw in Chapter 18, shows how the system’s temperature changes. At first, the temperature increases linearly. This is not hard to understand because Equation 19.16 can be written

The slope of the graph depends inversely on the system’s specific heat. A constant specific heat implies a constant slope and thus a linear graph. In fact, you can measure $c$ from such a graph.

FIGURE 19.13 The temperature of a system that is heated at a steady rate.

But there are times, shown as horizontal line segments, during which heat is being transferred to the system but the temperature isn’t changing. These are phase changes. The thermal energy continues to increase during a phase change, but the additional energy goes into breaking molecular bonds rather than speeding up the molecules. A phase change is characterized by a change in thermal energy without a change in temperature.

The amount of heat energy that causes 1 kg of a substance to undergo a phase change is called the heat of transformation of that substance. For example, laboratory experiments show that 333,000 J of heat are needed to melt 1 kg of ice at $0\text{°}\text{C}$. The symbol for heat of transformation is L. The heat required for the entire system of mass $M$ to undergo a phase change is

Heat of transformation is a generic term that refers to any phase change. Two specific heats of transformation are the heat of fusion $L_{\text{f}}$, the heat of transformation between a solid and a liquid, and the heat of vaporization $L_{\text{v}}$, the heat of transformation between a liquid and a gas. The heat needed for these phase changes is

where the $\pm$ indicates that heat must be added to the system during melting or boiling but removed from the system during freezing or condensing. You must explicitly include the minus sign when it is needed.

TABLE 19.3 gives the heats of transformation of a few substances. Notice that the heat of vaporization is always much larger than the heat of fusion. We can understand this. Melting breaks just enough molecular bonds to allow the system to lose rigidity and flow. Even so, the molecules in a liquid remain close together and loosely bonded. Vaporization breaks all bonds completely and sends the molecules flying apart. This process requires a larger increase in the thermal energy and thus a larger quantity of heat.

EXAMPLE 19.4 Melting wax

An insulated jar containing $200\text{g}$ of solid candle wax is placed on a hot plate that supplies heat energy to the wax at the rate of $220\text{J}\text{/}\text{s}$. The wax temperature is measured every $30\text{s,}$ yielding the following data:

Time (s)	Temperature ($\text{°}C$)
0	20.0
30	31.7
60	42.2
90	55.0
120	64.7
150	70.4
180	70.5
210	70.5
240	70.6
270	70.5
300	70.4
330	74.5

What are the specific heat of the solid wax, the melting point, and the wax’s heat of fusion?

MODEL The wax is in an insulated jar, so assume that heat loss to the environment is negligible.

VISUALIZE Heat energy is being supplied at the rate of $220\text{J}\text{/}\text{s}$, so the total heat energy that has been transferred into the wax at time $t$ is $Q=220t\text{J}$. FIGURE 19.14 shows the temperature graphed against the cumulative heat $Q$, although notice that the horizontal axis is in kJ, not J. The initial linear slope corresponds to raising the wax’s temperature to the melting point. Temperature remains constant during a phase change, even though the sample is still being heated, so the horizontal section of the graph is when the wax is melting. The temperature increase at the end shows that the temperature of the liquid wax is beginning to rise after melting is complete.

FIGURE 19.14 The heating curve of the wax.

FIGURE 19.14 Full Alternative Text

SOLVE From $Q=Mc\text{Δ}T$, the slope of the $T\text{-}\text{versus}\text{-}Q$ graph is $\text{Δ}T\text{/}Q=1\text{/}Mc$. The experimental slope of the best-fit line is $1.708\text{°}\text{C}\text{/}\text{kJ}=0.001708\text{K}\text{/}\text{J}$. Thus the specific heat of the solid wax is

$$c=\frac{1}{M\times \text{slope}}=\frac{1}{(0.200\text{kg})\left(0.001708\text{K}\text{/}\text{J}\right)}=2930\text{J}/\text{kg}\text{K}$$

From the table, we see that the melting temperature—which remains constant during the phase change—is $70.5\text{°}\text{C}$. The heat required for the phase change is $Q=M{L}_{\text{f}}$, so the heat of fusion is $L_{\text{f}}=Q\text{/}M$. With data recorded only every $30\text{s,}$ it’s not exactly clear when the melting began and when it ended. The extension of the initial slope shows that the temperature reached the melting point about halfway between $120\text{s and}150\text{s}$, so the melting started at about $135\text{s}$. We’ll assume it was complete about halfway between 300 s and 330 s, or at about $315\text{s}$. Thus the melting took $180\text{s,}$ during which, at $220\text{J}\text{/}\text{s},39,600\text{J}$ of heat energy was transferred from the hot plate to the wax. With this value of $Q$, the heat of fusion is

$$L_{\text{f}}=\frac{Q}{M}=\frac{39,600\text{J}}{0.200\text{kg}}=2.0\times {10}^5\text{J}\text{/}\text{kg}$$

REVIEW Both the specific heat and the heat of fusion are similar to values in Tables 19.2 and 19.3, which gives us confidence in our results.