Melting vaporization and latent heat

When a substance sits at its melting or boiling point, the heat supplied changes its phase instead of its temperature. The energy goes into weakening and breaking the bonds between molecules rather than increasing their average kinetic energy, so the thermometer reading stays fixed on a plateau. This energy is the latent heat, and it is proportional to the mass that changes phase: $Q = mL,$ where $L$ is the specific latent heat ($L_f$ for fusion, $L_v$ for vaporization). For water, $L_f = 3.34\times10^{5}\ \text{J/kg}$ and $L_v = 2.26\times10^{6}\ \text{J/kg}$. The sign follows the direction of the transfer: heat is absorbed during melting and vaporization ($Q>0$) and released during freezing and condensation ($Q<0$). To find the total heat of a multi-stage process, add the sensible-heat terms $mc\Delta T$ for each temperature change and the $mL$ terms for each plateau.

How much heat must be supplied to completely melt $m = 0.50\ \text{kg}$ of ice at $0\ ^\circ\text{C}$ without warming the resulting water? The specific latent heat of fusion of ice is $L_f = 3.34\times10^{5}\ \text{J/kg}$.

A kettle holds $m = 2.0\ \text{kg}$ of water at $20\ ^\circ\text{C}$. How much heat is needed to warm it to $100\ ^\circ\text{C}$ and then vaporize all of it? The specific heat capacity of water is $c = 4186\ \text{J/(kg}\cdot\text{K)}$ and $L_v = 2.26\times10^{6}\ \text{J/kg}$.

Into a thermally insulated vessel holding $m_w = 0.40\ \text{kg}$ of water at $40\ ^\circ\text{C}$ is dropped $m_i = 0.10\ \text{kg}$ of ice at $0\ ^\circ\text{C}$. Find the final equilibrium temperature. Take $c = 4186\ \text{J/(kg}\cdot\text{K)}$, $L_f = 3.34\times10^{5}\ \text{J/kg}$, and neglect the heat capacity of the vessel.

Goal: observe the temperature plateau during melting and connect it to $Q = mL$.

1. Open a phase-change simulation (for example PhET "States of Matter" or "Energy Forms and Changes") and choose water. Work in the simulation rather than with boiling water: open steam is hazardous to skin and airways.
2. Set up ice at $0\ ^\circ\text{C}$ and supply energy at a constant power $P$, recording the temperature at equal time intervals. Plot temperature versus energy added ($Q = Pt$).
3. Mark the plateaus on the graph and confirm that the melting and boiling temperatures stay constant even though energy keeps arriving.
4. Compare the length of the vaporization plateau with that of the melting plateau and explain the difference using the ratio $L_v/L_f$.
5. Safe optional check: put an ice cube into a glass of room-temperature water, stir gently, and read a thermometer every $30\ \text{s}$. While ice remains, the temperature stays near $0\ ^\circ\text{C}$. Do not heat the water to boiling and do not cover the vessel.

Show that temperature stays constant during a phase change while the heat supplied goes into latent heat; teach how to distinguish sloped segments from plateaus.

A sketch of the temperature-versus-heat-added graph for water, starting from ice at $-20\ ^\circ\text{C}$:

• segment 1 (slope up): warming the ice, $T$ rises from $-20\ ^\circ\text{C}$ to $0\ ^\circ\text{C}$, $Q = m c_{\text{ice}}\Delta T$;
• segment 2 (horizontal plateau at $0\ ^\circ\text{C}$): melting, $Q = m L_f$;
• segment 3 (slope up): warming the water from $0\ ^\circ\text{C}$ to $100\ ^\circ\text{C}$, $Q = m c_{\text{water}}\Delta T$;
• segment 4 (horizontal plateau at $100\ ^\circ\text{C}$): boiling, $Q = m L_v$ (the longest plateau, since $L_v \approx 6.8\,L_f$);
• segment 5 (slope up): warming the steam.

Label the plateaus "phase change: $T = \text{const}$" and the sloped segments "$T$ changes: $Q = mc\Delta T$"; mark with arrows the direction in which heat is absorbed.

On a kitchen hob $1.5\ \text{L}$ of water ($m = 1.5\ \text{kg}$) is heated from $20\ ^\circ\text{C}$ to $100\ ^\circ\text{C}$: $Q = c m \Delta T = 4200 \cdot 1.5 \cdot 80 \approx 5.0 \times 10^5\ \text{J}$. The water then boils at a steady $100\ ^\circ\text{C}$ while its level slowly falls: evaporating just $0.1\ \text{kg}$ takes $Q = L_v m = 2.26 \times 10^6 \cdot 0.1 \approx 2.3 \times 10^5\ \text{J}$.

Two different stages of one task appear here. While the temperature rises, the specific heat does the work: the heat supplied raises the internal energy of the water in proportion to $\Delta T$. At $100\ ^\circ\text{C}$ the temperature stops changing and every joule supplied goes into the specific latent heat of vaporization $L_v = 2.26 \times 10^6\ \text{J/kg}$ — into breaking the bonds between molecules rather than increasing their kinetic energy. The reverse process, steam condensing on a cold lid, gives back exactly the same amount of heat.

A runner covers $42\ \text{km}$ in about $3\ \text{h}$ and evaporates roughly $0.8\ \text{L}$ of sweat ($m = 0.8\ \text{kg}$). At skin temperature the specific latent heat of vaporization of water is close to $2.4 \times 10^6\ \text{J/kg}$, so evaporation carries away $Q = L_v m \approx 1.9 \times 10^6\ \text{J}$. Spread over $10800\ \text{s}$ this is an average cooling power of about $180\ \text{W}$.

Evaporation is a phase transition that requires energy, and the sweat takes that energy from the skin, so heat is transferred from the body to the evaporating liquid. The example shows that during the liquid-to-vapour transition the temperature does not change, and the heat absorbed equals $L_v m$. That is why in humid air, where evaporation is slower, the same heat is not carried off and the body overheats.

Freezing rain has left a $6\ \text{mm}$ ice sheet on $1\ \text{m}^2$ of road. With a density of $\rho = 917\ \text{kg/m}^3$ the mass of ice is $m = \rho V = 917 \cdot 0.006 \approx 5.5\ \text{kg}$, and melting it at $0\ ^\circ\text{C}$ requires $Q = L_f m = 3.34 \times 10^5 \cdot 5.5 \approx 1.8 \times 10^6\ \text{J}$ per square metre. Road salt acts differently: it lowers the melting point and forms brine instead of supplying that energy.

This case drills the specific latent heat of fusion $L_f = 3.34 \times 10^5\ \text{J/kg}$. While the ice melts its temperature stays at $0\ ^\circ\text{C}$: the heat delivered by the warmer air and asphalt is spent not on heating but on breaking the crystal lattice. At night the reverse transition occurs: freezing water releases the same $3.34 \times 10^5\ \text{J}$ per kilogram into the cold air.