What a change of state is
Matter normally exists in three states — solid, liquid and gas. A transition from one of these states to another is a change of state, and it happens when heat is exchanged between a substance and its surroundings. The two everyday changes are solid to liquid and liquid to gas, together with their reverses. What makes a change of state distinctive is that, while it proceeds, the temperature of the substance does not change even though heat is being continuously supplied or removed.
The reason is microscopic. In a solid the molecules are locked in a rigid lattice; in a liquid they touch but slide past one another; in a gas they are far apart and nearly free. Converting one state to another means rearranging these bonds. The heat supplied during the transition is spent breaking bonds — raising the molecules' potential energy — rather than speeding them up. Since temperature reflects average kinetic energy, it holds steady until the rearrangement is complete.
The six phase transitions
Three pairs of transitions connect the three states. Each forward process absorbs heat; each reverse process releases the same amount. The table fixes the vocabulary NEET expects you to use precisely.
| Transition | Name | Heat | Everyday example |
|---|---|---|---|
| Solid → Liquid | Melting (fusion) | Absorbed | Ice melting to water |
| Liquid → Solid | Freezing | Released | Water freezing to ice |
| Liquid → Gas | Vaporisation (boiling) | Absorbed | Water boiling to steam |
| Gas → Liquid | Condensation | Released | Steam fogging on a cold lid |
| Solid → Gas | Sublimation | Absorbed | Dry ice (solid CO₂), iodine |
| Gas → Solid | Deposition | Released | Frost forming on a window |
Sublimation deserves a note. Not every substance passes through all three states in sequence. Some go directly from solid to vapour and back without ever becoming liquid. Dry ice (solid carbon dioxide) and iodine are the standard examples; during sublimation the solid and vapour states coexist in thermal equilibrium, exactly as solid and liquid do during melting.
Melting point and boiling point
The melting point is the temperature at which the solid and liquid states of a substance coexist in thermal equilibrium. It is characteristic of the substance but also depends on pressure; the melting point at standard atmospheric pressure is the normal melting point. Likewise the boiling point is the temperature at which the liquid and vapour states coexist, and the boiling point at standard atmospheric pressure is the normal boiling point. For water at 1 atm these are 0 °C and 100 °C.
Both points hold steady throughout the transition: the temperature remains constant until the entire solid has melted, or the entire liquid has vaporised. This is the experimental signature that a change of state, and not mere heating, is in progress.
Effect of pressure: regelation and the pressure cooker
Pressure shifts both the melting point and the boiling point. The two effects sit behind a pair of classic NEET examples.
Regelation. Loop a metallic wire over a slab of ice and hang heavy blocks from its ends. The wire slowly passes through the slab — yet the slab does not split. Just below the wire, the high pressure lowers the melting point of the ice, so it melts; once the wire has moved past, the pressure is relieved and the water refreezes above it. This melting-under-pressure-then-refreezing is regelation. Ice is unusual: for it, increased pressure lowers the melting point. The same effect lets a skater glide on a thin film of meltwater formed under the blades.
Boiling and pressure. Boiling point rises with pressure. Trap the steam above boiling water — for instance by closing a flask's outlet — and the boiling stops; more heat is then needed before boiling resumes. This is exactly how a pressure cooker works: it raises the pressure so water boils above 100 °C, and food cooks faster at the higher temperature. The reverse explains why cooking is slow on hills: lower atmospheric pressure lowers water's boiling point below 100 °C.
The temperature plateau
Plot temperature against time as you heat ice on a constant source and the graph is not a straight climb. It rises while the ice warms, then flattens into a horizontal plateau while the ice melts at 0 °C, rises again as the water warms, and flattens once more at 100 °C while the water boils. Each plateau marks a change of state, and along it the temperature is pinned even though heat keeps flowing in.
The plateau is the heart of the topic. Heat added during a change of state goes entirely into rearranging molecular bonds — increasing potential energy — and none of it into kinetic energy. Because the thermometer reads only kinetic energy, the temperature cannot move until the transition finishes. Remove heat during a transition (freezing, condensation) and the same plateau appears, with energy now flowing out.
Latent heat and Q = mL
The heat absorbed or released per unit mass during a change of state, with no change in temperature, is the latent heat $L$ of the substance for that process. The word "latent" means hidden — the energy is stored in the new arrangement of molecules rather than appearing as a temperature rise. For a mass $m$ undergoing the transition,
$$ Q = mL, \qquad L = \frac{Q}{m}. $$
The SI unit of $L$ is J kg⁻¹. Its value depends on the substance and, weakly, on pressure, so it is quoted at standard atmospheric pressure. Two cases dominate NEET. The latent heat of fusion $L_f$ is the heat per unit mass to convert solid to liquid at the melting point. The latent heat of vaporisation $L_v$ is the heat per unit mass to convert liquid to gas at the boiling point. Both are reversibly released on the reverse transition.
When 0.15 kg of ice at 0 °C is mixed with 0.30 kg of water at 50 °C in a container, the resulting temperature is 6.7 °C. Calculate the heat of fusion of ice. ($s_\text{water} = 4186~\text{J kg}^{-1}\text{K}^{-1}$)
Heat lost by the warm water cooling from 50 °C to 6.7 °C: $Q_\text{lost} = m s_w \Delta T = (0.30)(4186)(50 - 6.7) = 54376~\text{J}$.
Heat gained by the ice has two parts — melting at 0 °C, then warming the meltwater to 6.7 °C: $Q_\text{gain} = m L_f + m s_w \Delta T = (0.15)L_f + (0.15)(4186)(6.7) = (0.15)L_f + 4207~\text{J}$.
Heat lost = heat gained: $54376 = (0.15)L_f + 4207 \Rightarrow L_f \approx 3.34 \times 10^{5}~\text{J kg}^{-1}$, matching the standard value for ice.
Mixing problems like this are the bread and butter of calorimetry — heat lost equals heat gained, with $mL$ terms for any phase change in the mix.
Fusion versus vaporisation
For any substance the latent heat of vaporisation far exceeds the latent heat of fusion. Melting only loosens molecules from a lattice into a liquid where they still touch; vaporisation pulls every molecule completely free of its neighbours and does extra work pushing back the atmosphere as the vapour expands. The factor-grid below contrasts the two for water using NCERT values.
| Property | Latent heat of fusion (Lf) | Latent heat of vaporisation (Lv) |
|---|---|---|
| Transition | Solid ⇌ liquid (at melting point) | Liquid ⇌ gas (at boiling point) |
| Value for water | 3.33 × 10⁵ J/kg | 22.6 × 10⁵ J/kg |
| Temperature for water | 0 °C | 100 °C |
| Meaning | Energy to melt 1 kg of ice at 0 °C | Energy to vaporise 1 kg of water at 100 °C |
| What the energy does | Loosens lattice into liquid | Frees molecules + work against atmosphere |
| Relative size | Smaller | ≈ 6.8 × larger than Lf |
The consequence is practical. Steam at 100 °C carries $22.6 \times 10^{5}$ J kg⁻¹ more energy than boiling water at the same 100 °C. When steam condenses on skin it dumps this entire latent heat before the water even begins to cool. That is precisely why burns from steam are far more serious than burns from boiling water — a fact NEET phrases as a reasoning question.
The heating curve of water
Track 1 kg of ice from below 0 °C all the way to steam above 100 °C and the temperature-versus-heat graph is built from five segments: two flat plateaus where the state changes, and three sloped lines where a single state warms. On the sloped portions $Q = mc\,\Delta T$; on the plateaus $Q = mL$. The slopes differ because the specific heats of ice, water and steam are not equal — water's sloped line is the gentlest because liquid water has the largest specific heat.
To compute the heat for the whole journey you add the segments. NCERT Example 10.5 does exactly this for 3 kg of ice taken from −12 °C to steam at 100 °C: warm the ice ($m s_\text{ice}\Delta T$), melt it ($m L_f$), warm the water to 100 °C ($m s_\text{water}\Delta T$), then vaporise it ($m L_v$). Each plateau contributes an $mL$ term; each slope an $mc\,\Delta T$ term. The boiling plateau is the longest segment because $L_v$ is so large.
How much heat converts 1 kg of ice at 0 °C into steam at 100 °C? Use $L_f = 3.33 \times 10^5$ J/kg, $c_\text{water} = 4186$ J kg⁻¹K⁻¹, $L_v = 22.6 \times 10^5$ J/kg.
Melt the ice: $Q_1 = m L_f = (1)(3.33\times10^5) = 3.33\times10^5~\text{J}$.
Warm water 0 → 100 °C: $Q_2 = m c \,\Delta T = (1)(4186)(100) = 4.19\times10^5~\text{J}$.
Vaporise the water: $Q_3 = m L_v = (1)(22.6\times10^5) = 2.26\times10^6~\text{J}$.
Total: $Q = Q_1 + Q_2 + Q_3 \approx 3.01 \times 10^{6}~\text{J}$. Notice the vaporisation term alone dwarfs the other two — the signature of $L_v \gg L_f$.
Triple point and phase diagram
Because the temperature of a substance stays constant during each change of state, the natural way to map its phases is a graph of pressure $P$ against temperature $T$ — a phase diagram, or P–T diagram. The diagram divides the plane into a solid region, a liquid region and a vapour region, separated by three curves: the fusion curve (solid–liquid coexistence), the vaporisation curve (liquid–vapour coexistence) and the sublimation curve (solid–vapour coexistence).
The single point where all three curves meet — where solid, liquid and vapour coexist in equilibrium — is the triple point. For water the triple point lies at 273.16 K and $6.11 \times 10^{-3}$ Pa (about 611 Pa). Because this point is fixed and exactly reproducible, it anchors the definition of the kelvin scale.
Change of state & latent heat in one breath
- Six transitions: melting/freezing, vaporisation/condensation, sublimation/deposition. Forward absorbs heat; reverse releases it.
- During any change of state the temperature is constant — heat goes to potential energy (breaking bonds), not kinetic energy.
- $Q = mL$ on a plateau; $Q = mc\,\Delta T$ on a slope. Never apply $mc\,\Delta T$ across a plateau.
- For water: $L_f = 3.33 \times 10^5$ J/kg at 0 °C, $L_v = 22.6 \times 10^5$ J/kg at 100 °C. $L_v \approx 6.8\,L_f$.
- Pressure lowers ice's melting point (regelation, skating) but raises water's boiling point (pressure cooker).
- Triple point of water: 273.16 K, 6.11 × 10⁻³ Pa — where solid, liquid and vapour coexist on the P–T diagram.