Physics · Thermal Properties of Matter

Change of State & Latent Heat

Heat a beaker of ice on a steady flame and the thermometer does something strange: it climbs, then stalls at 0 °C while the ice melts, climbs again, then stalls at 100 °C while the water boils. Those two flat stretches are where the physics lives. During a change of state the supplied heat stops raising temperature and instead does the silent work of pulling molecules apart. This deep-dive covers the transitions — melting, boiling, sublimation, regelation — the latent heat relation $Q = mL$, the ice→water→steam heating curve, and the NCERT values NEET tests every year.

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.

TransitionNameHeatEveryday example
Solid → LiquidMelting (fusion)AbsorbedIce melting to water
Liquid → SolidFreezingReleasedWater freezing to ice
Liquid → GasVaporisation (boiling)AbsorbedWater boiling to steam
Gas → LiquidCondensationReleasedSteam fogging on a cold lid
Solid → GasSublimationAbsorbedDry ice (solid CO₂), iodine
Gas → SolidDepositionReleasedFrost 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.

Figure 1 SOLID lattice locked LIQUID molecules touch GAS molecules free melting +Q freezing −Q vaporisation +Q condensation −Q sublimation +Q (solid → gas) deposition −Q (gas → solid)
The phase-change cycle. Each forward step (red/amber) absorbs heat; each reverse step (purple/green) releases the same amount of heat. Solid↔gas directly is sublimation/deposition.

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.

NCERT Example 10.4

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.

i
Related drill

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.

PropertyLatent heat of fusion (Lf)Latent heat of vaporisation (Lv)
TransitionSolid ⇌ liquid (at melting point)Liquid ⇌ gas (at boiling point)
Value for water3.33 × 10⁵ J/kg22.6 × 10⁵ J/kg
Temperature for water0 °C100 °C
MeaningEnergy to melt 1 kg of ice at 0 °CEnergy to vaporise 1 kg of water at 100 °C
What the energy doesLoosens lattice into liquidFrees molecules + work against atmosphere
Relative sizeSmaller≈ 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.

Figure 2 T(°C) Heat added → 0 100 ice melting (L_f) water boiling (L_v) steam T constant T constant
Temperature versus heat for water at 1 atm (not to scale). Two flat plateaus mark melting at 0 °C and boiling at 100 °C; the boiling plateau is far longer because $L_v \approx 6.8\,L_f$. Sloped lines use $Q = mc\,\Delta T$; plateaus use $Q = mL$.

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.

Worked total

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.

Figure 3 P T → triple point sublimation fusion vaporisation SOLID LIQUID VAPOUR
Qualitative P–T phase diagram (water). The fusion, vaporisation and sublimation curves meet at the triple point, where solid, liquid and vapour coexist. For water at 273.16 K and 6.11 × 10⁻³ Pa. (Schematic, not to scale.)
Quick recap

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.

NEET PYQ Snapshot — Change of State & Latent Heat

Latent heat is tested through energy-conservation framing. Each problem hinges on a $Q = mL$ term sitting beside an $mc\,\Delta T$ or mechanical-energy term.

NEET 2016

A piece of ice falls from a height $h$ so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of $h$ is: (Latent heat of ice is $3.4 \times 10^5$ J/kg, $g = 10$ N/kg)

  1. 544 km
  2. 136 km
  3. 68 km
  4. 34 km
Answer: (2) 136 km

Energy balance. Gravitational PE released $= mgh$. One-quarter is absorbed by the ice and melts it: $\tfrac{1}{4}(mgh) = mL_f$. So $h = \dfrac{4L_f}{g} = \dfrac{4 (3.4\times10^5)}{10} = 1.36 \times 10^{5}~\text{m} = 136~\text{km}$. The $mL_f$ term is the whole point — no temperature change appears.

NCERT-based · Steam burn

Why does $1~\text{g}$ of steam at 100 °C produce a more serious burn than $1~\text{g}$ of boiling water at 100 °C? Estimate the extra heat released by the steam as it condenses. ($L_v = 22.6 \times 10^5$ J/kg)

Answer: ≈ 2260 J extra per gram, released on condensation

Latent-heat reasoning. Both are at 100 °C, but steam must first condense to water before cooling, releasing $Q = mL_v = (10^{-3})(22.6\times10^5) = 2260~\text{J}$ per gram. Boiling water at 100 °C carries none of this. The steam therefore deposits far more energy on the skin — the classic NCERT illustration of $L_v$.

NCERT Example 10.4

0.15 kg of ice at 0 °C is mixed with 0.30 kg of water at 50 °C; the mixture settles at 6.7 °C. Find the latent heat of fusion of ice. ($s_\text{water} = 4186$ J kg⁻¹K⁻¹)

Answer: Lf ≈ 3.34 × 10⁵ J/kg

Heat lost = heat gained. Water cools: $(0.30)(4186)(50 - 6.7) = 54376$ J. Ice melts then warms: $(0.15)L_f + (0.15)(4186)(6.7)$. Equate: $54376 = 0.15\,L_f + 4207 \Rightarrow L_f \approx 3.34 \times 10^5$ J/kg — within rounding of the NCERT value $3.33 \times 10^5$ J/kg.

FAQs — Change of State & Latent Heat

Short answers to the phase-change questions NEET aspirants get wrong most often.

Why does temperature stay constant while ice melts even though heat is being supplied?
During melting the supplied heat goes into breaking the bonds that hold the solid lattice together — it raises the potential energy of the molecules, not their average kinetic energy. Temperature measures average kinetic energy, so it stays fixed at the melting point until the entire solid has become liquid. Only after the change of state is complete does further heat begin to raise the temperature again.
Why is latent heat of vaporisation much larger than latent heat of fusion?
For water Lf = 3.33 × 10⁵ J/kg while Lv = 22.6 × 10⁵ J/kg — about seven times larger. Melting only loosens molecules from a rigid lattice into a liquid where they still touch. Vaporisation must pull every molecule completely free of its neighbours and also do work pushing back the atmosphere as the vapour expands. Both effects demand far more energy, so Lv ≫ Lf for almost every substance.
Why are steam burns more dangerous than burns from boiling water?
Steam at 100 °C carries 22.6 × 10⁵ J per kilogram more energy than boiling water at the same 100 °C, because of the latent heat of vaporisation. When steam condenses on skin it releases this entire latent heat in addition to cooling from 100 °C, so it deposits far more energy than the same mass of water at 100 °C. That extra energy is why steam burns are more serious.
How does a pressure cooker cook food faster?
Boiling point rises with pressure. A pressure cooker traps steam and raises the pressure above the water, so water boils at a temperature higher than 100 °C. Food cooks at this higher temperature and therefore faster. The same physics, run in reverse, explains why water boils below 100 °C at high altitude where atmospheric pressure is lower, making cooking on hills slower.
What is regelation?
Regelation is the melting of ice under pressure followed by its refreezing once the pressure is removed. In the classic demonstration a loaded wire passes through a slab of ice: ice melts just below the wire because increased pressure lowers its melting point, and the water refreezes above the wire as the pressure is relieved. The slab is cut through yet does not split. Ice is unusual in that pressure lowers its melting point.
What is the triple point of water?
The triple point is the unique temperature and pressure at which the solid, liquid and vapour phases of a substance coexist in equilibrium — the point where the fusion, vaporisation and sublimation curves meet on the P–T phase diagram. For water it occurs at 273.16 K and 6.11 × 10⁻³ Pa (about 611 Pa). Because it is fixed and reproducible, it is used to define the kelvin scale.