Change of state

Leave an ice cube on the kitchen counter and its thermometer reading does something strange. It climbs steadily from the freezer temperature up to $0^\circ\text{C}$ — and then it stops. For long minutes the ice keeps melting, puddles keep forming, yet the thermometer sits stubbornly at $0^\circ\text{C}$ until the very last sliver of ice disappears. Only then does the water start warming again, all the way up to $100^\circ\text{C}$ , where the same trick repeats: the reading freezes even as the pan visibly boils dry. This is not a broken thermometer. It is one of the most quietly important facts in thermal physics: heat and temperature are not the same thing. Heat is energy in transit; temperature is a measure of a substance's average molecular agitation. Most of the time the two move together — pour in heat, temperature rises. But during a change of state — melting, freezing, boiling, condensing, or the direct solid-to-gas leap of sublimation — every joule of heat is spent reorganising how tightly molecules are bound to their neighbours, and none of it goes into speeding the molecules up. Temperature parks itself at a fixed value, the melting point or the boiling point, for as long as the transition lasts. This article builds the physical picture behind that plateau: why a heating curve for ice-to-steam has the shape it does, how a $P$ – $T$ phase diagram encodes melting and boiling points as curves, what makes the triple point and critical point special, and how the ordinary calorimetry formula $Q = mc\,\Delta T$ combines with the coming idea of latent heat to let you calculate exactly how much heat a multi-stage heating curve needs, stage by stage.

Key takeaways

Matter changes between solid, liquid and gas through six named transitions: fusion (melting), freezing/solidification, vaporisation (boiling), condensation, sublimation, and deposition.
During any change of state at fixed pressure, temperature stays constant while heat keeps flowing — the heat goes into breaking or forming intermolecular bonds, not into raising kinetic energy.
A heating curve (temperature plotted against heat added) alternates between sloped segments, where $Q = mc\,\Delta T$ governs a single phase warming up, and flat plateaus, where the substance is changing state.
A $P$ – $T$ phase diagram maps which state is stable at each pressure and temperature; the boundary curves are the melting-point and boiling-point lines.
The triple point is the one $(P,T)$ pair where solid, liquid and gas coexist together (for water, $273.16\ \text{K}$ at $611.73\ \text{Pa}$ ); the critical point is where the liquid–gas distinction disappears altogether.
Boiling point depends on external pressure — lower pressure (a mountaintop) means a lower boiling point, which is why food takes longer to cook at altitude.
Key idea: at a fixed pressure, temperature holds constant at the melting or boiling point throughout a phase change because the heat supplied restructures intermolecular bonds rather than raising kinetic energy, and this plateau is exactly what a heating curve and a phase diagram both encode.

The core insight

Every joule of heat absorbed or released during a change of state goes into rearranging intermolecular bonds at a fixed melting or boiling point, not into raising temperature, which is why a heating curve shows flat plateaus and a $P$ – $T$ phase diagram shows sharp boundary lines rather than a smooth blend between states.

Where it fits in the syllabus

Change of state sits inside the Thermal Properties of Matter unit of CBSE Class 11 Physics, coming right after calorimetry ( $Q = mc\,\Delta T$ ) and feeding directly into latent heat ( $Q = mL$ ), the article that follows this one. The prescribed scope asks you to know the names of the six transitions, understand why temperature is constant during a phase change, read a heating curve for a substance like ice turning into steam, and interpret a simple $P$ – $T$ phase diagram including the triple point and critical point at a qualitative level. This article stays entirely inside that scope — the deeper mechanism of why latent heat has the value it does, and quantitative calorimetry problems mixing $Q=mc\Delta T$ with $Q=mL$ , belong to the next article; here the focus is the shape of the transition itself.

The physical picture

Picture a solid being heated at a slow, steady rate — heat flows in at a constant number of joules per second. As long as the substance stays a solid, each joule raises its temperature according to $Q = mc\,\Delta T$ , with $c$ the solid's specific heat capacity. The molecules vibrate a little faster and a little further from their lattice positions as $T$ climbs.

Q = mc\,\Delta T \qquad \text{(single phase, no state change)}

where: $Q$ is the heat supplied (joules), $m$ is the mass (kg), $c$ is the specific heat capacity of that particular phase (J/kg·K — ice, water and steam each have their own value of $c$ ), and $\Delta T$ is the temperature change. The physical reading: within one phase, heat and temperature rise together in direct proportion, with $c$ setting how many joules it costs to raise each kilogram by one kelvin. At the melting point, everything changes. The lattice of the solid is right at the edge of collapse — molecules have exactly enough thermal agitation to sit in their fixed positions, no more. Any extra heat now goes into pulling molecules out of that ordered lattice into the freer, disordered arrangement of a liquid — breaking bonds, not speeding molecules up. Because none of the incoming heat raises the average kinetic energy, the thermometer does not move. Only once every last bit of solid has been converted to liquid does additional heat resume raising the temperature, now governed by the liquid's own (usually different) specific heat. The identical story repeats at the boiling point, where heat is now spent pulling molecules entirely free of each other's attraction into a gas.

Figure

A heating curve for ice warming into steam plotted as temperature on the vertical axis against heat added on the horizontal axis, showing a rising line for solid ice warming, a flat plateau at zero degrees Celsius while ice melts, a rising line for liquid water warming, and a flat plateau at one hundred degrees Celsius while water boils, followed by a rising line for steam warming further

The same competition between order and thermal energy also depends on pressure, and the cleanest way to see that is a $P$ – $T$ phase diagram: a graph with pressure on one axis and temperature on the other, divided by three boundary curves into solid, liquid and gas regions. The solid–liquid boundary is the melting-point curve; the liquid–gas boundary is the boiling-point curve; crossing either one at fixed pressure is exactly the flat plateau you meet on a heating curve. Two points on this map deserve special names. The triple point is the single $(P,T)$ combination where all three boundary curves meet, so solid, liquid and gas coexist together in equilibrium; for water it sits at $T = 273.16\ \text{K}$ and $P = 611.73\ \text{Pa}$ , a pressure far below one atmosphere. The critical point lies at the far end of the liquid–gas boundary curve, where that curve simply stops — beyond the critical temperature and pressure, no amount of squeezing can force a gas to condense into a distinguishable liquid; the two states blend into one. Because the liquid–gas boundary on the phase diagram is a curve, not a horizontal line, the boiling point is not a fixed number for a substance — it is a function of the surrounding pressure. Read the boundary curve at a lower pressure and you find it crosses into the gas phase at a lower temperature. This is exactly why water boils below $100^\circ\text{C}$ on a mountain, where atmospheric pressure is lower, and why a pressure cooker — which raises the internal pressure above one atmosphere — pushes the boiling point of water above $100^\circ\text{C}$ , cooking food faster. Some substances — camphor, naphthalene, solid carbon dioxide (dry ice) — can jump directly from solid to gas without ever forming a liquid, a process called sublimation; the reverse, gas condensing straight to solid, is deposition. On the phase diagram, this happens whenever the substance's triple-point pressure lies above the ambient pressure the substance is exposed to, so a horizontal line at that ambient pressure never crosses the liquid region at all — it goes straight from solid territory to gas territory. Complete inventory of the six named transitions: fusion (solid to liquid), freezing (liquid to solid), vaporisation (liquid to gas), condensation (gas to liquid), sublimation (solid to gas), and deposition (gas to solid).

Derivation ATotal heat to turn a block of ice into steam, stage by stage

The goal is to build, from nothing more than the calorimetry formula $Q = mc\,\Delta T$ and the idea of latent heat $Q = mL$ introduced properly in the next article, a single expression for the total heat needed to carry a mass $m$ of ice at a temperature below freezing all the way to steam above boiling — the exact quantity a heating-curve question always asks for. The reasoning is simply additive: heat is a scalar that adds up across consecutive stages, and each stage is either a single-phase warm-up (sloped segment) or a phase change (flat plateau), never both at once, so the total is just the sum of five separate pieces read straight off the heating curve.

Step 1. Identify the five stages between ice at $T_1$ (below $0^\circ\text{C}$ ) and steam at $T_2$ (above $100^\circ\text{C}$ ): ice warming from $T_1$ to $0^\circ\text{C}$ , ice melting at $0^\circ\text{C}$ , water warming from $0^\circ\text{C}$ to $100^\circ\text{C}$ , water boiling at $100^\circ\text{C}$ , and steam warming from $100^\circ\text{C}$ to $T_2$ . Each stage corresponds to exactly one segment of the heating curve sketched above.

Step 2. Write the heat for each sloped (single-phase) segment using $Q = mc\,\Delta T$ , with the specific heat capacity appropriate to that phase: $Q_1 = mc_{\text{ice}}(0 - T_1)$ for warming the ice, $Q_3 = mc_{\text{water}}(100 - 0)$ for warming the water, and $Q_5 = mc_{\text{steam}}(T_2 - 100)$ for warming the steam. Each term is positive because heat is flowing in and temperature is rising.

Step 3. Write the heat for each flat (phase-change) segment using latent heat, $Q = mL$ : $Q_2 = mL_f$ for melting the ice, where $L_f$ is the specific latent heat of fusion, and $Q_4 = mL_v$ for boiling the water, where $L_v$ is the specific latent heat of vaporisation. These terms carry no $\Delta T$ at all, because temperature does not change during either plateau — the entire heat is latent heat, spent on the phase change itself.

Step 4. Add all five stages, since heat supplied in sequence simply accumulates: $Q_{\text{total}} = mc_{\text{ice}}(0-T_1) + mL_f + mc_{\text{water}}(100-0) + mL_v + mc_{\text{steam}}(T_2-100)$ . Reading this sum off a heating curve is mechanical once you recognise which segments are sloped (use $mc\Delta T$ with that segment's specific heat) and which are flat (use $mL$ with that segment's latent heat) — no segment ever needs both formulas at once, because a substance is never simultaneously changing temperature and changing state.

Worked examples

Board level — reading a heating curve for ice to steam

A $1.0\ \text{kg}$ block of ice, initially at $-10^\circ\text{C}$ , is heated at a constant rate until it becomes steam at $100^\circ\text{C}$ . List the distinct segments of the heating curve in order and state which segments show constant temperature.

Step 1. Segment 1: ice warms from $-10^\circ\text{C}$ to $0^\circ\text{C}$ — temperature rises, governed by $Q = mc_{\text{ice}}\Delta T$ .

Step 2. Segment 2: ice melts to water at $0^\circ\text{C}$ — temperature constant, the plateau of fusion, heat absorbed as latent heat.

Step 3. Segment 3: water warms from $0^\circ\text{C}$ to $100^\circ\text{C}$ — temperature rises, governed by $Q = mc_{\text{water}}\Delta T$ .

Step 4. Segment 4: water boils to steam at $100^\circ\text{C}$ — temperature constant, the plateau of vaporisation, heat absorbed as latent heat.

Answer: four segments in order — rise, plateau at $0^\circ\text{C}$ , rise, plateau at $100^\circ\text{C}$ — with segments 2 and 4 the constant-temperature phase-change plateaus, and segments 1 and 3 the rising sensible-heat stretches.

Competitive level — comparing boiling points at two altitudes on the phase diagram

At sea level, atmospheric pressure is about $1.013\times10^5\ \text{Pa}$ and water boils at $100^\circ\text{C}$ . On a high mountain, atmospheric pressure drops to roughly $7\times10^4\ \text{Pa}$ (about 70% of sea-level pressure). Using the shape of the liquid–gas boundary on water's phase diagram, explain qualitatively what happens to the boiling point, and estimate why boiling water on that mountain cooks food more slowly.

Step 1. Locate both pressures on the liquid–gas boundary curve of the phase diagram. This curve rises from the triple point ( $611.73\ \text{Pa}$ , $273.16\ \text{K}$ ) up toward the critical point, meaning higher pressure always corresponds to a higher point on the boundary — a higher boiling temperature.

Step 2. Reading the curve at the lower mountain pressure of about $7\times10^4\ \text{Pa}$ (roughly 70% of sea-level pressure) gives a crossing temperature below $100^\circ\text{C}$ ; standard reference data put this near $90$ – $92^\circ\text{C}$ for that pressure ratio.

Step 3. The physical consequence for cooking: heat transfer into food depends on the temperature difference between the boiling water and the food, roughly following Newton's law of cooling in spirit (heat transfer rate scales with $\Delta T$ ). Water at $90^\circ\text{C}$ delivers less thermal driving force than water at $100^\circ\text{C}$ , so the same food takes noticeably longer to cook, even though the water is still visibly at a rolling boil the whole time.

Answer: lowering ambient pressure moves the crossing point on the liquid–gas boundary to a lower temperature, so mountain water boils near $90$ – $92^\circ\text{C}$ instead of $100^\circ\text{C}$ , and food cooks slower because the boiling water itself is cooler, not because it is boiling any less vigorously. Timing note: under two minutes. The only working idea needed is "read the boiling point off the pressure axis of the phase diagram" — do not try to compute an exact temperature from a formula; qualitative reasoning about the shape of the boundary curve, plus a remembered anchor value like the sea-level $100^\circ\text{C}$ point, is exactly what the board expects at this level.

Common traps

Assuming temperature keeps rising throughout heating — the flat plateaus during melting and boiling are the single most frequently tested, most frequently forgotten detail in this topic.
Mixing up which segment of a heating curve corresponds to melting versus boiling when a question uses an unfamiliar substance with melting/boiling points that are not the familiar $0^\circ\text{C}$ / $100^\circ\text{C}$ of water.
Thinking sublimation only happens at very low pressure — while it is favoured below the triple-point pressure, common substances like camphor and naphthalene sublime readily at ordinary atmospheric pressure.
Treating the triple point as the same thing as the ordinary freezing point at one atmosphere — the triple point is one specific, usually very low, pressure unique to the substance ( $611.73\ \text{Pa}$ for water, far below atmospheric), not standard atmospheric pressure.
Forgetting that specific heat capacity is different in each phase — the slope of the ice segment, the water segment, and the steam segment on a heating curve are all different, because $c_{\text{ice}}$ , $c_{\text{water}}$ and $c_{\text{steam}}$ are different numbers.

You should be able to…

Name and distinguish all six state-change processes: fusion, freezing, vaporisation, condensation, sublimation, and deposition.
Sketch and interpret a temperature-versus-heat-added heating curve for a substance like water, correctly identifying which segments are sloped and which are flat plateaus.
Explain, in terms of intermolecular bonds versus kinetic energy, why temperature stays constant during a phase change despite continuous heat input.
Build the total-heat expression for a multi-stage heating curve by adding $Q=mc\Delta T$ terms for sloped segments and $Q=mL$ terms for flat segments.
Define the triple point and critical point and locate them qualitatively on a $P$ – $T$ phase diagram.
Explain, using the phase diagram, why boiling point falls with decreasing pressure (mountain altitude) and rises with increasing pressure (pressure cooker).

Ask the tutor: want a walk-through of the full water phase diagram with the triple point and critical point marked, or a practice heating-curve problem with numbers plugged into every stage?

If heat keeps flowing in during melting, where does all that energy actually go?

It goes into potential energy, not kinetic energy. In a solid, molecules sit in a fixed lattice held together by intermolecular attraction; melting means pulling them out of that ordered arrangement into the looser, more mobile arrangement of a liquid. Doing that "pulling apart" costs energy, in exactly the same way lifting a ball against gravity costs energy without speeding the ball up. Since temperature measures average kinetic energy (how fast molecules jiggle, not how far apart they are arranged), the thermometer does not move even though the substance is absorbing real energy the whole time. That stored energy is released again, unchanged in amount, when the substance re-freezes.

Why does a pressure cooker cook food faster if boiling water is always the same temperature?

Boiling water is not always the same temperature — that is exactly the point of this topic. A pressure cooker seals the pot so that steam cannot escape, letting internal pressure build up well above one atmosphere. On the phase diagram, higher pressure pushes the liquid-gas boundary crossing to a higher temperature, so water inside a pressure cooker can reach $110$ - $120^\circ\text{C}$ before it boils, rather than stopping at $100^\circ\text{C}$ . Since cooking speed depends on temperature, not merely on "is it boiling," that extra $10$ - $20$ degrees cooks food substantially faster, which is the entire engineering point of the appliance.

Is the boiling point of water always exactly 100 degrees Celsius?

Only at exactly one atmosphere of pressure, which is why $100^\circ\text{C}$ is quoted as the boiling point of water in most textbooks — sea-level atmospheric pressure is the implicit assumption. Change the pressure and the boiling point changes with it, following the liquid-gas boundary curve on water's phase diagram: lower pressure (a mountain, a vacuum chamber) gives a lower boiling point, and higher pressure (a pressure cooker, deep underground) gives a higher one. The melting point is far less sensitive to pressure for most substances, which is why $0^\circ\text{C}$ feels like a much more universal number than $100^\circ\text{C}$ actually is.

What is the practical difference between the critical point and the triple point?

They sit at opposite ends of the liquid-gas boundary curve on a phase diagram and mean opposite things. The triple point is where three phases meet and coexist together — a single sharply defined pressure and temperature, used in fact to define the kelvin scale itself (water's triple point is fixed at exactly $273.16\ \text{K}$ ). The critical point is where the liquid-gas boundary curve simply ends: beyond that pressure and temperature, the properties of liquid and gas become identical, and no amount of compressing a gas at that temperature will ever produce a visibly distinct liquid. One point marks where all three states meet; the other marks where two of them stop being distinguishable at all.