In short

A phase transition is the change of a substance from one physical state to another — solid to liquid (melting, or fusion in reverse), liquid to gas (vaporisation), solid to gas (sublimation), and the three reverse processes. Each transition happens at a definite temperature (for a given pressure) and absorbs or releases a definite amount of energy per kilogram of material — the latent heat:

\boxed{\; Q \;=\; m L \;}

where L is the specific latent heat of the transition (J/kg). For water near standard atmospheric pressure the two numbers to carry in your head are

L_f \;=\; 3.34 \times 10^{5}\,\text{J/kg} \quad (\text{fusion, solid}\leftrightarrow\text{liquid at } 0\,^\circ\text{C})
L_v \;=\; 2.26 \times 10^{6}\,\text{J/kg} \quad (\text{vaporisation, liquid}\leftrightarrow\text{gas at } 100\,^\circ\text{C}).

Latent heat is "hidden" because during the transition the temperature does not change. Every joule you add breaks intermolecular bonds — stretching the lattice apart in melting, tearing molecules out of the liquid in boiling — rather than speeding molecules up. The temperature plateau on a heating curve is the visible signature.

The boiling point is not a universal constant. It depends on pressure — that is why a pressure cooker in Pune boils dal at 120 °C instead of 100 °C, and why an open pot in Leh (3500 m altitude) boils only at about 87 °C and takes forever to soften rajma. Melting points shift with pressure too, though far more weakly; for water (ice contracts on melting) the shift is negative — higher pressure lowers the melting point, and that is what makes a skater's blade slide.

Pour a tall glass of nimbu pani in a Chennai kitchen in April. The juice sits at room temperature, about 32 °C. Drop in two ice cubes, each at −10 °C from the freezer. Stick a thermometer into the drink and watch it carefully for the next three minutes.

What you expect, if you remember only your calorimetry, is a smooth drop. Two cold bodies landed in a warm one; Q_\text{lost}=Q_\text{gained} should settle everything at some intermediate temperature, maybe 4 or 5 °C, depending on how much ice you added. And the final answer is indeed roughly that. But if you watch the thermometer on the way to equilibrium, you see something stranger. The nimbu pani cools fast while the ice is still solid and cold. Then, as the outside of each cube reaches 0 °C and begins to melt, the drink temperature stalls. For twenty or thirty seconds the thermometer barely moves while the cube visibly shrinks. Only once the last sliver of ice has dissolved does the temperature begin to fall toward its final value again.

Something swallowed an enormous amount of heat from the lemon juice without warming by a single degree. That something was the melting ice. The ice was already at 0 °C — the melting temperature of water at atmospheric pressure — and every joule that arrived at it for the next half-minute went not into raising its temperature but into tearing the solid crystal apart, molecule by molecule, converting ice at 0 °C to liquid water at 0 °C. The heat was latent — hidden in the sense that no thermometer could detect it as a temperature rise, even though the energy was absolutely flowing in. The same thing happens in reverse when you freeze ice trays in the freezer — the water releases a huge amount of heat (at 0 °C) as it crystallises, and your freezer motor has to pump all of that out.

This article makes the latent-heat plateau quantitative. It derives Q = mL from first principles, measures L_f and L_v for water, draws the entire heating curve from ice at −20 °C to steam at 120 °C, explains why pressure shifts boiling and melting points (and why water is strange — the shift is in the opposite direction to almost every other substance), and closes with two big consequences: why the pressure cooker on your stove is the single most effective piece of thermodynamic engineering in an Indian kitchen, and why the latent heat of water vapour carries more energy through the Indian monsoon than any other single mechanism on the subcontinent.

What a phase transition is — the microscopic picture

A phase is a state of matter in which molecules arrange themselves in a characteristic way. In a solid, molecules sit in roughly fixed positions — a crystalline lattice for ice, salt, or iron; a disordered frozen network for glass — held there by attractive intermolecular forces strong enough to overcome the thermal jiggling. In a liquid, molecules are still touching and still attracting each other, but they are no longer locked into position; they can slide past one another while staying in contact, which is what lets a liquid take the shape of its container while keeping roughly constant volume. In a gas, molecules are far apart, moving essentially freely between occasional collisions; the attractive forces between them are negligible, and the gas expands to fill whatever volume is available.

The three phases differ in how tightly bound the molecules are. Roughly speaking, the binding energy per molecule is of order kT (with k Boltzmann's constant) at the relevant transition temperature — because this is exactly the thermal energy scale at which random jiggling starts to win against the bonds.

A phase transition is the change from one phase to another. There are six of them, grouped into three reversible pairs:

Direction Name What it looks like
solid → liquid melting (fusion) ice cube turning to water
liquid → solid freezing (solidification) ice tray in the freezer
liquid → gas vaporisation (boiling/evaporation) water turning to steam in the pressure cooker
gas → liquid condensation steam depositing as droplets on a cold glass
solid → gas sublimation dry ice fog, naphthalene balls (mothballs) shrinking
gas → solid deposition frost forming on Ladakhi windshields, snow nucleating from moist air

Each forward transition absorbs energy. Each reverse transition releases the same amount of energy. The energy absorbed or released per kilogram of material is the latent heat of that transition; it is called "latent" — from the Latin for hidden — because it flows without changing the temperature while the transition is happening.

For water, the two transitions you meet most in Indian kitchens and Indian weather are:

Vaporisation requires about seven times more energy per kilogram than fusion for water. This will turn out to matter enormously.

Deriving Q = m L

Hand a chunk of material sitting exactly at its transition temperature a small amount of heat \delta Q. If you are below the transition, the heat raises the temperature: \delta Q = m c\, dT. But at the transition, dT = 0 — by definition the temperature is locked while the phase change is happening. Where does the heat go?

It goes into breaking bonds. Every molecule in the solid sits in a potential-energy well dug by its neighbours. To melt, the molecule must be lifted out of that well — its potential energy rises without its kinetic energy (and hence temperature) changing. Call the energy required per molecule \varepsilon. To melt the whole sample of mass m, which contains N = m/m_0 molecules of molecular mass m_0, you need

Q \;=\; N \varepsilon \;=\; \frac{m \varepsilon}{m_0}. \tag{1}

Why: every molecule in the lump has to be freed from its lattice position, and each one costs the same energy \varepsilon on average. So total energy is number-of-molecules times cost-per-molecule. Nothing in this step cares about temperature — it is just book-keeping.

Define the specific latent heat

L \;\equiv\; \frac{\varepsilon}{m_0} \;=\; \text{energy to change phase per unit mass}. \tag{2}

Why: \varepsilon is the cost per molecule and m_0 is the mass per molecule, so \varepsilon / m_0 is the cost per kilogram. This is the natural intensive version — a material property, independent of the amount you have.

Then equation (1) becomes

\boxed{\; Q \;=\; m L \;}. \tag{3}

Why: the latent heat needed to carry a mass m through a phase change is simply mass times the latent-heat-per-kilogram. Same shape as Q = mc\Delta T for warming — but with \Delta T replaced by the fact of a phase change, and c replaced by L. The SI unit of L is J/kg.

The physics hidden in the derivation is entirely in \varepsilon. What sets \varepsilon for a given substance? The strength of its intermolecular bonds. For water, the hydrogen-bond network is strong; \varepsilon is correspondingly large. For a noble gas like argon, there are only feeble van der Waals attractions, so \varepsilon is tiny and L is small.

Latent heat of fusion is always much less than latent heat of vaporisation for the same substance. In melting, you only have to partially unlock the lattice — molecules still touch and still attract; you are buying mobility, not separation. In vaporising, you have to completely pull molecules out of each other's reach — breaking every remaining bond as the molecule takes off into the gas phase. The bond-breaking bill is vastly larger. For water, L_v / L_f \approx 6.77; for most common substances this ratio is in the range 5 to 20.

The heating curve for water — end to end

The clearest picture of phase transitions is the heating curve: a plot of temperature versus total heat added, at constant pressure (atmospheric, unless stated otherwise). Take one kilogram of ice at −20 °C at sea level and add heat slowly, one kilojoule at a time. Plot the temperature.

What you get is not a straight line. It is a piecewise curve with flat plateaus at the phase transitions. The slope of each sloped segment is 1/(mc) — the smaller the heat capacity, the steeper the climb — and the length of each flat plateau is m L for that phase transition.

Heating curve for one kilogram of water from ice at minus twenty to steam at one-twentyTemperature on the vertical axis from minus twenty to one-twenty Celsius, heat added on the horizontal axis from zero to three point one megajoules. The curve rises steeply through ice from minus twenty to zero, then flattens at zero for three hundred thirty-four kilojoules while ice melts, then rises through liquid water from zero to one-hundred, then flattens at one hundred for two thousand two hundred sixty kilojoules while water vaporises, then rises through steam above one hundred. heat added Q (kJ, for 1 kg) temperature T (°C) −20 0 20 40 60 80 100 120 0 500 1000 1500 2000 2500 3000 warming ice (c = 2100) melting plateau L_f = 334 kJ warming water (c = 4186) vaporisation plateau L_v = 2260 kJ steam
Heating curve for 1 kg of water, from ice at −20 °C to steam at 120 °C, at 1 atm. The flat plateaus at 0 °C and 100 °C are the phase transitions — every joule added during these stretches breaks bonds, not speeds molecules. The vaporisation plateau is about 6.8 times as long as the melting plateau because $L_v \approx 6.77\,L_f$ for water.

Five distinct regimes, read left to right:

Regime 1: warming ice from −20 °C to 0 °C. The sample is solid throughout. Heat goes into lattice vibrations — T rises. Q = m c_\text{ice}\Delta T = 1 \times 2100 \times 20 = 42\,000 J = 42 kJ.

Regime 2: melting at 0 °C. Both solid and liquid coexist; temperature locked. Q = m L_f = 1 \times 3.34 \times 10^5 = 334 kJ.

Regime 3: warming liquid water from 0 °C to 100 °C. Sample is liquid. Q = m c_\text{water}\Delta T = 1 \times 4186 \times 100 = 418.6 kJ.

Regime 4: vaporising at 100 °C. Liquid and vapour coexist; temperature locked. Q = m L_v = 1 \times 2.26 \times 10^6 = 2260 kJ.

Regime 5: warming steam from 100 °C to 120 °C. Sample is gas. Q = m c_\text{steam}\Delta T = 1 \times 2010 \times 20 = 40.2 kJ.

Total: 42 + 334 + 418.6 + 2260 + 40.2 = 3094.8 kJ. About 73 % of the total energy budget is spent on the vaporisation plateau alone. Every other stage is a rounding error beside the latent heat of vaporisation. This is why a kettle takes thirty seconds to warm cold water from the tap up to boiling temperature but then takes ten minutes to boil it dry — ninety percent of the time is the vaporisation plateau, where the thermometer reads 100 °C and nothing seems to be happening.

The slope of each sloped segment

The slope of any sloped segment is

\frac{dT}{dQ} \;=\; \frac{1}{mc}.

Why: differentiate Q = mc\Delta T with respect to Q, treating m and c as constants, and rearrange. The steeper the climb, the smaller the heat capacity — it takes less heat per kelvin.

So among the three sloped segments, the steam segment is the steepest (smallest c), the water segment is the shallowest (largest c), and the ice segment is intermediate. You can read that off the drawing directly.

Why the plateau is truly flat

It is worth pausing to justify the flatness. Why doesn't the temperature climb a little while the ice melts? Microscopically, here is the story: every molecule in the solid is sitting in a well of depth \varepsilon. Thermal jiggling occasionally kicks a molecule over the wall into the liquid phase, and the reverse happens too (a molecule in the liquid occasionally rejoins the lattice). At the melting temperature T_m, the rates of forward and reverse hopping exactly balance — the two phases are in equilibrium. Add heat and you shift the balance: more molecules jump out of the lattice than rejoin it, and the sample melts. But as long as both phases are present, the temperature stays pinned at T_m — thermal equilibrium between solid and liquid enforces it. Only when the last bit of solid vanishes (or, in the reverse process, when the last bit of liquid freezes) is the temperature free to move again.

This is a first-order phase transition — the defining feature is the latent-heat plateau. There are also second-order transitions (the ferromagnetic Curie point, the superfluid lambda transition in helium-4) which are continuous — no plateau, no latent heat — but the first-order case is the one you meet in daily life.

Watch the plateau

The plot shows five regimes frozen in a single snapshot. To see the plateau, you need to watch the temperature of the sample evolve in real time while heat is pumped in at a steady rate. The animated simulation below does exactly that. Imagine a 1 kg block of ice at −20 °C being heated on a 1000-watt burner (delivering 1 kJ of heat every second). The equations for temperature vs time come from Q = mc\Delta T and Q = mL with Q = P\cdot t substituted in — real physics, not a visual hack.

Animated heating curve for one kilogram of water A dot traces the temperature of a 1 kg sample as a 1 kW burner pumps heat in at 1 kJ per second, shown at 310x time-lapse. The dot rises from minus twenty through the melting plateau at zero from physical t=42 to t=376 seconds, then climbs to one hundred, then pauses at one hundred until physical t=3054 seconds while the water boils, then rises briefly into steam. time on 1 kW burner (s) — equals heat in kJ temperature (°C) 0 °C 100 °C 500 1000 1500 2000 2500 3000
Watch the sample's temperature as the 1 kW burner pumps heat in. The dot flies through the ice-warming phase (42 s), then sticks at 0 °C for five and a half minutes while the ice melts, climbs to 100 °C in about seven minutes, then pauses at the boiling plateau for thirty-seven minutes while the water boils away, and finally rises briefly into the steam regime. The dashed grey curve in the background is the theoretical heating curve; the red trail shows the real-time trace of the sample. Click replay to watch again.

Latent heats of water — the two numbers to know

For water at 1 atm the two critical values are

L_f = 3.34 \times 10^5 \;\text{J/kg} = 334\,\text{kJ/kg} = 80\,\text{cal/g}
L_v = 2.26 \times 10^6 \;\text{J/kg} = 2260\,\text{kJ/kg} = 540\,\text{cal/g}.

A 1 g ice cube melting absorbs 80 calories of heat from whatever it touches. A 1 g drop of water evaporating from your skin (perspiration) carries away 540 calories — which is why sweating cools you so effectively. These two numbers are responsible for a shocking amount of everyday physics.

Some scale comparisons

Process Heat per kg
Warming water 1 K 4.186 kJ
Warming water 80 K (0 → 80 °C) 335 kJ
Melting ice at 0 °C 334 kJ
Warming water 100 K (0 → 100 °C) 418.6 kJ
Vaporising water at 100 °C 2260 kJ
Warming steam 100 K 201 kJ

Look at the coincidence on rows 2 and 3: melting 1 kg of ice takes almost exactly as much heat as warming 1 kg of liquid water from 0 to 80 °C. This is why ice in nimbu pani is so much more effective at cooling than cold water would be. A cube of ice at 0 °C, melting and then warming up to the drink's temperature, absorbs dramatically more heat per gram than a cube of cold water at 0 °C. That extra budget is the latent heat of fusion. Ice cubes are thermal batteries; cold water is just cold water.

Values for other substances (at 1 atm)

Substance L_f (kJ/kg) Melts at L_v (kJ/kg) Boils at
Water 334 0 °C 2260 100 °C
Ethanol 109 −114 °C 841 78 °C
Mercury 11.3 −39 °C 293 357 °C
Lead 23 327 °C 870 1744 °C
Aluminium 396 660 °C 10 500 2467 °C
Nitrogen 26 −210 °C 199 −196 °C
Copper 205 1085 °C 4730 2562 °C

Water's L_v is the largest in this table, and one of the largest of any common substance on Earth — thanks again to hydrogen bonding. Nitrogen's L_v is small (only 199 kJ/kg), which is why liquid nitrogen boils away so vigorously when you pour it on the floor. Aluminium's vaporisation energy is enormous (10 500 kJ/kg), which is why aluminium is never deliberately vaporised in industry — it takes extraordinary energy input to do it.

Evaporation vs boiling — both are vaporisation

Vaporisation does not require the liquid to reach 100 °C. A puddle on a Jaipur sidewalk in May evaporates completely at 35 °C, and a wet uniform shirt dries on a clothesline even in winter. Both evaporation and boiling are the same physical process — molecules escaping from the liquid phase into the gas phase — but they happen under different conditions.

Evaporation happens at the surface, at any temperature, whenever the air above the liquid is not already saturated with that liquid's vapour. At any moment, molecules in the liquid have a range of speeds; a few at the top tail of the distribution have enough kinetic energy to break the surface tension and escape into the air. This drains kinetic energy away from the liquid — every escaped molecule carries energy roughly L_v \cdot m_0 with it — so the remaining liquid cools. This is exactly why wetting your forehead cools you: the evaporating water carries away its latent heat of vaporisation, and the supplier of that heat is your skin.

Boiling happens throughout the bulk of the liquid, at the temperature where the saturated vapour pressure of the liquid equals the external atmospheric pressure. At this temperature, bubbles of pure vapour can form inside the liquid without being crushed by the surrounding pressure; the bubbles grow and rise, which is what you see as a pot coming to the boil. Once boiling starts, any further heat input goes entirely into vaporising more liquid — the temperature stays locked at the boiling point.

The same L_v governs both. The difference is where the vaporisation happens (surface vs bulk) and at what temperature (any vs the boiling point).

Pressure dependence — why boiling point is not a constant

Boiling point is not written into a substance — it depends on the pressure against which the vapour must push. The quantitative statement is the Clausius-Clapeyron equation, derived in the going deeper section. The qualitative result:

At 1 atm water boils at 100 °C; at 2 atm water boils at about 121 °C; at 0.5 atm it boils at 82 °C.

Two Indian consequences:

Pressure cooker in Pune. The weight on top traps steam until the internal pressure is about 2 atm, at which point the regulator lifts and hisses. The water inside boils at ~120 °C, not 100 °C. Rajma that would take two hours to soften at 100 °C softens in thirty minutes at 120 °C — cooking rates depend on temperature very steeply (roughly doubling for every 10 K rise, by Arrhenius), so 20 extra kelvins is a factor of roughly 4 on the rate. Every Indian pressure cooker is a small applied-thermodynamics exam.

Open pot in Leh. At 3500 m altitude the atmospheric pressure is about 0.65 atm, so water boils at ~87 °C. You can put a pot of rajma on a fire in Leh and watch it boil for hours without ever softening properly — the water is not hot enough. This is why treks above 3000 m carry pressure cookers, and it is also why the quintessential Ladakhi dish, thukpa, is a noodle-and-broth dish — something that cooks well at 87 °C — rather than an Indian-mainland dal that requires higher temperatures.

Boiling temperature of water as a function of pressureA rising curve of boiling temperature versus pressure. Three marked points: Leh at 0.65 atm and 87 Celsius, Mumbai at 1 atm and 100 Celsius, and pressure cooker at 2 atm and 121 Celsius. external pressure (atm) boiling temperature (°C) 0.5 1.0 1.5 2.0 2.5 80 100 120 Leh (0.65 atm, 87 °C) Mumbai (1 atm, 100 °C) pressure cooker (2 atm, 121 °C)
The boiling point of water as a function of external pressure, with three locations marked. The curve follows the Clausius-Clapeyron equation (derived in the going-deeper section). A pressure cooker raises the boiling temperature by ~21 K; Leh lowers it by ~13 K. Cooking rates are exponentially sensitive to temperature, which is why these shifts matter so much in everyday kitchens.

Pressure dependence of the melting point — water is upside down

For most substances, higher pressure raises the melting point too (for the same reason as boiling): denser conditions favour the denser phase. But water is strange. Ice is less dense than liquid water — that is why ice cubes float in nimbu pani. So for water, higher pressure favours the liquid, which means the melting point decreases with pressure.

The shift is small — about -7.4 \times 10^{-3} K per atm — but it is real. A skater gliding across a frozen lake in Leh exerts enormous pressure through a thin blade edge; the pressure locally depresses the melting point just enough that a thin film of liquid water forms under the blade, and the skater slides on that liquid film. (Modern research shows that friction heating and surface pre-melting also play roles, but pressure melting is a real contributor.)

For almost every other substance — benzene, lead, iron — the melting point rises with pressure. Water is the exception, and the exception is what makes ice float, which is what makes freshwater lakes freeze from the top down (if ice sank, winter would wipe out aquatic life in every lake).

Explore: drag the pressure

The curve above shows three points. The interactive plot below lets you drag the pressure from 0.3 atm to 2.5 atm and watch the boiling temperature trace the curve in real time. The formula is the Clausius-Clapeyron result derived in the going-deeper section; the relevant latent heat is L_v = 2.26\times 10^6 J/kg, the molar mass is 0.018 kg/mol, and R = 8.314 J/(mol·K).

Interactive: boiling temperature as a function of pressure A curve showing boiling temperature of water rising with pressure. A draggable red dot moves along the pressure axis and a computed point follows the curve, with the current boiling temperature displayed. external pressure (atm) boiling T (°C) 70 80 100 120 0.5 1.0 1.5 2.0 2.5 drag the red point along the axis
Drag the red dot along the horizontal axis to change the external pressure. The curve is the Clausius-Clapeyron prediction for the boiling point of water; the readout shows the new boiling temperature. At 1 atm the answer is 100 °C; at 2 atm it is about 121 °C; at 0.5 atm it is about 82 °C. The shift readout expresses the change as a percentage of the absolute boiling temperature.

Worked examples

Example 1: Ice cubes in nimbu pani

Two ice cubes (total mass 60 g) at −10 °C are dropped into 250 g of nimbu pani at 32 °C in an insulated glass. Specific heat of ice is 2100 J/(kg·K); specific heat of water (and nimbu pani) is 4186 J/(kg·K); latent heat of fusion is L_f = 3.34\times 10^5 J/kg. Find the final temperature. Assume all the ice melts (check this at the end).

Energy flow diagram: ice cubes melting in a glass of nimbu pani A labelled box shows an ice cube at minus ten going through three stages: warm ice to zero (step A), melt ice at zero (step B), warm meltwater from zero to the final temperature (step C). A separate arrow shows the nimbu pani cooling from thirty-two to the final temperature. The total heat gained by the ice path equals heat lost by the pani path. Ice (60 g, −10 °C) A: warm ice to 0 °C Q_A = m·c_ice·(0 − (−10)) B: melt ice at 0 °C Q_B = m·L_f C: warm water to T_f Q_A+B+C Nimbu pani (250 g, 32 °C) cools from 32 °C → T_f Q_lost = M·c_water·(32 − T_f) Q_gained = Q_lost (energy conservation)
Three stages of heating for the ice (warm to 0 °C, melt at 0 °C, warm the meltwater to $T_f$) balanced against the cooling of the pani. The total heat gained by the ice path equals the heat lost by the pani.

Step 1. Set up the heat budgets.

The ice goes through three stages:

Q_A = m_\text{ice}\, c_\text{ice}\, (0 - (-10)) = 0.060 \times 2100 \times 10 = 1260\,\text{J}

Why: first, the ice warms up from −10 °C to its melting point, using its solid specific heat. Ice is still solid on this leg — nothing has melted yet.

Q_B = m_\text{ice}\, L_f = 0.060 \times 3.34\times 10^5 = 20\,040\,\text{J}

Why: at 0 °C the ice melts. The temperature does not change during this stage; all the heat goes into breaking up the crystal lattice. This is the latent-heat plateau.

Q_C = m_\text{ice}\, c_\text{water}\, (T_f - 0) = 0.060 \times 4186 \times T_f = 251.16\, T_f\,\text{J}

Why: once melted, the new liquid water warms from 0 °C up to the final mixture temperature. Specific heat switched from ice's value (2100) to water's (4186).

Step 2. Set up the heat lost by the pani.

Q_\text{lost} = m_\text{pani}\, c_\text{water}\, (32 - T_f) = 0.250 \times 4186 \times (32 - T_f) = 1046.5\,(32 - T_f)\,\text{J}

Why: the pani is liquid throughout; it just cools. Its specific heat is that of water because nimbu pani is mostly water.

Step 3. Apply energy conservation.

Q_A + Q_B + Q_C = Q_\text{lost}
1260 + 20\,040 + 251.16\, T_f \;=\; 1046.5\,(32 - T_f)
21\,300 + 251.16\, T_f \;=\; 33\,488 - 1046.5\, T_f
1297.66\, T_f \;=\; 12\,188
T_f \;=\; 9.39\,^\circ\text{C}

Why: the heat gained by the ice (which ends up as meltwater at T_f) has to equal the heat lost by the pani. Solve the resulting linear equation for T_f.

Step 4. Check the assumption that all the ice melted.

For all the ice to melt, the pani must deliver at least Q_A + Q_B = 21\,300 J. The pani cooling all the way from 32 °C to 0 °C can deliver 1046.5 \times 32 = 33\,488 J — comfortably more than 21 300 J. So yes, all the ice melts, and the final temperature is indeed above 0 °C. Consistent.

Result: The final temperature is 9.4 °C.

What this shows: The 60 g of ice cooled 250 g of pani from 32 °C to 9.4 °C — a drop of 22.6 K. If you had instead used 60 g of cold water at 0 °C (no latent heat, same final temperature), the same calculation would give a final temperature of about 25.8 °C — barely any cooling. The extra 16 K of cooling power comes entirely from the latent heat of fusion of the ice.

Example 2: Boiling rajma in a pressure cooker

A pressure cooker in a Chennai kitchen contains 1.5 L of water at 25 °C initially. The gas stove delivers heat at 1.8 kW into the cooker. The cooker regulator lifts at an internal pressure of 2 atm, at which water boils at 121 °C. How long does it take for the regulator to first lift, and then how long would the cooker take to boil away 10 % of the water as steam (a small but safe amount of "cooking steam")? Use c_\text{water} = 4186 J/(kg·K); at 2 atm, L_v(2\,\text{atm}) \approx 2.20\times 10^6 J/kg.

Pressure cooker schematic A cylindrical vessel with a lid and pressure regulator. Heat flows in at the bottom at 1.8 kilowatts; steam collects at the top raising internal pressure to 2 atmospheres; water inside at 121 Celsius when regulator lifts. regulator (2 atm) 1.5 kg water steam space 1.8 kW in
A 1.5 L pressure cooker receiving 1.8 kW from the stove. Before the regulator lifts, heat goes into raising the water temperature. After it lifts, heat vaporises water at 121 °C.

Step 1. Compute the heat to warm the water from 25 °C to 121 °C.

Q_1 = m\, c_\text{water}\, \Delta T = 1.5 \times 4186 \times (121 - 25)
Q_1 = 1.5 \times 4186 \times 96 = 602\,784\,\text{J} \approx 603\,\text{kJ}

Why: before the regulator lifts, the cooker is effectively sealed at atmospheric pressure until the vapour pressure inside builds up to 2 atm. During this phase, most of the heat goes into warming the liquid water up to its new (pressurised) boiling point.

Step 2. Compute the time to lift the regulator.

t_1 = \frac{Q_1}{P} = \frac{602\,784}{1800} = 335\,\text{s} \approx 5.6\,\text{min}

Why: the stove delivers power P = 1800 W into the cooker; the time to accumulate Q_1 is heat divided by power. Five and a half minutes of preheat before the regulator starts whistling — which is about what you measure in a kitchen, once you subtract heat lost to the cooker walls.

Step 3. Compute the heat to boil away 10 % of the water at 121 °C.

Q_2 = (0.10 \times m) \times L_v(2\,\text{atm}) = 0.15 \times 2.20 \times 10^6 = 330\,000\,\text{J} = 330\,\text{kJ}

Why: once the regulator is lifting, the cooker's internal pressure and temperature are both pinned; every further joule of heat input goes into vaporising water into steam. The mass to vaporise is 10 % of 1.5 kg = 0.15 kg, and each kilogram costs L_v joules.

Step 4. Compute the time for the boiling stage.

t_2 = \frac{Q_2}{P} = \frac{330\,000}{1800} = 183\,\text{s} \approx 3.1\,\text{min}

Step 5. Sum the stages.

t_\text{total} = t_1 + t_2 = 335 + 183 = 518\,\text{s} \approx 8.6\,\text{min}

Why: the stove does not care whether it is heating or vaporising — it delivers 1.8 kW either way. Total time is just the sum of the two stages' energy demands divided by the common power.

Result: The regulator first lifts at about 5.6 min. An additional 3.1 min gets 10 % of the water vaporised. Total: ~8.6 min.

What this shows: The latent-heat stage (Q_2 = 330 kJ) is astonishing — it takes half as much heat to vaporise 10 % of the water as it did to warm all 100 % of it by 96 kelvins. If you forget the cooker on the stove, vaporisation will eventually drain it dry; but it will do so much more slowly than naive intuition suggests, because each gram of water leaving as steam carries 2200 joules with it. That same latent-heat sink is why the cooker never overpressurises — excess heat simply converts more water to steam, and the steam vents out through the regulator, keeping the interior temperature pegged at 121 °C.

Example 3: Sweat evaporation — cooling a cyclist in Chennai

A cyclist riding through Chennai in April weighs 65 kg and loses 600 g of sweat in one hour. Assume all of the sweat evaporates from the skin (rather than dripping off) and that the specific heat of the human body is 3500 J/(kg·K) and the latent heat of vaporisation of water at skin temperature (~33 °C) is 2.42 × 10⁶ J/kg. (a) How much heat does the evaporation carry away? (b) By how much would the cyclist's body temperature rise in that hour if no sweat evaporation occurred and the heat had nowhere else to go?

Sweat evaporation energy balance A simplified figure shows a person with droplets evaporating from the skin, with an arrow labelled latent heat leaving the body and a label reading Q equals m L. 65 kg cyclist latent heat out Q = m·L_v air (absorbs) Q = 0.6 kg × 2.42 MJ/kg
Sweat leaves the skin as liquid and enters the air as vapour. The energy to vaporise it is drawn from the body, giving a cooling effect set by $Q = m L_v$.

Step 1. Compute the heat carried away by the evaporation.

Q = m_\text{sweat} \times L_v = 0.6 \times 2.42 \times 10^6 = 1.452 \times 10^6\,\text{J} \approx 1.45\,\text{MJ}

Why: each kilogram of sweat that evaporates from skin takes L_v joules with it. The skin supplies these joules by cooling down; they leave the body inside the H₂O molecules that are now in the air.

Step 2. Convert to a power.

Per hour, that is a rate of

P = \frac{Q}{t} = \frac{1.452 \times 10^6}{3600} \approx 403\,\text{W}.

Why: divide total heat by time. 403 W of cooling from sweat alone — comparable to a hair dryer running continuously on the body surface.

Step 3. Compute the temperature rise that would occur without evaporation.

\Delta T = \frac{Q}{M_\text{body}\, c_\text{body}} = \frac{1.452 \times 10^6}{65 \times 3500} = \frac{1.452 \times 10^6}{227\,500} \approx 6.4\,\text{K}.

Why: if the same 1.45 MJ had gone into warming the cyclist's body tissues instead of vaporising sweat, it would have raised core temperature by Q/(Mc). The body is about 60 % water (high c) and 40 % other tissue, so an averaged c of 3500 J/(kg·K) is reasonable.

Result: (a) Sweat evaporation carries away ~1.45 MJ in the hour. (b) Without that escape route, the cyclist's body temperature would rise by about 6.4 K — from 37 °C to 43 °C, which is fatal hyperthermia.

What this shows: Evaporative cooling is not a comfort feature of the human body; it is a survival mechanism. The latent heat of vaporisation of water, the same number that sets the pressure-cooker plateau, is what keeps you alive on a hot Chennai afternoon. Humid air is dangerous because it slows evaporation — sweat beads up instead of vaporising, and the latent-heat escape hatch closes. This is why a 40 °C dry afternoon in Jaisalmer is survivable but a 35 °C humid afternoon in Kolkata can be more dangerous.

Common confusions

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If you came here to understand phase transitions and use Q = mL to solve problems, you have what you need. What follows is for readers who want the quantitative theory of how the transition temperature itself depends on pressure — the Clausius-Clapeyron equation — and a quick tour of how latent heat enters the Indian monsoon and climate system.

The Clausius-Clapeyron equation — derivation

Start with a sealed piston-cylinder containing a liquid and its vapour in equilibrium at temperature T and pressure P. Both phases coexist — the liquid is on the verge of evaporating, the vapour on the verge of condensing. The question: if you change T by a small amount dT, how must P change to keep them in equilibrium?

The clean answer uses a reversible cycle through the two phases. Transfer 1 kg of liquid to vapour at (T, P) — this takes in heat L_v. Now raise the temperature by dT at constant pressure (the vapour expands slightly, doing work). Then condense the 1 kg back to liquid at (T+dT, P+dP), releasing heat L_v' (slightly different). Finally cool the liquid by dT at constant pressure. You have returned to the starting state — a closed Carnot-like cycle.

By the second law (Carnot's theorem), the efficiency of this tiny cycle equals dT/T. By the first law, the net work done in the cycle equals the area enclosed on a P-V diagram — which, for this Carnot-shaped cycle, is dP \cdot \Delta V, where \Delta V = V_\text{vap} - V_\text{liq} is the volume difference per kilogram between the two phases. The heat input is L_v. Thus:

\eta = \frac{W_\text{net}}{Q_\text{in}} = \frac{dP \cdot \Delta V}{L_v} = \frac{dT}{T}.

Why: the vaporisation step absorbs L_v at temperature T, and the condensation step releases nearly L_v at temperature T+dT. The difference between them drives the cycle's work output. Applying Carnot's efficiency formula \eta = dT/T (valid for a small-dT reversible cycle between two temperatures) equates this work to (dP)(\Delta V).

Rearrange:

\boxed{\;\frac{dP}{dT} \;=\; \frac{L_v}{T\,\Delta V}\;} \tag{*}

This is the Clausius-Clapeyron equation — the exact relationship between pressure and temperature along a phase-coexistence curve. The right-hand side is always positive for normal liquid-vapour transitions, so boiling pressure rises with boiling temperature.

Approximation for vapour: the exponential form

For liquid → vapour transitions below the critical point, V_\text{vap} \gg V_\text{liq}, so \Delta V \approx V_\text{vap}. Using the ideal gas law V_\text{vap} = RT/(MP) for one kilogram of an ideal gas (where M is the molar mass and R the universal gas constant):

\frac{dP}{dT} \;=\; \frac{L_v \, M \, P}{R T^2}.

Why: substitute V_\text{vap} = RT/(MP) into equation (*). The P in the denominator of V ends up in the numerator of dP/dT, which is what makes this equation separable and gives the characteristic exponential behaviour of vapour pressure.

Separate variables and integrate:

\int \frac{dP}{P} \;=\; \int \frac{L_v \, M}{R T^2}\, dT
\ln P \;=\; -\frac{L_v M}{R T} + C.

Choosing the constant so that P = P_0 at T = T_0:

\boxed{\;\ln\!\left(\frac{P}{P_0}\right) \;=\; -\frac{L_v M}{R}\!\left(\frac{1}{T} - \frac{1}{T_0}\right)\;} \tag{**}

Why: this is the integrated Clausius-Clapeyron equation for a liquid-vapour transition, treating L_v as roughly constant over the temperature range of interest. Exponentiating both sides gives P(T) = P_0 \exp[-L_v M / R \cdot (1/T - 1/T_0)] — vapour pressure increases exponentially with temperature.

For water, plugging in L_v = 2.26 \times 10^6 J/kg, M = 0.018 kg/mol, R = 8.314 J/(mol·K), T_0 = 373.15 K, P_0 = 1 atm, one obtains T_\text{boil}(2\,\text{atm}) = 393.6 K = 120.4 °C, in excellent agreement with the tabulated 121 °C. This is where the pressure cooker's "121 °C" number comes from.

The melting curve — dP/dT for water is negative

The same equation (*) applies to solid-liquid transitions with L_v replaced by L_f and \Delta V = V_\text{liq} - V_\text{sol}.

For most substances the solid is denser than the liquid (so V_\text{sol} < V_\text{liq}, \Delta V > 0, and dP/dT > 0) — meaning higher pressure raises the melting point. For water, ice is less dense than liquid water: at 0 °C, the specific volume of ice is 1.091 \times 10^{-3} m³/kg and of liquid water is 1.000 \times 10^{-3} m³/kg, so \Delta V = -9.1 \times 10^{-5} m³/kg — negative. Plugging into (*):

\frac{dP}{dT} \;=\; \frac{3.34 \times 10^5}{273.15 \times (-9.1 \times 10^{-5})} \;\approx\; -1.34 \times 10^7\,\text{Pa/K}.

Equivalently, \frac{dT}{dP} \approx -7.4 \times 10^{-8} K/Pa = -7.4 \times 10^{-3} K/atm. Increase the pressure by 1 atm and water's melting point drops by about 7.4 mK — small, but measurable, and the sign is what matters.

Latent heat in the Indian monsoon

Every kilogram of water vapour that condenses back to liquid releases L_v \approx 2.26 \times 10^6 J to the surrounding air. The 2021 monsoon delivered roughly 10^{12} tonnes of rain over the Indian subcontinent. The latent heat released by condensing that much water vapour is

Q \;=\; 10^{15}\,\text{kg} \times 2.26 \times 10^6\,\text{J/kg} \;=\; 2.26 \times 10^{21}\,\text{J}.

For comparison, India's entire annual electricity generation is about 1.4 \times 10^{19} J. The latent heat released by one monsoon is more than 150 times India's yearly electricity output. This heat is what drives the monsoon circulation — air rises violently over the heated subcontinent, condenses into clouds, releases latent heat that drives further convection, and powers the cyclonic storms that occasionally sweep in from the Bay of Bengal.

The rain that drenches Mumbai in July is not just water — it is a pressurised stream of stored solar energy, evaporated off the Indian Ocean over the previous months, now reversing course and dumping its energy back onto land. L_v is the conversion factor.

The triple point and the phase diagram

Combining the solid-liquid and liquid-vapour curves into a single (P, T) plot gives the phase diagram of a substance. The curves intersect at the triple point — the unique temperature and pressure where all three phases coexist. For water, the triple point sits at T_\text{tp} = 273.16 K, P_\text{tp} = 611.657 Pa. By international agreement the kelvin was defined (until 2019) so that the triple point of water was exactly 273.16 K — that is, the kelvin itself was pinned to a phase transition.

Above the triple point, the liquid-vapour and solid-liquid curves separate, with a solid region, a liquid region, and a vapour region occupying distinct areas of the plot. Below the triple point, the liquid phase does not exist at any pressure — solid sublimates directly to vapour. This is why dry ice (solid CO₂) sublimes at atmospheric pressure without passing through a liquid stage: CO₂'s triple-point pressure is 5.18 atm, well above atmospheric, so at 1 atm you are always below the triple-point curve and solid-to-vapour is the only accessible transition.

Where this leads next