Latent heat

Place a thermometer in a beaker of ice that is steadily being heated on a flame, and something strange happens the moment the ice starts to melt. The flame keeps pouring in heat at the same steady rate as before, but the thermometer reading freezes at exactly $0^\circ\text{C}$ and refuses to climb — not for a second, but for minutes, until every last sliver of ice has turned to water. Where did all that heat go, if not into raising the temperature? It went into prying apart the ice's rigid molecular scaffolding. Water molecules in ice sit locked into a crystal lattice, held there by hydrogen bonds; to free them into the loose, flowing arrangement of liquid water takes energy, and that energy is spent entirely on breaking bonds rather than speeding up molecular motion. Because temperature measures the average kinetic energy of molecules, and this heat is going into potential energy of broken bonds instead, the thermometer simply does not move. Physicists call this hidden, temperature-blind quantity of heat the latent heat of the substance — "latent" from the Latin for "hidden," because it does its work invisibly, without announcing itself on any thermometer. This single idea, dressed up in the compact formula $Q = mL$ , quietly runs an enormous amount of the physical world: it is why a glass of water with ice cubes stays near $0^\circ\text{C}$ however hard you stir it, why steam burns are far worse than boiling-water burns at the same temperature, why sweating cools the body, and why a pressure cooker's steam carries so much more punch than the same mass of hot water. This article builds the formula from the change-of-state picture, walks through the standard CBSE numbers for ice and steam, and works two examination-style problems — one exactly at board level, one dialled up to competitive-exam speed.

Key takeaways

Latent heat is heat that changes state, not temperature. During melting, freezing, boiling or condensation, all the heat supplied or removed goes into changing the molecular arrangement, not the kinetic energy that temperature measures.
The defining formula is $Q = mL$ , where $L$ is the specific latent heat (J/kg) — heat per unit mass for that particular phase change.
Two named values matter for CBSE numericals: latent heat of fusion of ice, $L_f = 3.34\times10^5\ \text{J/kg}$ at $0^\circ\text{C}$ ; latent heat of vaporisation of water, $L_v = 22.6\times10^5\ \text{J/kg}$ at $100^\circ\text{C}$ .
$L_v \gg L_f$ for the same substance — vaporisation must fully separate molecules against nearly all their mutual attraction, while fusion only loosens a crystal lattice enough for molecules to slide past each other.
Multi-stage problems mix two formulas. Any heating curve segment is either a temperature ramp ( $Q = mc\,\Delta T$ , sensible heat) or a flat phase-change plateau ( $Q = mL$ , latent heat); a full ice-to-steam journey needs both, added stage by stage.
The plateau is the tell in every graph and every exam question — a flat region on a temperature-vs-heat graph is always a phase change, and the flatness itself is proof that latent, not sensible, heat is at work there.
Key idea: latent heat is the heat absorbed or released at a fixed temperature during a phase change, quantified by $Q = mL$ , and every multi-stage heating problem is just sensible-heat and latent-heat terms added stage by stage along the heating curve.

The core insight

Latent heat $L$ is the heat per unit mass absorbed or released at constant temperature during a change of state ( $Q = mL$ ), distinct from the sensible heat $Q = mc\,\Delta T$ that changes temperature within a single phase.

Where it fits in the syllabus

Latent heat sits inside the CBSE Class 11 Thermal Properties of Matter chapter, immediately after change of state, and is explicitly prescribed as $Q = mL$ for fusion and vaporisation — one of the chapter's core numerical tools alongside calorimetry ( $Q = mc\,\Delta T$ ). It is examined almost every year in combination with specific heat capacity, since realistic problems (ice at a sub-zero temperature turning into steam, mixtures reaching a common final temperature) require chaining sensible-heat and latent-heat terms together. The chapter's out-of-scope items — Newton's law of cooling as a derived exponential, and any radiation-law derivations — do not intrude here; latent heat stays a clean, self-contained arithmetic topic at this level, with no calculus and no microscopic derivation expected beyond the qualitative bond-breaking picture above.

The physical picture

Every substance, at a fixed pressure, changes state at a fixed temperature: ice always melts at $0^\circ\text{C}$ , water always boils at $100^\circ\text{C}$ (at one atmosphere). Supplying heat to a solid at its melting point, or to a liquid at its boiling point, does not raise the temperature further until the change of state is complete — all of that heat is "spent" on the change of state itself. The heat needed per unit mass for that change is the specific latent heat of the substance for that particular transition.

Q = mL

where: $Q$ is the total heat absorbed (melting, boiling) or released (freezing, condensing), measured in joules; $m$ is the mass undergoing the change of state, in kilograms; and $L$ is the specific latent heat of the substance for that transition, in $\text{J kg}^{-1}$ , a fixed property of the substance and the particular change (fusion or vaporisation) at the given pressure. The physical reading is direct: double the mass of ice you want to melt, and you double the heat required, with no temperature term anywhere in the formula, because temperature simply does not change during the process. Two named latent heats are used throughout the chapter. The latent heat of fusion $L_f$ is the heat needed to convert unit mass of a solid to liquid at its melting point (equally, the heat released when the same mass of liquid freezes back to solid at the same temperature); for ice at $0^\circ\text{C}$ , $L_f = 3.34\times10^5\ \text{J kg}^{-1}$ . The latent heat of vaporisation $L_v$ is the heat needed to convert unit mass of a liquid to gas at its boiling point (equally, the heat released when the same mass of vapour condenses back to liquid); for water at $100^\circ\text{C}$ , $L_v = 22.6\times10^5\ \text{J kg}^{-1}$ . Notice immediately that $L_v$ is nearly seven times $L_f$ for the same substance — vaporisation must completely separate molecules from one another, overcoming almost all of the intermolecular attraction that holds liquid water together, while fusion only needs to loosen the rigid crystal lattice of ice enough that molecules can slide past one another while still touching. The clearest picture of where latent heat sits is the heating curve — a graph of temperature against heat supplied, at a steady rate, to a fixed mass of a substance starting as a solid well below its melting point.

Figure

A temperature versus heat supplied graph for a fixed mass of ice being heated steadily from below zero degrees to steam, showing a rising ramp for solid ice, a flat plateau at zero degrees during melting, a rising ramp for liquid water, a flat plateau at one hundred degrees during boiling, and a final rising ramp for steam, with the two flat plateaus labelled as latent heat regions and the three ramps labelled as sensible heat regions

The two flat plateaus in that graph, at $0^\circ\text{C}$ and $100^\circ\text{C}$ , are where $Q = mL$ applies exclusively — the temperature genuinely does not move no matter how much heat is added, until the phase change is complete. The three rising ramps are where the sensible-heat formula $Q = mc\,\Delta T$ applies instead, with $c$ the specific heat capacity for whichever phase (solid ice, liquid water, or steam) the substance is currently in. A composite problem spanning several of these segments is simply the sum of the correct formula, correctly applied, segment by segment.

Derivation ATotal heat for a multi-stage change from sub-zero ice to steam

The goal here is not to derive $Q = mL$ itself from first principles — at this level it is a defined, measured property of a substance, exactly parallel to how specific heat capacity $c$ is a defined, measured property — but to derive the correct total-heat expression for a substance that starts as a solid below its melting point and ends as a gas above its boiling point, the standard CBSE composite-numerical setup. The subtlety students must get right is recognising exactly how many segments the journey has, and applying $Q = mc\,\Delta T$ or $Q = mL$ correctly to each one without mixing them up.

Step 1. Identify every segment the substance passes through. Starting at temperature $T_1$ below the melting point $T_m$ and ending at temperature $T_2$ above the boiling point $T_b$ , the substance passes through exactly five segments in order: solid warming from $T_1$ to $T_m$ ; melting at $T_m$ ; liquid warming from $T_m$ to $T_b$ ; boiling at $T_b$ ; gas warming from $T_b$ to $T_2$ . Each segment is either a temperature ramp (sensible heat) or a fixed-temperature plateau (latent heat), never both at once — this is the key structural fact that makes the whole calculation additive.

Step 2. Write the heat for each ramp segment using $Q = mc\,\Delta T$ , with the specific heat appropriate to the phase occupying that ramp: $Q_{\text{solid}} = mc_{\text{solid}}(T_m - T_1)$ for warming the solid, $Q_{\text{liquid}} = mc_{\text{liquid}}(T_b - T_m)$ for warming the liquid, and $Q_{\text{gas}} = mc_{\text{gas}}(T_2 - T_b)$ for warming the gas. The mass $m$ stays the same throughout, since no mass is lost, only rearranged in state.

Step 3. Write the heat for each plateau segment using $Q = mL$ , with the latent heat appropriate to that transition: $Q_{\text{fusion}} = mL_f$ for melting at $T_m$ , and $Q_{\text{vap}} = mL_v$ for boiling at $T_b$ . Neither term carries any temperature difference, because by definition the temperature does not change across a phase-change plateau.

Step 4. Add all five terms, because heat supplied in one segment cannot be "recovered" or reused in another — the total heat delivered by the source is simply the sum of what each segment consumed. This gives the general total-heat expression $Q_{\text{total}} = mc_{\text{solid}}(T_m - T_1) + mL_f + mc_{\text{liquid}}(T_b - T_m) + mL_v + mc_{\text{gas}}(T_2 - T_b)$ . Any real problem is this same sum with only the segments that are actually present in that problem's temperature range, since a problem might, for instance, stop at $T_b$ and never include the final gas-warming term. Verified: the five-term sum reduces exactly to the four-stage arithmetic used in Worked example — competitive level below, where the range is $-10^\circ\text{C}$ ice to $100^\circ\text{C}$ steam and the final gas-warming ramp is absent because the journey stops exactly at $T_b$ .

Worked examples

Board level - melting a known mass of ice

Find the total heat required to convert $0.50\ \text{kg}$ of ice at $0^\circ\text{C}$ completely into water at $0^\circ\text{C}$ . Take $L_f = 3.34\times10^5\ \text{J kg}^{-1}$ .

Step 1. Recognise the type of problem. Both the initial and final temperatures given are $0^\circ\text{C}$ , so no sensible heating is involved anywhere in this problem — it is a pure phase-change from solid to liquid, sitting entirely on the melting plateau of the heating curve.

Step 2. Apply $Q = mL_f$ directly, since this is exactly the fusion transition at the melting point.

Q = mL_f = (0.50)(3.34\times10^5)

Step 3. Compute the product.

Q = 1.67\times10^5\ \text{J} = 167\ \text{kJ}

So the heat required is $167\ \text{kJ}$ . Notice this is the entire answer in one line once the problem is correctly recognised as a pure latent-heat case — the most common board-exam mistake here is reaching for $Q = mc\,\Delta T$ out of habit and getting confused when $\Delta T = 0$ makes that formula give zero.

Competitive level - full ice-to-steam conversion with timing pressure

Find the total heat required to convert $0.20\ \text{kg}$ of ice at $-10^\circ\text{C}$ into steam at $100^\circ\text{C}$ . Use $c_{\text{ice}} = 2100\ \text{J kg}^{-1}\text{K}^{-1}$ , $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$ , $L_f = 3.34\times10^5\ \text{J kg}^{-1}$ , $L_v = 22.6\times10^5\ \text{J kg}^{-1}$ .

Step 1. Map the journey onto the heating curve. This range spans four segments, not five, because the problem stops exactly at $100^\circ\text{C}$ steam with no further gas-warming ramp: ice warms from $-10^\circ\text{C}$ to $0^\circ\text{C}$ ; ice melts at $0^\circ\text{C}$ ; water warms from $0^\circ\text{C}$ to $100^\circ\text{C}$ ; water boils at $100^\circ\text{C}$ .

Step 2. Compute the ice-warming ramp.

Q_1 = mc_{\text{ice}}\Delta T = (0.20)(2100)(10) = 4200\ \text{J}

Step 3. Compute the melting plateau.

Q_2 = mL_f = (0.20)(3.34\times10^5) = 66{,}800\ \text{J}

Step 4. Compute the water-warming ramp.

Q_3 = mc_{\text{water}}\Delta T = (0.20)(4186)(100) = 83{,}720\ \text{J}

Step 5. Compute the boiling plateau.

Q_4 = mL_v = (0.20)(22.6\times10^5) = 452{,}000\ \text{J}

Step 6. Add all four segments; do not average or take the largest term alone.

Q = Q_1+Q_2+Q_3+Q_4 = 4200 + 66{,}800 + 83{,}720 + 452{,}000

Q = 606{,}720\ \text{J} \approx 6.07\times10^5\ \text{J}

So the total heat needed is $606.7\ \text{kJ}$ . Notice the boiling step $Q_4$ alone is nearly three-quarters of the whole answer — a fast sanity check in a timed exam is to glance at which term is largest and confirm it is the vaporisation term, since $L_v \gg L_f$ always makes the boiling plateau dominate a full ice-to-steam sum. Timing note: in a competitive exam, do not compute all four terms to full precision and then add — round intermediate products to 3 significant figures as you go ( $4200$ , $66{,}800$ , $83{,}700$ , $452{,}000$ ), sum mentally in tens of kilojoules, and cross-check only the final total; this problem should take under 90 seconds once the four-segment structure is recognised, since each individual term is a single multiplication.

Common traps

Reaching for $Q = mc\,\Delta T$ during an actual phase change — on a plateau, $\Delta T = 0$ by definition, so only $Q = mL$ is valid there; using the wrong formula either gives zero heat (wrong) or silently drops the latent-heat term from a composite answer.
Trying to force a single formula across a multi-stage range instead of breaking the journey into its separate ramp-and-plateau segments — a full ice-to-steam problem has no single formula; it has exactly the sum of the segments actually present.
Swapping or misremembering $L_f$ and $L_v$ , or forgetting that $L_v$ is roughly seven times $L_f$ for water — this is exactly the kind of numerical slip that a quick order-of-magnitude sanity check (boiling should dominate) catches immediately.
Treating evaporation and boiling as the same process. Evaporation happens only at a liquid's surface, at any temperature below the boiling point, and proceeds slowly; boiling happens throughout the liquid's volume at one sharply defined temperature (for a given pressure). Both draw on the same latent heat of vaporisation, but only boiling has that fixed temperature — evaporative cooling of skin, for instance, happens well below $100^\circ\text{C}$ .
Forgetting that latent heat is released, not absorbed, during freezing and condensation — the same $Q = mL$ formula applies with heat flowing out of the substance, which matters in problems about mixtures reaching a common final temperature (calorimetry).

You should be able to…

State the definition of latent heat and explain, in terms of molecular bonds, why it does not change temperature.
Write and apply $Q = mL$ correctly, choosing $L_f$ or $L_v$ as appropriate to the transition.
Identify, on a temperature-versus-heat graph, which segments are sensible heat and which are latent heat.
Break a multi-stage heating problem into its correct sequence of ramp and plateau segments and sum them.
Explain, physically, why $L_v$ is much larger than $L_f$ for the same substance.
Distinguish evaporation from boiling, and explain evaporative cooling in terms of latent heat.

Ask the tutor: want to work through a mixture (calorimetry plus latent heat) problem step by step, or double-check a heating-curve sketch you drew?

Why does the thermometer stay stuck at 0 degrees while ice is melting, even if I keep the flame on high?

Because temperature measures average molecular kinetic energy, and during melting the incoming heat is not going into kinetic energy at all — it is going into breaking the hydrogen bonds that hold water molecules in the rigid ice lattice. Only once every molecule has been freed from that lattice does additional heat start speeding molecules up again, and the thermometer resumes climbing. Turning the flame up higher only makes the melting finish faster; it does not make the thermometer move sooner.

Is the heat used to melt ice actually gone, or can I get it back somehow?

It is not gone, it is stored — as the potential energy of a looser molecular arrangement rather than as kinetic energy. You get it back the moment the process reverses: when water freezes back into ice, that same amount of heat, $mL_f$ , is released back into the surroundings. This is exactly why a box of ice cubes in a cooler keeps drinks cold for so long, and why farmers sometimes spray water on crops before a frost — the water releases latent heat as it freezes, which can protect the crop from dropping below zero degrees.

Why does a steam burn hurt so much more than a boiling-water burn, even though both are at 100 degrees Celsius?

Because steam carries an enormous extra dose of latent heat that liquid water at the same temperature does not. When steam touches your skin and condenses back to water, it dumps its full latent heat of vaporisation, $L_v = 22.6\times10^5\ \text{J kg}^{-1}$ , into your skin before it even starts cooling as liquid water. Gram for gram, that is nearly seven times as much heat as the same mass of boiling water can deliver just by cooling down a little, which is exactly why steam burns are so much more dangerous.

In a mixture problem, like ice dropped into warm water, how do I know whether to use latent heat at all?

Check whether the final state of the ice, once everything settles to a common temperature, is still ice, is fully melted water, or is somewhere in between. If the final temperature works out above 0 degrees Celsius, the ice must have fully melted, so a latent-heat term $mL_f$ belongs in your heat-balance equation alongside the sensible-heat terms. If your equation predicts a final temperature that would require melting more ice than is actually present, that is a signal the final state is actually a solid-liquid mixture sitting at exactly 0 degrees Celsius, and you need to redo the balance with that constraint instead.