Latent Heat & Change of State

Here is a fact that stops students cold the first time they meet it: hold a beaker of ice-water over a flame, stir, and watch the thermometer. It climbs to $0\ \mathrm{^\circ C}$ and then — while the flame roars and energy pours in — it stops. For minutes the reading refuses to budge, even though heat is plainly flowing. Only once the last sliver of ice vanishes does the temperature start rising again. Where did all that energy go, if not into making things hotter? It went into breaking the ice apart — pulling water molecules out of their rigid crystal and into the free-flowing liquid. Energy that changes a substance's state without changing its temperature is called latent heat, from the Latin for "hidden," because it hides from the thermometer. This one idea explains why steam scalds far worse than boiling water, why a fever is broken by a sweat, why a refrigerator can pump heat out of your food at all, and why the humble equation $Q = mL$ appears in every board's calorimetry section. This article builds the heating curve from scratch, separates the two latent heats you must know cold, and drills the mixed $mc\Delta T + mL$ chains that examiners love — the ones that only balance when you stop assuming heat always raises temperature.

Key takeaways

Latent heat is the energy absorbed or released when a substance changes state (solid↔liquid↔gas) at constant temperature.
The heat for a change of state is $Q = mL$ , where $L$ is the specific latent heat (J kg⁻¹): $L_f$ for fusion (melting/freezing), $L_v$ for vaporisation (boiling/condensing).
For water, $L_f = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$ and $L_v = 2.26\times10^{6}\ \mathrm{J\,kg^{-1}}$ — vaporisation costs about seven times as much as fusion.
On a heating curve, temperature ramps while a single phase warms ( $Q = mc\Delta T$ ) and stays flat during a phase change ( $Q = mL$ ).
During melting or boiling the temperature is constant because the incoming energy breaks intermolecular bonds (raises potential energy), not molecular speed.
Key idea: a change of state swallows or releases energy at constant temperature — so heating problems are ramps ( $mc\Delta T$ ) joined by plateaus ( $mL$ ), and you must account for both.

The core insight

Latent heat is the energy a substance absorbs or releases to change state at constant temperature: $Q = mL$ , with $L_f$ for melting/freezing and $L_v$ for boiling/condensing; on a temperature–time graph it is the flat plateau between the sloping ramps.

Where this lives today

Latent heat runs the machinery of everyday life and the planet. A refrigerator pumps heat out of your food by letting a liquid coolant evaporate inside (absorbing $L_v$ ) and condense outside (releasing it) — the same trick your body uses when sweat evaporates to cool you on a hot day. Steam heating and power stations exploit the enormous $L_v$ of water to carry huge energy in a small flow. On the largest scale, the water cycle is a planet-sized latent-heat engine: sunshine evaporates the oceans (storing energy), and when that vapour condenses into clouds and rain it dumps that same energy high in the atmosphere — the power source of every thunderstorm and hurricane. And the notorious steam burn is the same $L_v$ in reverse: condensing on skin, each gram of steam delivers its full latent heat before it even starts to cool.

The physical picture

To follow a substance from cold solid to hot gas, watch its molecules. In a solid they are locked in a lattice, vibrating in place. Adding heat makes them vibrate harder — the temperature rises — until the vibration is violent enough to shake the lattice apart. At that point every extra joule goes not into faster vibration but into tearing molecules loose from their neighbours: the solid melts at constant temperature. The energy is stored as molecular potential energy (bonds pulled apart), invisible to a thermometer, which only senses molecular kinetic energy. Keep heating the liquid and its molecules speed up again (temperature rises) until, at the boiling point, they have enough energy to break free of the liquid entirely — and again the temperature stalls while that liberation is paid for.

Figure

The heating curve of water — ramps where one phase warms, plateaus where a phase changes at constant temperature

The two flat stretches are the latent heats, and they are not equal — the boiling plateau is far longer than the melting one. Melting only has to loosen the lattice into a liquid where molecules still touch; boiling must pull them completely apart into a gas, against the full attraction between them and doing work to push back the atmosphere. That is why, for water, vaporisation costs roughly seven times as much energy as fusion — a ratio you can feel every time a pot takes seconds to reach the boil but minutes to boil dry.

Figure

Melting loosens the lattice; boiling separates molecules completely — so latent heat of vaporisation greatly exceeds latent heat of fusion

The latent-heat equation

Joseph Black's ice experiments established the basic law: the heat a change of state swallows scales with the amount of substance changing. Double the ice, double the energy needed to melt it; the temperature of the transition never enters. The energy to change the state of a mass $m$ is therefore simply proportional to that mass:

Q = mL \qquad (1)

where: $Q$ — heat absorbed (melting/boiling) or released (freezing/condensing), in joules; $m$ — mass changing state (kg); $L$ — the specific latent heat of that transition (J kg⁻¹). There is no $\Delta T$ in the formula, and that absence is the whole point: the transition happens at a single temperature. Use $L_f$ , the specific latent heat of fusion, for melting and freezing, and $L_v$ , the specific latent heat of vaporisation, for boiling and condensing. For water at standard pressure,

L_f = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}, \qquad L_v = 2.26\times10^{6}\ \mathrm{J\,kg^{-1}} \qquad (2)

Freezing and condensing release exactly the same latent heat that melting and boiling absorb — which is why steam heating systems work and why frost forming on crops actually warms the air around them slightly.

Ramps plus plateaus: costing the whole journey

Real problems rarely stop at a single melt: they march a substance across several warming stages and several changes of state, then ask for the total energy. The rule is pure bookkeeping — one ramp term for every stretch where a single phase warms, one plateau term for every change of state, all summed:

Q_{\text{total}} = \sum mc\,\Delta T + \sum mL \qquad (3)

where: each ramp takes the specific heat capacity $c$ of the phase actually warming — $c_{\text{ice}} = 2100\ \mathrm{J\,kg^{-1}\,K^{-1}}$ while it is ice, $c_w = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$ once it is liquid water — and each plateau takes the latent heat $L$ of the transition being crossed. Because $c_{\text{ice}}$ is about half of $c_w$ , the same heater drives the ice ramp up roughly twice as steeply as the water ramp — a slope change the heating-curve questions love to probe.

Derivation AMelting ice, then warming the melt — a two-stage energy chain

We compute the total heat to turn $200\ \mathrm{g}$ of ice at $0\ \mathrm{^\circ C}$ into water at $20\ \mathrm{^\circ C}$ . The point is to show why a single formula never suffices once a phase change is involved: the journey splits into a plateau (melting at constant temperature) followed by a ramp (warming the liquid), and the two contributions must be summed. Watch how the constant-temperature melting step uses $mL$ with no $\Delta T$ , while the warming step uses $mc\Delta T$ with no phase change.

Step 1. Melt the ice at $0\ \mathrm{^\circ C}$ (a plateau): $Q_1 = mL_f = 0.200\times3.34\times10^{5} = 66\,800\ \mathrm{J}$ . Why no $\Delta T$ : the temperature is pinned at $0\ \mathrm{^\circ C}$ throughout melting; energy goes into breaking the crystal, not into warming.

Step 2. Warm the resulting water from $0\ \mathrm{^\circ C}$ to $20\ \mathrm{^\circ C}$ (a ramp), with $c_w = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$ : $Q_2 = mc_w\Delta T = 0.200\times4180\times20 = 16\,720\ \mathrm{J}$ .

Step 3. Add the stages: $Q = Q_1 + Q_2 = 66\,800 + 16\,720 = 83\,520\ \mathrm{J}$ .

Step 4. Notice the split: melting the ice cost four times as much as warming the melt by a full $20\ \mathrm{^\circ C}$ . Latent heat dominates — which is exactly why ice is such an effective coolant, and why treating the whole process as one $mc\Delta T$ (a common slip) undercounts the energy grossly. Verified: $0.200\times3.34\times10^{5} = 66\,800$ ; $0.200\times4180\times20 = 16\,720$ ; sum $= 83\,520\ \mathrm{J}$ in verify_hs2-latent-heat-change-of-state.py.

Go deeper · Derivation B · 1 minWhy the thermometer stops — the energy ledger of a plateau

Derivation BWhy the thermometer stops — the energy ledger of a plateau

We justify the plateau itself: why must the temperature hold constant while ice melts, however fiercely we heat? The argument is an energy ledger. Temperature tracks the average kinetic energy of the molecules, so if added heat were becoming speed the reading would climb; we show the numbers force every joule supplied during the change of state to a different destination — potential energy of separation.

Step 1. Feed heat into melting ice and track one gram. Each gram converted from crystal to liquid absorbs $L_f = 334\ \mathrm{J}$ — yet the thermometer stays pinned at $0\ \mathrm{^\circ C}$ .

Step 2. Test the alternative. Had that $334\ \mathrm{J}$ gone into kinetic energy, it would have warmed the gram of water by $\Delta T = L_f/c_w = 334/4.18 \approx 80\ \mathrm{K}$ — an enormous, unmissable rise. The stubborn zero on the thermometer proves the energy is not becoming speed.

Step 3. Follow the joules to their real home: they are spent pulling molecules out of the lattice against intermolecular attraction — raising molecular potential energy while the average kinetic energy, and hence $T$ , is unchanged. Any parcel of water that runs momentarily hot shares its excess with the ice still mixed through it: coexisting phases pin the system at the transition temperature.

Step 4. The plateau ends the instant the last crystal is gone: with no bonds left to break, incoming energy has nowhere to go but kinetic, and the temperature resumes its climb. Constant $T$ during a change of state is not an exam convention — it is energy conservation plus the two destinations energy can take. Verified: $334/4.18 = 79.9 \approx 80\ \mathrm{K}$ recomputed in verify_hs2-latent-heat-change-of-state.py.

Limiting cases

Case	Result	Meaning
exactly at the melting/boiling point	temperature constant, $Q = mL$	all energy goes to the phase change, none to warming
latent heat $L \to 0$ (hypothetical)	curve has no plateau	a substance that changed state "for free" would jump phases instantly
tiny mass changing state	tiny $Q$	latent heat scales with mass, like every heat term
freezing / condensing	$Q$ released $= mL$	the reverse transition gives back exactly what melting/boiling took
$L_v$ vs $L_f$ (water)	$L_v/L_f \approx 6.8$	separating molecules fully costs far more than merely loosening them

Common trap“The temperature rises steadily as you heat ice into steam”

It does not — and this is the single most-tested misconception in the chapter. A temperature–time graph from ice to steam is not a straight climb; it is two ramps interrupted by two flat plateaus. While ice melts and while water boils, the thermometer holds dead steady even though heat keeps arriving, because that energy is breaking bonds, not increasing molecular speed. Students who "know" heat always raises temperature will forget the $mL$ plateaus and undercount the energy — sometimes by a factor of several, since the boiling plateau alone can dwarf every ramp combined. Whenever a problem crosses $0\ \mathrm{^\circ C}$ or $100\ \mathrm{^\circ C}$ , stop and ask: is there a phase change hiding here that needs its own $mL$ term?

Worked examples

Board level — ice into warm water (CBSE/ICSE calorimetry pattern)

How much heat is needed to convert $0.50\ \mathrm{kg}$ of ice at $0\ \mathrm{^\circ C}$ into water at $0\ \mathrm{^\circ C}$ ? Then, separately, what is the least mass of ice at $0\ \mathrm{^\circ C}$ that could be melted by $1.0\times10^{5}\ \mathrm{J}$ ? ( $L_f = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$ .)

Step 1. Pure melting at constant temperature uses $Q = mL_f$ (no $\Delta T$ ). For the first part, $Q = 0.50\times3.34\times10^{5} = \boxed{1.67\times10^{5}\ \mathrm{J}}$ (167 kJ).

Step 2. For the second part, invert the same formula: $m = Q/L_f = (1.0\times10^{5})/(3.34\times10^{5}) = \boxed{0.30\ \mathrm{kg}}$ .

Step 3. Sanity: melting half a kilogram of ice takes as much energy as heating that half-kilogram of water by $80\ \mathrm{^\circ C}$ — proof of how "hidden" this heat is. State on the script: "Melting occurs at constant temperature, so only $mL_f$ applies, no $mc\Delta T$ ."

Board level — the complete ice-to-steam chain (CBSE four-segment classic)

Find the total heat needed to turn $1.0\ \mathrm{kg}$ of ice at $-10\ \mathrm{^\circ C}$ all the way into steam at $100\ \mathrm{^\circ C}$ . Take $c_{\text{ice}} = 2100$ and $c_w = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$ , $L_f = 3.34\times10^{5}$ and $L_v = 2.26\times10^{6}\ \mathrm{J\,kg^{-1}}$ . The journey crosses two phase changes, so the heating curve has four pieces — two ramps and two plateaus — and each needs its own term. Take them in order and add.

Step 1. Warm the ice from $-10\ \mathrm{^\circ C}$ up to its melting point at $0\ \mathrm{^\circ C}$ — a ramp, using the specific heat of ice: $Q_1 = mc_{\text{ice}}\Delta T = 1.0\times2100\times10 = 21\,000\ \mathrm{J}$ .

Step 2. Melt all of it at $0\ \mathrm{^\circ C}$ — a plateau, no $\Delta T$ : $Q_2 = mL_f = 1.0\times3.34\times10^{5} = 334\,000\ \mathrm{J}$ .

Step 3. Warm the melt-water from $0\ \mathrm{^\circ C}$ to $100\ \mathrm{^\circ C}$ — a ramp, now using the specific heat of water: $Q_3 = mc_w\Delta T = 1.0\times4180\times100 = 418\,000\ \mathrm{J}$ .

Step 4. Boil it all to steam at $100\ \mathrm{^\circ C}$ — the long plateau, no $\Delta T$ : $Q_4 = mL_v = 1.0\times2.26\times10^{6} = 2\,260\,000\ \mathrm{J}$ . Summing the four segments, $Q = 21\,000 + 334\,000 + 418\,000 + 2\,260\,000 = \boxed{3.033\times10^{6}\ \mathrm{J}}$ . The single boiling term ( $2.26\ \mathrm{MJ}$ ) is about three-quarters of the entire bill — more than the other three segments put together. Any answer that skips a plateau, or reuses $c_w$ for the ice ramp, lands far from $3.033\ \mathrm{MJ}$ .

Competitive level — steam to warm a beaker (JEE-Main pattern)

Dry steam at $100\ \mathrm{^\circ C}$ is passed into $200\ \mathrm{g}$ of water at $20\ \mathrm{^\circ C}$ until the water reaches $40\ \mathrm{^\circ C}$ . What mass of steam condenses? ( $L_v = 2.26\times10^{6}\ \mathrm{J\,kg^{-1}}$ , $c_w = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$ .)

Step 1. Heat gained by the beaker water: $Q_{\text{gain}} = 0.200\times4180\times(40-20) = 16\,720\ \mathrm{J}$ .

Step 2. Each kilogram of steam gives up energy in two stages: it first condenses at $100\ \mathrm{^\circ C}$ ( $L_v$ ), then the resulting hot water cools from $100\ \mathrm{^\circ C}$ to the final $40\ \mathrm{^\circ C}$ ( $c_w\Delta T$ ). Energy released per kg $= L_v + c_w(100-40) = 2.26\times10^{6} + 4180\times60 = 2.26\times10^{6} + 2.508\times10^{5} = 2.511\times10^{6}\ \mathrm{J\,kg^{-1}}$ .

Step 3. Set released = gained: $m_{\text{steam}} = 16\,720 / (2.511\times10^{6}) = \boxed{6.66\times10^{-3}\ \mathrm{kg} = 6.66\ \mathrm{g}}$ .

Step 4. The lesson: a mere $6.7\ \mathrm{g}$ of steam — a wisp — delivers as much heat as $200\ \mathrm{g}$ of water losing $20\ \mathrm{^\circ C}$ , almost all of it from the condensation latent heat. Forgetting the $L_v$ term (counting only the cooling) would demand ten times more steam and lose the question. Timing note: about 2.5 minutes. The trap is treating the steam as if it only cools; the dominant term is the condensation $L_v$ . Write both stages explicitly before dividing.

Exam traps

Dropping the $mL$ plateau. Any process crossing $0\ \mathrm{^\circ C}$ or $100\ \mathrm{^\circ C}$ hides a latent-heat term. Forgetting it is the number-one calorimetry error — and because $L_v$ is huge, the answer is not just slightly wrong but wildly so.

Sticking a $\Delta T$ onto a phase change. During melting or boiling the temperature is constant; $Q = mL$ has no $\Delta T$ . Writing $mL\Delta T$ is dimensional nonsense that markers spot instantly.

Confusing $L_f$ and $L_v$ . Fusion for the solid↔liquid step, vaporisation for the liquid↔gas step. Their values differ by nearly an order of magnitude, so a swap is catastrophic.

Using water's $c$ on the ice ramp. Each ramp takes the specific heat of the phase that is warming: $c_{\text{ice}} = 2100$ , about half of $c_w = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$ . Grabbing $4180$ for a below-zero ice segment inflates that term by a factor of two — a silent error that survives into the final answer.

Units of $L$ . Specific latent heat is J kg⁻¹ (energy per unit mass), not J kg⁻¹ K⁻¹ — there is no per-kelvin, precisely because temperature does not change. Some books quote J g⁻¹; convert carefully ( $334\ \mathrm{J\,g^{-1}} = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$ ).

Ignoring direction. Freezing and condensing release latent heat; melting and boiling absorb it. In a mixture problem, get the signs of gained/released right or the balance fails.

Which values? Boards quote $L_f$ as $334$ or $336\ \mathrm{kJ\,kg^{-1}}$ and $L_v$ as $2256$ , $2260$ or $2.26\times10^{3}\ \mathrm{kJ\,kg^{-1}}$ ; NEET/CBSE also use calorie forms ( $80\ \mathrm{cal\,g^{-1}}$ for fusion, $540\ \mathrm{cal\,g^{-1}}$ for vaporisation). Use exactly what the paper supplies and keep one system throughout.

Origins · 1 minÉmile Clapeyron and how this was discovered

ÉC

Émile Clapeyron

1799–1864 · Paris

Émile Clapeyron was the French engineer who turned Sadi Carnot's abstract reasoning about heat into working equations — and in doing so found the law that governs latent heat. In 1834 he wrote down the relation, set on firm thermodynamic footing by Clausius in 1850, that links the latent heat of a phase change to the slope of the phase boundary on a pressure–temperature diagram. In plain terms: raise the pressure and molecules need more energy to escape, so water boils above $100\ \mathrm{^\circ C}$ (a pressure cooker); lower it and water boils cooler — around $70\ \mathrm{^\circ C}$ near Everest's summit. His equation ties the whole phase diagram — melting, boiling, sublimation — to the single quantity $L$ , and remains the daily tool of refrigeration and steam engineers.

History

1761

— Joseph Black identifies "latent heat," measuring the hidden energy that melts ice and boils water at constant temperature.

1765

— James Watt, who had measured steam's latent heat himself, exploits it in his separate-condenser engine, launching the steam age.

1834

— Émile Clapeyron formulates the pressure–temperature relation later named Clausius–Clapeyron, linking latent heat to the phase boundary.

1850

— Rudolf Clausius places the relation on a firm thermodynamic footing within the emerging science of heat.

1900

s — latent heat becomes the basis of refrigeration, steam power and, later, climate science (the vast heat stored and released by evaporating and condensing water).

Going further

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) — the CBSE/JEE statement of latent heat and calorimetry · Cambridge IGCSE Physics 0625, Thermal Physics 2.2.3 (specific latent heat) · CIE A-Level 9702 (specific latent heat of fusion and vaporisation) · Halliday, Resnick & Walker, "Temperature, Heat and the First Law of Thermodynamics."

Provenance. Standard results and tabulated constants. A verifier should recompute: $0.50\times3.34\times10^{5} = 1.67\times10^{5}\ \mathrm{J}$ ; $m = 1.0\times10^{5}/3.34\times10^{5} = 0.30\ \mathrm{kg}$ ; $0.200\times3.34\times10^{5} + 0.200\times4180\times20 = 83\,520\ \mathrm{J}$ ; full chain $2100\times10 + 3.34\times10^{5} + 4180\times100 + 2.26\times10^{6} = 3.033\times10^{6}\ \mathrm{J}$ with boiling share $2.26\times10^{6}/3.033\times10^{6} = 74.5\%$ ; ratio $L_v/L_f = 6.77$ ; steam mass $16\,720/(2.26\times10^{6}+4180\times60) = 6.66\times10^{-3}\ \mathrm{kg}$ ; plateau ledger $334/4.18 = 79.9\ \mathrm{K}$ ; steam-burn gram $0.001\times4180\times63 = 263\ \mathrm{J}$ of cooling vs $2260 + 263 = 2523\ \mathrm{J}$ for condensing steam ( $\approx 9.6\times$ ); calorie forms $3.34\times10^{5}/4186 \approx 80$ and $2.26\times10^{6}/4186 \approx 540\ \mathrm{cal\,g^{-1}}$ .

Prerequisites. Specific Heat & Heat Capacity · Calorimetry & the Method of Mixtures. Leads to. Evaporation, Boiling & Melting: The Molecular View. Go deeper (UG). The thermodynamics of phase changes — latent heat as a discontinuity in entropy, and the full Clausius–Clapeyron relation — is developed in Latent Heat and Phase Coexistence in the university Statistical Mechanics track.

Why does the temperature stay constant during melting and boiling?

Because the incoming energy is spent breaking intermolecular bonds, not speeding molecules up. Temperature measures the average kinetic energy of the molecules; during a phase change the added energy becomes potential energy (molecules pulled apart) instead. So the thermometer — which senses only kinetic energy — holds steady until the change of state is complete, after which further heating once again raises the speed and the temperature climbs.

Where does the latent heat "go"? Is energy conserved?

Absolutely conserved. The latent heat is stored in the substance as increased molecular potential energy — the molecules are now farther apart, having been pulled against their mutual attraction. You get every joule back on the reverse change: freezing and condensing release exactly the latent heat that melting and boiling absorbed. Nothing is lost; it is just temporarily "hidden" from the thermometer, which is precisely what "latent" means.

Why is latent heat of vaporisation so much bigger than fusion?

Melting only has to jostle molecules out of their fixed lattice positions into a liquid, where they are still packed close and still strongly attracting one another. Boiling has to do far more: pull each molecule completely free of all its neighbours against the full intermolecular attraction, and also push back the surrounding atmosphere to make room for the gas. Both effects demand a lot of energy, so for water $L_v$ is about 6.8 times $L_f$ .

Why does steam at 100 °C burn far worse than water at 100 °C?

They are at the same temperature, so the difference is pure latent heat. A gram of boiling water landing on skin (near $37\ \mathrm{^\circ C}$ ) can release at most $mc\Delta T \approx 0.001\times4180\times63 \approx 263\ \mathrm{J}$ as it cools. A gram of steam must first condense, surrendering $mL_v = 0.001\times2.26\times10^{6} = 2260\ \mathrm{J}$ — and only then cools like hot water, delivering about $2520\ \mathrm{J}$ in all: nearly ten times the energy ( $\approx 9.6\times$ ) to the same patch of skin. That factor is $L_v$ at work, and it is why steam scalds are treated as far more serious than splashes of boiling water.

If I keep heating water that is boiling, does it get hotter?

No — not while it is boiling at a fixed pressure. Once water reaches its boiling point, any extra heat goes into converting liquid to vapour at constant temperature; the thermometer stays at about $100\ \mathrm{^\circ C}$ and the water simply boils away faster. To make the steam hotter (superheated steam) you must first boil all the liquid off, or trap it under higher pressure — which raises the boiling point itself.

Does water always boil at exactly 100 °C?

Only at standard atmospheric pressure. Boiling happens when the liquid's molecules can push out against the surrounding pressure, so the boiling point depends on that pressure. In a pressure cooker (higher pressure) water boils above $100\ \mathrm{^\circ C}$ , cooking food faster; high on a mountain (lower pressure) it boils below $100\ \mathrm{^\circ C}$ , which is why food takes longer to cook there. This is exactly what Clapeyron's equation describes — the phase boundary between liquid and gas has a definite slope on a pressure–temperature graph.

In a calorimetry problem, how do I know when to include a latent-heat term?

Track the temperature journey and flag every crossing of a melting or boiling point. If the process starts as ice below $0\ \mathrm{^\circ C}$ and ends as water, it must pass through melting at $0\ \mathrm{^\circ C}$ — insert an $mL_f$ term there. If it reaches or crosses $100\ \mathrm{^\circ C}$ and vapour forms, insert $mL_v$ . Draw the heating curve mentally: every plateau you cross is an $mL$ term, every slope is an $mc\Delta T$ term, and the total heat is their sum.

Are the melting point and freezing point the same temperature?

Yes, for a pure substance at a given pressure they are identical — water melts and freezes at $0\ \mathrm{^\circ C}$ . The difference is only the direction of energy flow: melting absorbs the latent heat of fusion, freezing releases the very same amount. The transition temperature is a property of the substance; whether energy goes in or comes out just tells you which way the change is running.