ARCHETYPE: Heating Curves & Latent-Heat Numericals

Your kettle is a heating curve you can hear. A $2\ \mathrm{kW}$ element takes $1\ \mathrm{kg}$ of water from $20\ \mathrm{^\circ C}$ to boiling with $334\,400\ \mathrm{J}$ — about $167\ \mathrm{s}$, or $2.8$ minutes, which matches your kitchen experience. To then boil it dry would need the full $2.26\times10^{6}\ \mathrm{J}$: another $1130\ \mathrm{s}$, or $18.8$ minutes, which is $6.76$ times as long as the heating-up took. That ratio is why a kettle boils quickly but a pan of pasta water simmers away for ages, and it is why boil-dry protection exists. The same plateau runs modern power stations. A thermal or nuclear plant boils water to spin a turbine, then must condense the exhaust steam back to water — and condensation gives back every joule of that latent heat, which is why power stations need rivers, cooling towers or the sea to dump it into. Watt's separate condenser was the first engineering exploitation of exactly this plateau, and the principle is unchanged; his own unit for engine output, the horsepower, is $746\ \mathrm{W}$ in the units we now name after him. Refrigeration engineers think in plateaus too. Freezing $1\ \mathrm{kg}$ of water in a food product dumps $334\ \mathrm{kJ}$ into the freezer, while merely chilling it from $20\ \mathrm{^\circ C}$ to $0\ \mathrm{^\circ C}$ dumps only $83.6\ \mathrm{kJ}$ — the latent load is $4.0$ times the sensible one, and it dominates the sizing of every blast freezer. The same arithmetic is why an ice pack outperforms a bottle of chilled water, why "phase-change materials" (waxes that melt near room temperature) are now built into building walls to flatten the daily temperature swing, and why the aircraft de-icing industry exists at all. And in the analytical lab, the instrument called a differential scanning calorimeter is nothing but this article automated: heat a milligram of sample at constant power, watch for a plateau, and read the material's identity and purity straight off the curve.

Geometry first, because in an archetype the geometry is the physics — and here the "geometry" is a graph. A single substance of mass $m$ sits in a container. A heater of constant power $P$ pours energy in at a steady rate: in every second it delivers $P$ joules, no more, no less. A $500\ \mathrm{W}$ immersion heater hands over exactly $500\ \mathrm{J}$ per second, forever, whatever the substance happens to be doing with it. We assume no heat escapes to the surroundings and none is soaked up by the container, so all of it reaches the substance. That single assumption is the master key. Total energy delivered in time $t$ is $Q = Pt$, which is a straight-line relation: the time axis of the graph is the energy axis in disguise, rescaled by $P$. Twice the time, twice the joules. So when you look at a heating curve, you are looking at a picture of where the energy went. Wide segment, lots of energy. Narrow segment, little energy. Nothing else on the graph carries information. Now watch the substance spend its budget. While it is a single phase — all solid, or all liquid, or all gas — the incoming energy speeds the molecules up, the kinetic energy rises, and the thermometer climbs: a ramp. But at a melting or boiling point, the incoming energy is spent pulling molecules apart against their mutual attraction. That raises their potential energy, not their speed. The thermometer, which only senses molecular kinetic energy, sees nothing at all: a plateau, flat as a table, while the heater roars on.

Look hard at that figure, because it contains the four facts that answer most heating-curve questions before you write anything. The ice ramp is a sliver — warming ice from $-20\ \mathrm{^\circ C}$ to $0\ \mathrm{^\circ C}$ costs $42\ \mathrm{kJ}$, a mere $1.4\%$ of the trip. The melting plateau is modest at $334\ \mathrm{kJ}$ ($10.9\%$). The water ramp is comparable to it, $418\ \mathrm{kJ}$ ($13.7\%$) — slightly more than melting, which surprises everybody. And the boiling plateau is a monster: $2260\ \mathrm{kJ}$, or $74.0\%$ of the entire journey from cold ice to hot steam. Everything solid — warming the ice and melting it — is just $12.3\%$ of the story. Two idealizations make the staircase clean, and both deserve an explicit line in a board answer. The power is constant and fully absorbed: no losses to the room, no heat stored in the beaker. The substance is pure, so it melts and boils at a single sharp temperature rather than over a range — that is what makes a plateau perfectly flat rather than gently sloped.

Every heating-curve problem is built from three equations, and you should be able to write them from memory in ten seconds. The first is the definition of power, rearranged: energy delivered equals power multiplied by time. It is the bridge between the clock (what the graph shows) and the joules (what the physics costs).

where: $Q$ — energy delivered, in joules; $P$ — the heater's constant power, in watts (joules per second); $t$ — the time it was on, in seconds. Read it physically: at $500\ \mathrm{W}$, every second of the graph is worth $500\ \mathrm{J}$, so a segment twice as wide swallowed twice the energy. This is what licenses the phrase "the time axis is an energy axis." On a ramp the substance stays in one phase and gets hotter. The energy needed is the sensible-heat formula — mass, specific heat capacity, temperature change:

where: $m$ — mass in kilograms; $c$ — specific heat capacity of that phase in $\mathrm{J\,kg^{-1}K^{-1}}$; $\Delta T$ — the temperature rise. Note the phrase that phase: ice, water and steam have three different values ($c_{\text{ice}} = 2100$, $c_{\text{water}} = 4180$, $c_{\text{steam}} = 2010\ \mathrm{J\,kg^{-1}K^{-1}}$), and using $4180$ on the ice ramp is one of the most common single-mark losses in the chapter. Because $\Delta T$ is a difference, a change of $1\ \mathrm{^\circ C}$ is a change of $1\ \mathrm{K}$: either unit is safe here. On a plateau the substance changes phase at a fixed temperature. The energy is the latent-heat formula, and there is no $\Delta T$ in it at all — because there is no $\Delta T$:

where: $L$ — the specific latent heat of the transition, in $\mathrm{J\,kg^{-1}}$: $L_f = 3.34\times10^{5}$ for fusion (melting or freezing at $0\ \mathrm{^\circ C}$) and $L_v = 2.26\times10^{6}$ for vaporization (boiling or condensing at $100\ \mathrm{^\circ C}$). Read it physically: this is the price of rearranging the molecules rather than speeding them up. The absence of $\Delta T$ in equation (3) is not an oversight — it is the whole point of a plateau.

The three equations above tell you the energy. To turn them into the shape of the curve — which is what an exam graph actually gives you — combine each with $Q = Pt$. Take a ramp first. In a small time $\Delta t$ the heater delivers $P\,\Delta t$, and that energy raises the temperature by $\Delta T$ according to $m c \Delta T = P \Delta t$. Divide, and the gradient of the ramp appears:

where: slope is in kelvin per second. Read it physically: a ramp is steep when the heater is strong, and shallow when there is a lot of substance to heat or when that substance is stubborn (large $c$). Water, with its famously large $c$, gives a lazy ramp; the same mass of ice climbs $4180/2100 = 1.99$ times as fast, almost exactly twice as steep. Rearranged, equation (4) is a measuring instrument: $c = P/(m \times \text{slope})$. Now a plateau. During a phase change the graph is flat, so its length is the observable. Setting $P\,\Delta t = mL$ and solving for the time gives

where: $\Delta t_{\text{plateau}}$ is the width of the flat stretch, in seconds. Read it physically: the plateau is long when there is a lot of substance, when the latent heat is large, or when the heater is feeble. Notice what is absent: $c$ does not appear. The specific heat capacity has no say whatever in how long a plateau lasts — a favourite MCQ trap. Rearranged, this too is a measuring instrument: $L = P\,\Delta t/m$, and that single rearrangement is the entire IGCSE/A-Level practical on latent heat. Those two rearrangements are the ARC skill in one sentence: plateau length gives you $L$; ramp slope gives you $c$. The figure below shows both being read off a real lab curve — $30\ \mathrm{g}$ of ice under a $50\ \mathrm{W}$ heater.

String the segments together and you get the archetype's headline result. To take a mass $m$ of ice from below zero all the way to steam, you pay four separate bills — warm the ice, melt it, warm the water, boil it — and the total is simply their sum, because energy is conserved and each segment is independent of the others.

where: $\Delta T_1$ — the rise from the starting temperature up to $0\ \mathrm{^\circ C}$; $\Delta T_2$ — the full $100\ \mathrm{^\circ C}$ rise of the liquid. Read it physically: it is a shopping list, and the receipt is dominated by one item. At constant power the time for each stage is that stage's energy divided by $P$, so equation (6) doubles as a timetable. If the steam is then superheated above $100\ \mathrm{^\circ C}$, a fifth term $m\,c_{\text{steam}}\,\Delta T_3$ joins the list — for $1\ \mathrm{kg}$ pushed on to $120\ \mathrm{^\circ C}$ that is a further $40\,200\ \mathrm{J}$, small change next to the boiling bill.

Every one of these follows from equations (4) and (5) alone, and each is a ready-made MCQ. Learn the direction of each effect and you can eliminate three options without arithmetic.

The two "no change" rows are where marks are won. A substance with a huge specific heat capacity and a tiny latent heat has lazy ramps and stubby plateaus; a substance with the reverse has steep ramps and vast flat stretches. The graph separates the two constants for you — which is exactly why it is a measuring instrument.

1 · Board level — melt it, then warm it (CBSE/ICSE pattern, 3 marks)

A $500\ \mathrm{W}$ immersion heater is placed in $0.50\ \mathrm{kg}$ of ice at $0\ \mathrm{^\circ C}$. Assuming all the heat reaches the ice, find (a) the time to melt it completely, (b) the further time to warm the melt-water to $50\ \mathrm{^\circ C}$, and (c) the total time. Take $L_f = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$ and $c_{\text{water}} = 4180\ \mathrm{J\,kg^{-1}K^{-1}}$.

Step 1. Identify the segments. The ice is already at its melting point, so there is no solid ramp: the journey is one plateau followed by one liquid ramp. (Spotting the missing segment is the whole point of the "at $0\ \mathrm{^\circ C}$" in the question.)

Step 2. Plateau energy, from equation (3): $Q_1 = mL_f = 0.50 \times 3.34\times10^{5} = 167\,000\ \mathrm{J}$. Time from equation (1): $t_1 = Q_1/P = 167\,000/500 = \boxed{334\ \mathrm{s}}$, i.e. $5\ \mathrm{min}\ 34\ \mathrm{s}$.

Step 3. Ramp energy, from equation (2): $Q_2 = m c_{\text{water}} \Delta T = 0.50 \times 4180 \times 50 = 104\,500\ \mathrm{J}$. Time: $t_2 = 104\,500/500 = \boxed{209\ \mathrm{s}}$, i.e. $3\ \mathrm{min}\ 29\ \mathrm{s}$.

Step 4. Total: $Q = 271\,500\ \mathrm{J}$ and $t = 334 + 209 = \boxed{543\ \mathrm{s}}$ — $9\ \mathrm{min}\ 3\ \mathrm{s}$. Sanity: melting takes $1.60$ times as long as the whole $50\ \mathrm{^\circ C}$ climb, and the ramp's slope is $P/(mc) = 500/(0.5 \times 4180) = 0.239\ \mathrm{K\,s^{-1}}$ — which is just $50\ \mathrm{K}$ over $209\ \mathrm{s}$, as it must be. And if you left the heater on to boil the water away, that plateau would run for $mL_v/P = 2260\ \mathrm{s}$, or $37.7$ minutes: $6.77$ times the melting plateau, exactly the ratio $L_v/L_f$.

2 · Board level — extract the latent heat from a printed curve (IGCSE/A-Level pattern, 4 marks)

A $50\ \mathrm{W}$ heater is switched on inside a $30\ \mathrm{g}$ block of a pure solid at $-20\ \mathrm{^\circ C}$. The temperature–time graph shows a flat region at $0\ \mathrm{^\circ C}$ lasting $200\ \mathrm{s}$; after it, the temperature rises by $20\ \mathrm{^\circ C}$ in $50\ \mathrm{s}$. Find the specific latent heat of fusion and the specific heat capacity of the liquid, and identify the substance.

Step 1. Convert the mass at once: $30\ \mathrm{g} = 0.030\ \mathrm{kg}$. (Leaving it in grams multiplies your latent heat by a thousand and is worth zero marks.)

Step 2. The plateau. Energy supplied during it: $Q = P\Delta t = 50 \times 200 = 10\,000\ \mathrm{J}$. All of it melted the sample, so $L_f = Q/m = 10\,000/0.030 = \boxed{3.33\times10^{5}\ \mathrm{J\,kg^{-1}}}$.

Step 3. The ramp above it. Slope $= 20/50 = 0.40\ \mathrm{K\,s^{-1}}$, so from equation (4), $c = P/(m \times \text{slope}) = 50/(0.030 \times 0.40) = \boxed{4167\ \mathrm{J\,kg^{-1}K^{-1}}} \approx 4.2\times10^{3}\ \mathrm{J\,kg^{-1}K^{-1}}$.

Step 4. Identify. A solid that melts at $0\ \mathrm{^\circ C}$ with $L_f \approx 3.3\times10^{5}\ \mathrm{J\,kg^{-1}}$, whose liquid has $c \approx 4.2\times10^{3}\ \mathrm{J\,kg^{-1}K^{-1}}$: this is water (accepted values $3.34\times10^{5}$ and $4180$, so the experiment is out by $0.20\%$ and $0.32\%$ respectively). The examiner's follow-up is always the same: why is the measured value slightly low? Because a real heater also warms the beaker and loses heat to the room, so not all of $Pt$ reaches the ice — the plateau would have to be longer to deliver the true $mL_f$. State that and take the mark.

3 · Competitive level — the ratio question, no calculator (JEE-Main/NEET pattern, ~1 minute)

A constant-power heater is switched on in a beaker of ice at $0\ \mathrm{^\circ C}$. The ice takes $4.0$ minutes to melt completely. Estimate (a) how long the water then takes to reach $100\ \mathrm{^\circ C}$, and (b) how long it takes to boil away entirely. The mass is unchanged throughout; take $c_{\text{water}} = 4180$, $L_f = 3.34\times10^{5}$, $L_v = 2.26\times10^{6}$.

Step 1. Refuse to compute the mass. Every segment's time is (its energy)/$P$, and $m$ and $P$ are the same in every segment — so they cancel in every ratio. Times are in the ratio of the energies per kilogram: $c\Delta T$ against $L$.

Step 2. Water ramp versus melting plateau: $\dfrac{c_{\text{water}} \times 100}{L_f} = \dfrac{418\,000}{334\,000} = 1.25$. So the $0 \to 100\ \mathrm{^\circ C}$ climb takes $1.25 \times 4.0 = \boxed{5.0\ \text{minutes}}$.

Step 3. Boiling plateau versus melting plateau: $\dfrac{L_v}{L_f} = \dfrac{2.26\times10^{6}}{3.34\times10^{5}} = 6.77$. So boiling it dry takes $6.77 \times 4.0 = \boxed{27.1\ \text{minutes}}$.

Step 4. Total from ice-at-zero to the last wisp of steam: $4.0 + 5.0 + 27.1 = \boxed{36.1\ \text{minutes}}$ — just over nine times the melting plateau. Timing note: the two ratios $1.25$ and $6.77$ are worth committing to memory — with them this is a sixty-second question, and the mass and power never need to be found at all. Students who insist on computing $m$ from the first plateau (impossible: $P$ is not given) burn four minutes and then panic. If an MCQ offers "the boiling plateau is about seven times the melting one," that is your answer.

4 · Competitive level — does the process even finish? (JEE-Advanced flavour, ~4 minutes)

A $1.5\ \mathrm{kW}$ heater is switched on in $0.30\ \mathrm{kg}$ of ice at $-10\ \mathrm{^\circ C}$ and left running for exactly $10.0$ minutes. All the heat is absorbed by the sample. Describe the final state completely: the temperature, and the mass of each phase present. ($c_{\text{ice}} = 2100$, $c_{\text{water}} = 4180$, $L_f = 3.34\times10^{5}$, $L_v = 2.26\times10^{6}$.)

Step 1. Compute the energy budget first — the total the heater can possibly deliver: $Q_{\text{in}} = Pt = 1500 \times 600 = 900\,000\ \mathrm{J}$. Every later step is a question of what this budget can afford.

Step 2. Price each segment of the chain, in order, and keep a running total. Warm the ice: $0.30 \times 2100 \times 10 = 6\,300\ \mathrm{J}$. Melt it: $0.30 \times 3.34\times10^{5} = 100\,200\ \mathrm{J}$. Warm the water to boiling: $0.30 \times 4180 \times 100 = 125\,400\ \mathrm{J}$. Running total to reach water at $100\ \mathrm{^\circ C}$: $231\,900\ \mathrm{J}$ — comfortably affordable, and it happens in $231\,900/1500 = 154.6\ \mathrm{s}$, i.e. $2.58$ minutes.

Step 3. Now test the last segment before assuming it completes. Boiling all $0.30\ \mathrm{kg}$ away would cost $0.30 \times 2.26\times10^{6} = 678\,000\ \mathrm{J}$, but only $900\,000 - 231\,900 = 668\,100\ \mathrm{J}$ of budget remains. Since $668\,100 < 678\,000$, the boiling does not finish — the heater runs out of time on the plateau, and the answer is a mixture pinned at $100\ \mathrm{^\circ C}$.

Step 4. Split the phases with equation (3). The leftover budget vaporizes $m_{\text{vap}} = 668\,100/2.26\times10^{6} = 0.2956\ \mathrm{kg}$. So the final state is $\boxed{100\ \mathrm{^\circ C}}$, with $\boxed{295.6\ \mathrm{g}\ \text{of steam and}\ 4.4\ \mathrm{g}\ \text{of water}}$ still in the beaker.

Step 5. Sanity-check with the clock. The full chain costs $909\,900\ \mathrm{J}$, needing $606.6\ \mathrm{s} = 10.11$ minutes. We ran for $10.0$ minutes — short by $9\,900\ \mathrm{J}$, which is $6.6\ \mathrm{s}$ of heater time. The problem was engineered to stop you $6.6$ seconds from a clean answer, and the segment times ($4.2\ \mathrm{s}$, $66.8\ \mathrm{s}$, $83.6\ \mathrm{s}$, $452\ \mathrm{s}$) show exactly where the time went. Timing note: budget four minutes and spend the first thirty seconds writing the four segment costs in a column. The trap here is the assumption — "ten minutes is loads, it must all be steam" — and it is punished with zero. Whenever a competitive question gives a specific run time or a specific quantity of heat, it is almost always because some segment fails to complete: test the last one explicitly. This "does it finish?" reflex is the same one you use in method-of-mixtures problems when asking whether all the ice melts.

Dropping a ramp from the chain (CBSE/ICSE, 5-mark question). The classic wreck is a three-term answer for a four-term journey — usually the ice ramp, because "it's only $20$ degrees." Write the four segments as a column before computing anything: warm, melt, warm, boil. The $42\,000\ \mathrm{J}$ you forgot is a whole mark.

Using $c_{\text{water}}$ on the ice or steam ramp (all boards). Three phases, three constants: $2100$, $4180$, $2010\ \mathrm{J\,kg^{-1}K^{-1}}$. The liquid value is nearly twice the solid value, so this slip visibly deforms the graph — a water ramp is lazy, an ice ramp is $1.99$ times steeper, and a steam ramp is steeper still ($c_{\text{water}}/c_{\text{steam}} = 2.08$).

Putting a $\Delta T$ into a plateau, or an $L$ into a ramp (all boards). $Q = mL$ has no temperature in it; $Q = mc\Delta T$ has no latent heat in it. If your working contains $mL\Delta T$, you have invented physics. Tag each segment "ramp" or "plateau" on the script and the formulas assign themselves.

Assuming the last segment completes (JEE-Main/NEET/JEE-Advanced). When a question fixes the time, the power, or the heat supplied, suspect that some segment is cut short. Compute the running total, compare it with the budget, and if the budget dies on a plateau, the answer is a mixture at the plateau temperature — as in Example 4, where $295.6\ \mathrm{g}$ of steam coexists with $4.4\ \mathrm{g}$ of water at $100\ \mathrm{^\circ C}$.

Forgetting that the boiling plateau dwarfs the melting one (NEET/JEE-Main MCQ). The ratio is $L_v/L_f = 6.77$. If a graph shows two plateaus of roughly equal length for the same substance and the same heater, the graph is wrong — or the masses differ. This ratio is also the fastest sanity check on any total you compute: boiling should be about three-quarters of the whole ice-to-steam bill ($74.0\%$).

Reading the height of a plateau instead of its length (IGCSE/A-Level). The plateau's height tells you the melting or boiling point; only its width carries the latent heat. Similarly, a ramp's height gives $\Delta T$, but you need its gradient — height over width — before $c$ appears.

Grams, and the missing kilogram (all boards). $L$ is quoted in $\mathrm{J\,kg^{-1}}$ and $c$ in $\mathrm{J\,kg^{-1}K^{-1}}$. Convert masses to kilograms before you start, or your $L_f$ will come out a thousand times too big. Minutes are the same trap: $P$ is in watts, so time must be in seconds.

Treating the time axis as an energy axis when the power is not constant (JEE-Advanced, A-Level). Everything in this article assumes a steady $P$. If a question says the heater is switched off, or the flame is turned down, or the sample loses heat to the room, then $Q = Pt$ applies only to the intervals where $P$ genuinely held constant — do the bookkeeping interval by interval.

Which value should I use? CBSE and NCERT quote $L_f = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$ and $L_v = 2.26\times10^{6}\ \mathrm{J\,kg^{-1}}$; IGCSE and A-Level papers usually print the constants they want on the paper itself. Use the printed values, state the ones you used, and never mix conventions inside one solution.

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) — the CBSE/JEE/NEET statement of $Q = mc\Delta T$, $Q = mL$ and the heating curve · Cambridge IGCSE Physics 0625, Thermal Physics 2.2.3 (measuring specific latent heat with a constant-power heater) · CIE A-Level 9702, Thermal Physics · HC Verma, Concepts of Physics Vol. 2, "Calorimetry" · Halliday, Resnick & Walker, "Temperature, Heat and the First Law of Thermodynamics."

Provenance. All results are standard; the constants are the SI values used throughout this track. A verifier should recompute: the four segments $42\,000 + 334\,000 + 418\,000 + 2\,260\,000 = 3.054\times10^{6}\ \mathrm{J}$ and their shares $1.4\%$, $10.9\%$, $13.7\%$, $74.0\%$; the $2\ \mathrm{kW}$ chain time $1527\ \mathrm{s}$; the $500\ \mathrm{W}$ anchors $334\ \mathrm{s}$ and $209\ \mathrm{s}$ (total $543\ \mathrm{s}$, slope $0.239\ \mathrm{K\,s^{-1}}$, boil-away $2260\ \mathrm{s}$); the graph extraction $L_f = 50 \times 200/0.030 = 3.33\times10^{5}\ \mathrm{J\,kg^{-1}}$ ($0.20\%$ error) and $c = 50/(0.030 \times 0.40) = 4167\ \mathrm{J\,kg^{-1}K^{-1}}$ ($0.32\%$ error); the ratios $L_v/L_f = 6.77$, $(c_{\text{water}} \times 100)/L_f = 1.25$, $L_v/(c_{\text{water}} \times 100) = 5.41$ and $c_{\text{water}}/c_{\text{ice}} = 1.99$; Example 3's $5.0$, $27.1$ and $36.1$ minutes; Example 4's $900\,000\ \mathrm{J}$ budget, $231\,900\ \mathrm{J}$ running total, $668\,100\ \mathrm{J}$ remainder, $295.6\ \mathrm{g}$ of steam and $4.4\ \mathrm{g}$ of water, and the $606.6\ \mathrm{s}$ full-chain time.

Prerequisites. Latent Heat & Change of State · Specific Heat & Heat Capacity. Sibling archetype. ARCHETYPE: Method-of-Mixtures Problems — the same $mc\Delta T$ versus $mL$ bookkeeping, but with two bodies exchanging heat instead of one heater supplying it; the "does the phase change complete?" test is shared. Go deeper (UG). Why a plateau exists at all — latent heat as a jump in entropy between two coexisting phases, and the Clausius–Clapeyron relation that fixes where the plateau sits — is developed in Latent Heat and Phase Coexistence in the university Statistical Mechanics track.