Phase Changes & Heating Curves
Learn to read a heating curve segment by segment: use q = mc times change in temperature on the sloped parts, and q = m times heat of fusion or vaporization on the flat plateaus.
What you'll do in this lesson
A voice-first session with the Crimsora tutor on Phase Changes & Heating Curves, then targeted practice and FRQs — with the tutor adapting to where you get stuck.
What this lesson covers
Put a beaker of ice on a hot plate and take the temperature every thirty seconds. At first the reading climbs steadily. Then, right at 0 °C, it stops climbing — the thermometer sits frozen at zero for minutes while the burner keeps pouring in energy. Only after the last crystal disappears does the temperature start rising again.
That stubborn plateau is the heart of this lesson. Energy is always flowing in, but it isn't always raising the temperature. Sometimes it is speeding molecules up (temperature rises), and sometimes it is prying them apart (temperature holds steady while the phase changes). A heating curve is the graph that shows both behaviors at once, and each part of the curve gets its own equation. By the end you'll be able to look at any segment, decide which formula applies, and add the pieces to find the total energy needed to take a substance from solid to gas.
That stubborn plateau is the heart of this lesson. Energy is always flowing in, but it isn't always raising the temperature. Sometimes it is speeding molecules up (temperature rises), and sometimes it is prying them apart (temperature holds steady while the phase changes). A heating curve is the graph that shows both behaviors at once, and each part of the curve gets its own equation. By the end you'll be able to look at any segment, decide which formula applies, and add the pieces to find the total energy needed to take a substance from solid to gas.
Reading the Shape of a Heating Curve
A heating curve plots temperature (y-axis) against energy added or time of steady heating (x-axis). For a pure substance heated at a constant rate, the graph has a distinctive staircase shape: sloped rises separated by horizontal plateaus.
Water from below freezing to above boiling gives five segments:
Two details are worth noticing. First, the sloped segments have different steepnesses. Ice warms faster than liquid water for the same energy input because ice has a smaller specific heat ( J/g·°C versus J/g·°C). A smaller means a steeper slope.
Second, the boiling plateau is much longer than the melting plateau. Vaporizing water takes about J/g while melting it takes only about J/g — nearly seven times more energy. If a graph shows a melting plateau longer than the boiling plateau for a normal substance, something is off with the reading.
During a plateau, both phases are present at once. At the melting plateau you have a slushy mixture of ice and water; at the left edge it is all solid, at the right edge all liquid.
Water from below freezing to above boiling gives five segments:
| Segment | What's happening | Temperature | Equation |
|---|---|---|---|
| 1 | Solid ice warms | rises to 0 °C | |
| 2 | Ice melts | flat at 0 °C | |
| 3 | Liquid water warms | rises to 100 °C | |
| 4 | Water boils | flat at 100 °C | |
| 5 | Steam warms | rises above 100 °C |
Second, the boiling plateau is much longer than the melting plateau. Vaporizing water takes about J/g while melting it takes only about J/g — nearly seven times more energy. If a graph shows a melting plateau longer than the boiling plateau for a normal substance, something is off with the reading.
During a plateau, both phases are present at once. At the melting plateau you have a slushy mixture of ice and water; at the left edge it is all solid, at the right edge all liquid.
Why the Temperature Stalls
Temperature measures the average kinetic energy of particles — how fast they move. But the energy you add to a substance can go two places: into kinetic energy (faster motion, higher temperature) or into potential energy (overcoming the attractions holding particles together).
On a sloped segment, added energy speeds particles up. Kinetic energy rises, so the thermometer rises.
On a plateau, every joule goes into breaking intermolecular attractions. Molecules in ice sit in a rigid hydrogen-bonded lattice; to become liquid they must escape those fixed positions, and that costs energy without making them move any faster on average. Potential energy rises, kinetic energy stays constant, and the temperature holds still. Boiling is the same idea taken further — the molecules must escape their neighbors entirely, so vaporization costs far more than melting.
A very common misconception is that a plateau means no energy is being absorbed. The opposite is true: the plateau is where the substance is soaking up the most energy per gram. Another frequent error is saying the molecules "stop moving" during melting. They keep moving at the same average speed; they simply rearrange.
This also explains why an ice-water drink stays at 0 °C until the ice is gone. Heat leaking in from the room is consumed melting ice rather than warming liquid. Once the last ice melts, the temperature climbs immediately. The same logic explains why steam burns are worse than hot-water burns: condensing steam releases about J per gram onto your skin before the liquid even begins to cool.
On a sloped segment, added energy speeds particles up. Kinetic energy rises, so the thermometer rises.
On a plateau, every joule goes into breaking intermolecular attractions. Molecules in ice sit in a rigid hydrogen-bonded lattice; to become liquid they must escape those fixed positions, and that costs energy without making them move any faster on average. Potential energy rises, kinetic energy stays constant, and the temperature holds still. Boiling is the same idea taken further — the molecules must escape their neighbors entirely, so vaporization costs far more than melting.
A very common misconception is that a plateau means no energy is being absorbed. The opposite is true: the plateau is where the substance is soaking up the most energy per gram. Another frequent error is saying the molecules "stop moving" during melting. They keep moving at the same average speed; they simply rearrange.
This also explains why an ice-water drink stays at 0 °C until the ice is gone. Heat leaking in from the room is consumed melting ice rather than warming liquid. Once the last ice melts, the temperature climbs immediately. The same logic explains why steam burns are worse than hot-water burns: condensing steam releases about J per gram onto your skin before the liquid even begins to cool.
Choosing and Using the Two Equations
Everything on a heating curve comes down to one decision: is the temperature changing, or is the phase changing?
If temperature changes, usewhere is mass in grams, is the specific heat of that particular phase, and in °C (or K — the size of a degree is the same).
If the phase changes, usewhere is the heat of fusion (melting or freezing) and is the heat of vaporization (boiling or condensing). There is no in this equation, because is zero — writing for a plateau gives , which is the single most common mistake in this unit.
Three cautions. First, each phase has its own specific heat; using J/g·°C to warm ice or steam is wrong. Second, values come in two flavors: per gram (J/g) and per mole (kJ/mol, such as kJ/mol for fusion of water). Match the unit to whether you have grams or moles, and convert if needed. Third, watch signs and direction. Absorbing energy (melting, boiling, warming) gives positive ; releasing energy (freezing, condensing, cooling) gives negative of the same magnitude. Freezing 10 g of water releases exactly as much energy as melting 10 g absorbs.
If temperature changes, usewhere is mass in grams, is the specific heat of that particular phase, and in °C (or K — the size of a degree is the same).
If the phase changes, usewhere is the heat of fusion (melting or freezing) and is the heat of vaporization (boiling or condensing). There is no in this equation, because is zero — writing for a plateau gives , which is the single most common mistake in this unit.
Three cautions. First, each phase has its own specific heat; using J/g·°C to warm ice or steam is wrong. Second, values come in two flavors: per gram (J/g) and per mole (kJ/mol, such as kJ/mol for fusion of water). Match the unit to whether you have grams or moles, and convert if needed. Third, watch signs and direction. Absorbing energy (melting, boiling, warming) gives positive ; releasing energy (freezing, condensing, cooling) gives negative of the same magnitude. Freezing 10 g of water releases exactly as much energy as melting 10 g absorbs.
Multi-Step Problems and Where Students Slip
Any problem that crosses a phase change must be broken into segments, and each segment gets its own calculation. Total energy is the sum:A reliable procedure: sketch the curve, mark the starting and ending temperatures, list every melting or boiling point crossed in between, then number the segments. If you go from °C ice to 50 °C water, you cross only 0 °C, so there are three steps: warm the ice, melt it, warm the liquid. You do not include a boiling step, because you never reach 100 °C.
Where students actually go wrong:
Stopping too early or going too far — adding a vaporization term for a problem that ends at 60 °C, or forgetting the final warming step after melting.
Using the wrong for a segment. Warming ice from °C runs only to 0 °C, so , not 60. Each segment ends at the phase-change temperature, not at the final temperature of the whole problem.
Mixing units. Specific heat gives joules while some heat-of-fusion tables give kilojoules. Convert everything to joules before adding, then report a sensible unit at the end.
Forgetting that mass stays the same throughout. The same appears in every term, so you can sometimes factor it out.
A cooling curve is simply this process run backward: gas cools, condenses at a plateau, liquid cools, freezes at a plateau, solid cools. The magnitudes are identical; only the sign of flips.
Where students actually go wrong:
Stopping too early or going too far — adding a vaporization term for a problem that ends at 60 °C, or forgetting the final warming step after melting.
Using the wrong for a segment. Warming ice from °C runs only to 0 °C, so , not 60. Each segment ends at the phase-change temperature, not at the final temperature of the whole problem.
Mixing units. Specific heat gives joules while some heat-of-fusion tables give kilojoules. Convert everything to joules before adding, then report a sensible unit at the end.
Forgetting that mass stays the same throughout. The same appears in every term, so you can sometimes factor it out.
A cooling curve is simply this process run backward: gas cools, condenses at a plateau, liquid cools, freezes at a plateau, solid cools. The magnitudes are identical; only the sign of flips.
Key terms
- Heating curve.
- A graph of temperature versus energy added (or time at constant heating) for a substance, showing sloped warming segments separated by flat phase-change plateaus.
- Specific heat capacity (c).
- The energy required to raise the temperature of 1 gram of a substance by 1 °C. Each phase has its own value; for water it is about 4.18 J/g·°C liquid, 2.09 J/g·°C ice, 2.03 J/g·°C steam.
- Heat of fusion (ΔH_fus).
- The energy needed to melt one gram (or one mole) of a solid at its melting point with no temperature change. For water, about 334 J/g or 6.02 kJ/mol.
- Heat of vaporization (ΔH_vap).
- The energy needed to boil one gram (or one mole) of a liquid at its boiling point with no temperature change. For water, about 2260 J/g or 40.7 kJ/mol.
- Plateau.
- A horizontal section of a heating curve where two phases coexist and all added energy goes into changing phase rather than raising temperature.
- Kinetic vs. potential energy.
- Kinetic energy is particle motion and determines temperature; potential energy is stored in the attractions between particles and increases during melting and boiling.
- Cooling curve.
- The reverse of a heating curve, showing temperature dropping with plateaus at the freezing and condensation points, where energy is released rather than absorbed.
Worked example
How much energy is required to convert 25.0 g of ice at -10.0 °C into steam at 110.0 °C? Use c(ice) = 2.09 J/g·°C, c(water) = 4.18 J/g·°C, c(steam) = 2.03 J/g·°C, ΔH_fus = 334 J/g, ΔH_vap = 2260 J/g.
Sketch the path: the substance starts below 0 °C and ends above 100 °C, so it crosses both phase-change temperatures. That means five segments.
Step 1 — warm the ice from °C to °C: J
Step 2 — melt the ice at 0 °C (no here): J
Step 3 — warm the liquid water from °C to °C: J
Step 4 — boil the water at 100 °C: J
Step 5 — warm the steam from °C to °C: J
Add them:Rounded to three significant figures, J, or about kJ.
Sanity check: step 4 alone is roughly 74 percent of the total. That fits the physical picture — pulling molecules completely apart into a gas is by far the most expensive step, and any answer where boiling is a small contribution should be re-examined.
Step 1 — warm the ice from °C to °C: J
Step 2 — melt the ice at 0 °C (no here): J
Step 3 — warm the liquid water from °C to °C: J
Step 4 — boil the water at 100 °C: J
Step 5 — warm the steam from °C to °C: J
Add them:Rounded to three significant figures, J, or about kJ.
Sanity check: step 4 alone is roughly 74 percent of the total. That fits the physical picture — pulling molecules completely apart into a gas is by far the most expensive step, and any answer where boiling is a small contribution should be re-examined.
Practice questions
A pure substance is heated at a constant rate. During the flat plateau on its heating curve, which statement is correct?
- No energy is being absorbed, so the temperature cannot change.
- Energy is absorbed and used to increase the average kinetic energy of the particles.
- Energy is absorbed and used to overcome attractions between particles, so the average kinetic energy stays constant.
- The particles stop moving until the phase change is complete.
Answer: Energy is absorbed and used to overcome attractions between particles, so the average kinetic energy stays constant.
Energy flows in the entire time the heater is on. On a plateau that energy raises potential energy by separating particles rather than speeding them up. Since temperature measures average kinetic energy, the thermometer holds steady. The first option confuses a flat temperature with zero energy transfer, and the last one contradicts the fact that liquid particles move at the same average speed as the solid particles they came from at the melting point.
Calculate the energy released when 40.0 g of liquid water at 25.0 °C is cooled and completely frozen to ice at 0.0 °C. Use c(water) = 4.18 J/g·°C and ΔH_fus = 334 J/g.
Answer: About 17,540 J (roughly 17.5 kJ) released, so q ≈ -1.75 × 10^4 J.
Two segments are needed. Cooling the liquid: J. Freezing at 0 °C: J. Total magnitude J. Because the water is losing energy to its surroundings, is negative from the water's point of view. Note that freezing uses the same as melting — direction changes the sign, not the size. A frequent error is trying to use for the freezing step, which would give zero since the temperature never changes.
Two identical 50.0 g samples, one of substance A and one of substance B, are heated with the same burner. Substance A's melting plateau lasts 4 minutes; substance B's lasts 12 minutes. What can you conclude, and what can you not conclude?
Answer: Substance B has a heat of fusion about three times larger than substance A's; you cannot conclude anything about their melting points or specific heats from plateau length alone.
With equal masses and equal heating rates, energy delivered is proportional to time. Three times the time means three times the energy per gram to melt, so . Plateau length depends only on mass and heat of fusion. The height of the plateau on the temperature axis tells you the melting point, and the steepness of the sloped segments tells you about specific heats — those are separate features of the graph.
FAQ
- Why doesn't the temperature rise while ice is melting, even though heat is still being added?
- Because the added energy is doing a different job. Temperature reflects how fast particles are moving on average. During melting, incoming energy is spent breaking the hydrogen bonds that lock water molecules into the ice lattice, which raises potential energy instead of speed. Once every bit of ice has melted, the next energy that arrives goes back into speeding molecules up and the temperature climbs again.
- When do I use q = mcΔT and when do I use q = mΔH?
- Look at the temperature. If the temperature is changing and the substance stays in one phase, use with the specific heat of that phase. If the temperature is fixed at a melting or boiling point while the substance changes phase, use . Never put a into a phase-change step — it is zero, and you would calculate zero energy for the step that actually needs the most.
- Why is the heat of vaporization so much bigger than the heat of fusion?
- Melting only loosens the particles enough to let them slide past one another; they stay in contact and most attractions remain. Vaporizing requires separating them completely so they no longer interact. For water that difference is dramatic: about 334 J per gram to melt but about 2260 J per gram to boil, which is why the boiling plateau on a heating curve is so much longer.
- Do I need moles, or can I just use grams?
- Either works as long as your units match. If your table lists heat of fusion in J/g, use mass in grams. If it lists kilojoules per mole, such as 40.7 kJ/mol for vaporizing water, convert your mass to moles first. Mixing the two — grams times a per-mole value — is one of the most common sources of answers that are off by a large factor.
Learn this with a teacher, not a page
The Crimsora tutor teaches Phase Changes & Heating Curves live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.