D. Energy and Society · Thermal and Nuclear Energy

Heating and Cooling Curves and Multi-Step Heat Problems

D2.11 D2.10 D3.7 A1.6 A1.11About 70 minutes

Learning goals

I can draw and label a heating curve and a cooling curve, including the plateaus for changes of state.
I can explain what is happening to the particles in each section of a heating or cooling curve.
I can calculate the total energy for a process that includes temperature changes and changes of state.
I can solve mixture problems that involve melting ice.

Words to know

heating curve
A graph of temperature against energy added (or time, with steady heating) as a substance is heated through its changes of state.

cooling curve
A graph of temperature against time as a substance cools and changes state.

plateau
A flat (horizontal) section of a heating or cooling curve, where a change of state is happening and the temperature is constant.

melting point / freezing point
The temperature at which a solid melts and a liquid freezes. For a pure substance they are the same temperature.

multi-step heat problem
A problem in which the total heat is the sum of several Q = mcΔT and Q = mL steps.

The long road from snowball to steam

Imagine scooping 1 kg of snow at −20 °C into a pot on the stove and leaving it there until it is steam at 120 °C. How long does each stage take? Most people guess that warming from 0 °C to 100 °C would take the longest. In fact, the stage where the water is boiling takes over five times as long as the warming stage.

A heating curve shows this whole journey in one picture. In this lesson you will draw and read heating and cooling curves and use them to solve problems with several steps.

Reading a heating curve

A heating curve plots temperature (vertical axis) against energy added or time (horizontal axis). If energy is added at a steady rate, energy and time are proportional, so the curve has the same shape either way.

A heating curve for water has five sections:

A, sloped: solid ice warms. Q = m c_ice ΔT. Particles vibrate faster; kinetic energy and temperature rise.
B, flat (plateau) at 0 °C: ice melts. Q = mL_f. Energy increases the particles' potential energy; temperature is constant.
C, sloped: liquid water warms. Q = m c_water ΔT.
D, flat at 100 °C: water boils. Q = mL_v. This plateau is the longest because L_v is so large.
E, sloped: steam warms. Q = m c_steam ΔT.

Two things to notice:

The flat parts are changes of state. The length of a plateau tells you the latent heat (length = mL).
The slope of a sloped part depends on c. On a temperature-versus-energy graph, slope = ΔT/Q = 1/(mc). A smaller c gives a steeper slope. Ice (c = 2100) warms about twice as fast as liquid water (c = 4180), so section A is about twice as steep as section C.

Heating curve for 1.0 kg of water from −20 °C to 120 °C, drawn to scale. The boiling plateau (D, 2260 kJ) is far longer than the melting plateau (B, 334 kJ) and the warming of liquid water (C, 418 kJ).

Section	Process	Equation	Energy for 1.0 kg (kJ)
A	ice −20 °C → 0 °C	Q = mcΔT = (1.0)(2100)(20)	42
B	melting at 0 °C	Q = mL_f = (1.0)(3.34 × 10⁵)	334
C	water 0 °C → 100 °C	Q = mcΔT = (1.0)(4180)(100)	418
D	boiling at 100 °C	Q = mL_v = (1.0)(2.26 × 10⁶)	2260
E	steam 100 °C → 120 °C	Q = mcΔT = (1.0)(2000)(20)	40
Total		add all five	3094

The energy for each section of the heating curve above.

Worked example

Example 1: Ice to steam

How much energy is needed to change 0.50 kg of ice at −10 °C into steam at 110 °C?

Break the journey into steps. Each step is either Q = mcΔT (temperature changes) or Q = mL (state changes).
1. Warm the ice: Q₁ = (0.50)(2100)(10) = 10 500 J.
2. Melt the ice: Q₂ = (0.50)(3.34 × 10⁵) = 167 000 J.
3. Warm the water: Q₃ = (0.50)(4180)(100) = 209 000 J.
4. Boil the water: Q₄ = (0.50)(2.26 × 10⁶) = 1 130 000 J.
5. Warm the steam: Q₅ = (0.50)(2000)(10) = 10 000 J.
Total: Q = 10 500 + 167 000 + 209 000 + 1 130 000 + 10 000 = 1 526 500 J ≈ 1.53 × 10⁶ J.

About 1.53 × 10⁶ J (1.53 MJ). Boiling alone is about 74% of it.

Cooling curves

A cooling curve is the same idea run backwards: temperature against time as a substance loses heat to its surroundings. A liquid cools, then levels off at its freezing point while it solidifies (it is releasing latent heat, which keeps the temperature steady), then the solid cools.

For a pure substance, the freezing point on the cooling curve equals the melting point on the heating curve. That is one way chemists test purity: an impure substance freezes over a range of temperatures instead of at one sharp temperature.

Cooling curves are usually curved rather than straight, because a hot object loses heat faster when it is much hotter than the room (larger temperature difference) and more slowly as it approaches room temperature.

Cooling curve for lauric acid, a substance often used in school labs (freezing point about 44 °C). The plateau shows where it is freezing. Sample data; your lab results may differ.

Worked example

Example 2: Ice in a drink

0.050 kg of ice at 0 °C is dropped into 0.300 kg of water at 25 °C in an insulated cup. Find the final temperature.

Check first: can the water melt all the ice? Melting needs Q = (0.050)(3.34 × 10⁵) = 16 700 J. Cooling the water all the way to 0 °C would give (0.300)(4180)(25) = 31 350 J. That is more than enough, so all the ice melts and T_f is above 0 °C.
Write Q_ice + Q_meltwater + Q_warm water = 0:
16 700 + (0.050)(4180)(T_f − 0) + (0.300)(4180)(T_f − 25) = 0
16 700 + 209T_f + 1254T_f − 31 350 = 0
1463T_f = 14 650, so T_f ≈ 10 °C.

T_f ≈ 10 °C. (If the check had shown not enough energy, the final temperature would be 0 °C with some ice left.)

Worked example

Example 3: Reading the graph with steady power

A 500 W heater warms 0.40 kg of a solid. Its heating curve has a flat section that lasts from t = 80 s to t = 300 s. Find the specific latent heat of fusion of the substance.

The flat section is melting. It lasts 300 − 80 = 220 s.
Energy supplied = power × time = 500 J/s × 220 s = 110 000 J.
L_f = Q ÷ m = 110 000 ÷ 0.40 = 2.75 × 10⁵ J/kg (2.8 × 10⁵ J/kg to 2 significant digits).

L_f ≈ 2.8 × 10⁵ J/kg.

Just for fun

Why did the heating curve take a long nap at 100 °C?

It had a lot on its plate-au. (About 2260 kJ per kilogram, to be exact.)

Hands-on activity

Make your own cooling curve

You need: a bowl, lots of ice, about 1/3 cup of table salt, a small plastic cup or plastic test tube, water, a kitchen thermometer (a probe thermometer is ideal), a timer and graph paper or a spreadsheet. Grown-up note: a salt-ice mixture can get colder than −10 °C and can hurt skin if handled for a long time; wear mittens or use a spoon to stir it.

Fill the bowl with ice, sprinkle the salt over it and stir. Pour in a little cold water to make a slushy bath.
Pour about 2 cm of room-temperature water into the small cup and put the thermometer in it.
Set the cup into the salt-ice bath so the water is below the level of the slush. Start the timer.
Read the temperature every 30 s for at least 15 minutes. Gently move the thermometer from time to time. Note when ice first appears and when the water seems frozen solid.
Graph temperature (vertical) against time (horizontal). Label the liquid-cooling section, the freezing plateau and the solid-cooling section.

What to look for: the temperature falls, then stays near 0 °C for several minutes while the water freezes, then falls again. Sometimes water dips a little below 0 °C before it suddenly starts freezing and jumps back up to 0 °C. That is called supercooling, and if you see it, you have done a careful experiment!

Your turn

On a heating curve, what is happening during a flat section?

On a heating curve, why is the boiling plateau much longer than the melting plateau?

How much energy is needed to warm 0.20 kg of ice from −15 °C to 0 °C and melt it?

How much energy is released when 0.10 kg of water at 30 °C is cooled to 0 °C and frozen?

On a temperature-versus-energy graph, section A (ice) is steeper than section C (water). Explain.

A cooling curve for an unknown pure liquid levels off at 63 °C, then falls again. What does this tell you about the substance?

More practice: warm-up

What two equations are used in multi-step heat problems?

What is plotted on the axes of a cooling curve?

For a pure substance, how do its melting point and freezing point compare?

During a plateau on a cooling curve, is latent heat absorbed or released?

Name the five sections of the heating curve of water, in order.

More practice: on your own

How much energy is needed to turn 0.25 kg of water at 40 °C into steam at 100 °C?

A 1000 W heater supplies energy to 0.50 kg of ice at 0 °C. How long does it take to melt it all (ignore losses)?

0.020 kg of ice at 0 °C is added to 0.200 kg of water at 30 °C. Find the final temperature.

Sketch (describe) the cooling curve of steam at 120 °C cooled to ice at −10 °C.

In a lab, a heating curve's melting plateau is slightly slanted rather than flat. Give two possible reasons.

Challenge

0.10 kg of ice at 0 °C is added to 0.20 kg of water at 20 °C. Does all the ice melt? If not, how much is left and what is the final temperature?

How much heat must be removed by a freezer to turn 1.5 kg of water at 20 °C into ice at −18 °C?

A heating curve is drawn with temperature against time for a sample heated by a constant 250 W source. The solid section rises from −30 °C to −10 °C in 40 s, and the sample's mass is 0.20 kg. Find c for the solid.

Check your understanding

Quiz

On a heating curve of water, what is happening during the plateau at 100 °C?

Why is the boiling plateau of water longer than the melting plateau?

Which equation describes the energy for a sloped section of a heating curve?

How much energy is needed to melt 0.10 kg of ice at 0 °C and warm it to 10 °C?

On a temperature-versus-energy heating curve, a steeper sloped section means:

A cooling curve of a pure substance shows a plateau at 44 °C. What is happening there?

For parents and teachers

Big idea: expectation D2.11 asks students to draw and analyse heating and cooling curves; D2.10 asks them to solve the related problems. Every heating-curve problem breaks into steps: sloped sections use Q = mcΔT, flat sections use Q = mL.

How to help: Have your teen sketch the heating curve of water from memory and label the five sections and their equations. For problems, ask them to list the steps before calculating anything.

Common difficulties:

Skipping a step (often forgetting to warm the ice up to 0 °C first).
Using c(water) for ice or steam. Ice is about 2100 and steam about 2000 J/(kg·°C).
In ice-in-water problems, not checking whether all the ice melts (Challenge 1).
Thinking a flat section means no energy is being added.

Safety: salt-ice baths get very cold; use mittens or a spoon.