Changes of State and Mixtures

How much energy to melt, to boil, and where does it all end up?

Physics · delta (3A) · 20 October 2026

Open question

Why so much ice in a cocktail?

Bartenders put in a lot of ice: to give you less drink, or is there a physical reason? Take a mocktail (a non-alcoholic cocktail) of 20 cl, about 0.20 kg, at room temperature: 27 °C (300 K). The ice comes from the freezer at −18 °C, in cubes of about 20 g. You want the ice to cool the drink without all melting, or the drink gets watered down. How many cubes so that some ice survives?

Ignore the glass and the room.

1–2 cubes
3 cubes
4 cubes
8 cubes or more

Why it is here: the answer needs both formulas of the lesson (temperature changes AND state changes) and the budget idea of the second half. Ask for a hand count for each option and write the counts on the board. Do NOT reveal the answer: it comes back on the last slides. If someone asks "how much does the drink weigh / how big is a cube?", that is the right question: the answer depends on those two assumptions (0.20 kg, 20 g). Answer for you: 4 cubes (3 cubes of 60 g all melt, the drink ends at about 0.6 °C; 4 cubes leave about 20 g of ice). The bigger point, for the end: beyond 4 cubes the drink does not get more watery, the extra ice just survives. Not in the model: how fast the ice melts (big cubes melt more slowly), and the alcohol of a real cocktail, which lowers the freezing point: hence the mocktail.

Heating at constant power

1 kg of ice at −20 °C, all the way to steam

Energy supplied Q0 kJ

Compare the slopes of the ice line and the water line. Which is steeper? Why?

1 kg, to scale. Ice 42 kJ · melting 330 kJ · water 420 kJ · boiling 2300 kJ · steam 40 kJ. At 1000 W, 1 kJ takes 1 s.

Why the interaction: the curve is to scale, so the students can SEE that the plateaus are not decoration. Before moving the slider, ask: (1) when the ice reaches 0 °C, does it start to rise in temperature again immediately? (2) which plateau is longer, melting or boiling? (3) the heater is on, delivering 1000 J every second, yet during a plateau the thermometer does not move: where does the energy go? Leave (3) open and answer it on the next slide. Then drag. Use Play for the non-interactive version. Use Zoom to show the first three segments clearly: ice and steam have the same slope, about twice as steep as water, because c(ice) = 2100 and c(steam) = 2000 are about half of c(water) = 4200 J/kg·K: the same energy raises the temperature twice as much. Constants: c ice 2100, c water 4200, c steam 2000 J/kg·K; L_f = 3.3×10^5 J/kg, L_v = 2.3×10^6 J/kg (rounded, as on the exercise sheet). Boiling takes about 73% of the whole horizontal axis; melting about 10%.

One question decides the formula

T is changing

Q=mc\Delta T

Energy goes into the motion of the particles. c in J/(kg·°C): different for ice, water, steam.

State is changing, T is fixed

Q=mL

Energy goes into the arrangement of the particles. L in J/kg: the specific latent heat.

Before writing any term, ask: is T changing, or is the state changing?

Beyond the 0653 syllabus: quantitative calorimetry (extension).

How big are the latent heats?

Worked example

200 g of ice at −10 °C → water at 30 °C

How much energy is needed? Constants: c_ice = 2100, c_water = 4200 J/(kg·°C), L_f = 3.3×10⁵ J/kg.

Q_1=0.20\times2100\times10=4200\ \mathrm{J}

T changes: ice warms from −10 to 0 °C

Q_2=0.20\times3.3\times10^{5}=66\,000\ \mathrm{J}

State changes at 0 °C: melting

Q_3=0.20\times4200\times30=25\,200\ \mathrm{J}

T changes: water warms from 0 to 30 °C

Q=Q_1+Q_2+Q_3=95\,400\ \mathrm{J}\approx 95\ \mathrm{kJ}

Melting is by far the biggest term

Your turn · 1

One step at a time

1. How much energy heats 250 g of water from 20 °C to 70 °C?
2. How much energy melts 0.50 kg of ice already at 0 °C?
3. How much energy evaporates 100 g of water already at 100 °C?

Your turn · 2

Time, and a burn

4. A 1000 W heater warms 0.50 kg of ice from −20 °C and loses no energy. How long until: (a) the ice reaches 0 °C, (b) all of it has melted, (c) the water reaches 100 °C, (d) all of it has boiled away? Sketch T against t.
5. 10 g of steam at 100 °C, or 10 g of boiling water at 100 °C, lands on your skin and cools to 34 °C. Compare the energy released.

Mixtures · the reasoning

Two bodies touch

Nothing escapes

Q_{lost}=Q_{gained}

The heat leaving one body enters the other.

Body 1 warms up (the colder)

Q_{gained}=m_1c_1\,(\textcolor{#7B1E28}{T_{eq}}-T_1)

Final minus initial, as always.

Body 2 cools down (the hotter)

Q_{lost}=m_2c_2\,(T_2-\textcolor{#7B1E28}{T_{eq}})

\textcolor{#7B1E28}{T_{eq}}-T_2<0: flip it, so the heat lost is positive.

One temperature for both

m_1c_1\,(\textcolor{#7B1E28}{T_{eq}}-T_1)=m_2c_2\,(T_2-\textcolor{#7B1E28}{T_{eq}})

\textcolor{#7B1E28}{T_{eq}}: the equilibrium temperature (unknown).

Go block by block, slowly: this is the mental procedure, in the order a student should think it. Each click draws the path and brings in the next block. (1) Nothing escapes, so all the heat that leaves one body enters the other: lost = gained. (2) Start from the body that warms up, the colder one (body 1): its temperature rises from T_1 to the equilibrium temperature, so the heat gained is m_1c_1 (T_eq − T_1): final minus initial, exactly as in every formula so far. (3) The hotter body (body 2) cools from T_2 to T_eq. Final minus initial, T_eq − T_2, would be negative, but "heat lost" is a positive quantity: so we flip it and write T_2 − T_eq. Say it out loud: here the equilibrium temperature sits in the second place, and that is not a mistake, it is the flip. (4) Both bodies end at the SAME temperature, the equilibrium temperature T_eq: it appears in both brackets and is the only unknown. Put the two expressions in the equality and solve. Next slide: a single rule for every body, which removes the flip and works with three bodies. We call the final common temperature T_eq from now on, so that "final" is never confused with the T_final of ΔT.

Mixtures · one rule for every body

Always final minus initial

The rule

Q=mc\,(\textcolor{#7B1E28}{T_{eq}}-T_{i})

Warms up: positive. Cools down: negative.

What it gives

Q_1=Q_{gained}\quad Q_2=-Q_{lost}

The minus is already inside \textcolor{#7B1E28}{T_{eq}}-T_2.

Then

-Q_2=Q_1\ \Rightarrow\ Q_1+Q_2=0

-Q_2 changes side, the sign flips: \textstyle\sum Q=0.

Three free blocks on a zigzag, joined by a dotted path that is drawn as you click. The previous slide used two different orders of subtraction, one for each body (and a flip for the hot one); with three bodies it is easy to slip. Use ONE rule, always final minus initial (block 1), with the equilibrium temperature as the final temperature of every body. For the cold body (body 1) nothing changes: Q_1 = Q_gained. For the hot body (body 2) T_eq − T_2 is negative, so Q_2 = m_2c_2 (T_eq − T_2) = −Q_lost: the energy lost, seen from the body that lost it. The minus sign is not added by us: it is already inside (T_eq − T_2) (block 2). Substitute Q_lost = −Q_2 into lost = gained and you get −Q_2 = Q_1; moving −Q_2 to the other side flips the sign again and leaves Q_1 + Q_2 = 0 (block 3). Students who write "hot minus cold" for one body and "cold minus hot" for the other get the right answer by luck with two bodies and the wrong one with three.

Worked example

Hot water into cold water

0.20 kg of water at 80 °C is poured into 0.30 kg of water at 20 °C. Equilibrium temperature?

0.20\cdot4200\,(80-T_{eq})=0.30\cdot4200\,(T_{eq}-20)

Lost = gained, each one positive. Both bodies end at the same T_eq.

0.30\cdot4200\,(T_{eq}-20)+0.20\cdot4200\,(T_{eq}-80)=0

Same equation, one rule (final minus initial): Q₁ > 0 (cold), Q₂ < 0 (hot), sum = 0.

0.30\,(T_{eq}-20)+0.20\,(T_{eq}-80)=0\ \Rightarrow\ 0.50\,T_{eq}=22

c is the same: it cancels

T_{eq}=44\ ^{\circ}\mathrm{C}

Check: between 20 and 80, closer to the larger mass. Q_lost = Q_gained = 30 240 J.

When ice is involved

Budget first, equation second

Budget. How much energy can the hot body give by cooling to 0 °C? How much does the ice need to warm up and to melt completely?

Equation. Three possible endings. All the ice melts: T_{eq}\gt0, a melting term in \sum Q=0. Only some melts: T_{eq}=0\ ^{\circ}\mathrm{C}, ice survives. None melts: T_{eq}\lt0, no melting term.

Check. Is T_{eq} inside the range that ending allows?

Try it

The mocktail and the ice: the budget in action

Mocktail (0.20 kg) at27 °C

Ice cubes (20 g each)4 cubes

Ice temperature−18 °C

Why the interaction: the budget check is visible. Teal bar = what the mocktail can give by cooling to 0 °C. Navy + bordeaux bar = what the ice needs to warm up to 0 °C and then to melt. If the teal bar is longer than the whole stacked bar: all the ice melts and T_eq > 0. If it is shorter: T_eq = 0 °C and some ice survives. Walk through the three buttons. 3 cubes: the ice needs 60 g × 367 800 J/kg = 22 068 J against 22 680 J: all of it melts, T_eq is about 0.6 °C, the drink is 23% meltwater. 4 cubes: not enough energy, 60 g melt and 20 g survive, T_eq = 0 °C. Before pressing 8 cubes, ask: more cubes, more watery drink? (No: 110 g survive, about 50 g melt, 20% meltwater, T_eq still 0 °C.) Then move the ice temperature and the drink temperature (a hot terrace: 35 °C) to see when the answer changes. Constants: c_water 4200, c_ice 2100 J/kg·K, L_f = 3.3×10^5 J/kg. The drink is treated as 0.20 kg of water; no energy is exchanged with the glass or the room.

Your turn · 3

Mixtures

6. A 0.30 kg block of metal at 100 °C is dropped into 0.50 kg of water at 20 °C. The equilibrium temperature is 25 °C. Find the specific heat capacity of the metal. Which metal might it be?
7. 0.20 kg of tea at 80 °C. Compare adding 50 g of ice at 0 °C with adding 50 g of water at 0 °C: equilibrium temperature in each case?

Your turn · 4

Does it all melt? Steam as a heater

8. 0.50 kg of water at 20 °C is poured onto 0.40 kg of ice at 0 °C. What is in the container at the end, and at what temperature?
9. A coffee machine blows 20 g of steam at 100 °C into 200 g of milk at 5 °C (treat the milk as water). Equilibrium temperature?

The one from Reels

Lava on a block of ice

1 kg of lava at 1100 °C lands on a 100 kg block of ice at −10 °C. Contact is perfect and no energy escapes. When nothing changes any more, how much ice has melted?

Lava: c ≈ 1000 J/(kg·°C). Ignore the solidification of the lava.

All 100 kg
Some: T_eq is 0 °C
None at all

Your turn · 5

Lava vs ice: the budget

10. Find the equilibrium temperature and how much ice melts. Start with the budget.

1\cdot1000\,(T_{eq}-1100)+100\cdot2100\,(T_{eq}+10)=0\ \Rightarrow\ T_{eq}\approx-4.7\ ^{\circ}\mathrm{C}

Back to the mocktail

What did we predict?

Back to the mocktail

How many cubes so that some survive?

Q_{drink}=0.20\times4200\times27=22\,680\ \mathrm{J}

What the mocktail can give on the way to 0 °C

2100\times18+3.3\times10^{5}=367\,800\ \mathrm{J/kg}

Each kg of ice, to melt completely: warm to 0 °C, then melt

m=\dfrac{22\,680}{367\,800}\approx0.062\ \mathrm{kg}\approx62\ \mathrm{g}\approx 3.1\ \text{cubes}

3 cubes (60 g) all melt; 4 cubes (80 g): about 60 g melt, 20 g survive

More ice, not more water. Beyond 4 cubes the drink still gives only 22.7 kJ: about 50–60 g melt, the rest stays ice.

8 cubes: 110 g of ice survive, 20% meltwater, T_eq = 0 °C

Answer to the opening question. The drink can give at most 22 680 J, which melts about 60 g of ice whatever the number of cubes; extra cubes do not end up in the glass, they stay ice and keep the drink at 0 °C for as long as they last. So a bartender who uses a lot of ice is not giving you more water: he is giving you a colder drink, for longer. With more cubes the dilution even drops slightly (23% with 4, 20% with 8) because there is more ice to warm up from −18 °C to 0 °C, which leaves less energy for melting. Honest limits of the model: it says how cold and how watery at equilibrium, not how fast; big cubes also melt more slowly (less surface), and a real cocktail contains alcohol, which lowers the freezing point. Ties together: ice warming (mcΔT), melting (mL), the budget check. No full equation here, by choice.

In summary

T changing: Q=mc\Delta T. State changing at constant T: Q=mL.
Latent heats are huge: melting 1 kg of ice costs as much as warming water by 79 °C; boiling costs 5.5 times warming water from 0 to 100 °C.
Mixtures: \sum Q=0 with Q=mc(T_{eq}-T_i) for every body.
With ice: budget first, equation second, check last.