🧊 States of Matter & Phase Transitions

Part 1 of 7 — The Big Picture

All matter exists in one of several phases (states): solid, liquid, or gas. In AP Physics 2, understanding how matter transitions between these phases — and the energy involved — is essential for thermodynamics problems.

The Three Common Phases

The key difference is the strength of intermolecular bonds relative to the kinetic energy of the molecules.

The Six Phase Transitions

Every phase change has a name — and a reverse process:

Energy Rules

• Melting, vaporization, sublimation → require energy input (endothermic)
• Freezing, condensation, deposition → release energy (exothermic)

During any phase change, the temperature stays constant even though energy is being added or removed. All the energy goes into breaking or forming intermolecular bonds rather than changing kinetic energy.

Phase Transitions Check 🔄

Phase Diagrams

A phase diagram maps out which phase a substance is in as a function of temperature (x-axis) and pressure (y-axis).

Key Features

• Phase regions: Large areas labeled solid, liquid, and gas where the substance exists in that phase
• Phase boundaries: Lines separating regions — along these lines, two phases coexist in equilibrium
• Triple point: The unique temperature and pressure where all three phases coexist simultaneously
• Critical point: The endpoint of the liquid-gas boundary — above this temperature and pressure, there is no distinction between liquid and gas (called a supercritical fluid)

Water's Phase Diagram

For water at standard atmospheric pressure ($1.01 \times 10^5$ Pa):

• Melting point: 0°C (273 K)
• Boiling point: 100°C (373 K)
• Triple point: 0.01°C at 611 Pa
• Critical point: 374°C at $2.21 \times 10^7$ Pa

Water is unusual because its solid-liquid boundary line slopes to the left (negative slope), meaning increasing pressure lowers the melting point. This is why ice skating works — pressure under the blade can melt ice.

Phase Diagram Concepts 📊

Phase Transition Identification 🎯

Identify the phase transition in each scenario:

🔥 Latent Heat

Part 2 of 7 — Energy Without Temperature Change

When a substance undergoes a phase change, energy is absorbed or released without any change in temperature. This energy is called latent heat (from Latin latere — "to lie hidden").

The Latent Heat Equation

$Q = mL$

where:

• $Q$ = heat energy absorbed or released (J)
• $m$ = mass of the substance (kg)
• $L$ = specific latent heat (J/kg) — depends on the substance and the type of phase change

Two Types of Latent Heat

Latent Heat of Fusion ($L_f$)

Energy per kilogram to melt a solid (or released when freezing):

$Q_{\text{fusion}} = m L_f$

For water: $L_f = 334{,}000$ J/kg = $3.34 \times 10^5$ J/kg

Latent Heat of Vaporization ($L_v$)

Energy per kilogram to vaporize a liquid (or released when condensing):

$Q_{\text{vaporization}} = m L_v$

For water: $L_v = 2{,}260{,}000$ J/kg = $2.26 \times 10^6$ J/kg

Why is $L_v \gg L_f$?

Vaporization requires much more energy than fusion because:

• Melting only loosens intermolecular bonds (molecules still stay close)
• Vaporization must completely separate molecules against attractive forces
• For water: $L_v / L_f \approx 6.8$ — vaporization takes almost 7× more energy!

Sign Convention

• Phase change requiring energy (melting, vaporization): $Q > 0$
• Phase change releasing energy (freezing, condensation): $Q < 0$

Latent Heat Concepts 🧠

Why Temperature Stays Constant — A Closer Look

Consider heating ice at exactly 0°C:

1. You add energy → some ice melts
2. The liquid water formed is also at 0°C
3. As long as both ice and water coexist, the temperature stays at 0°C
4. Only after ALL ice has melted does the temperature begin to rise

The Microscopic View

• Temperature measures the average kinetic energy of molecules
• During a phase change, added energy increases potential energy (breaking bonds)
• Since kinetic energy doesn't change → temperature doesn't change

How Long Does It Take?

If you supply heat at a constant rate $P$ (power in watts):

$t = \frac{Q}{P} = \frac{mL}{P}$

Example: How long to melt 2.0 kg of ice with a 500 W heater?

$t = \frac{(2.0)(334{,}000)}{500} = \frac{668{,}000}{500} = 1336 \text{ s} \approx 22.3 \text{ min}$

Latent Heat Calculations 🔢

Use: $L_f = 334{,}000$ J/kg, $L_v = 2{,}260{,}000$ J/kg for water.

1. Energy needed to melt 0.50 kg of ice at 0°C (in kJ)
2. Energy released when 3.0 kg of steam at 100°C condenses to water at 100°C (in MJ)
3. Mass of ice at 0°C that can be melted by 1.00 MJ of energy (in kg)

Round all answers to 3 significant figures.

Exit Quiz

📈 Heating Curves & Multi-Step Energy Problems

Part 3 of 7 — From Ice to Steam

A heating curve shows how the temperature of a substance changes as energy is continuously added. For water, there are five distinct stages — three with temperature changes and two flat sections (phase changes).

The Complete Heating Curve for Water

Starting with ice at $-20$°C and ending with steam at $120$°C, there are five stages:

Stage 1: Heating Ice ($-20$°C → $0$°C)

$Q_1 = m c_{\text{ice}} \Delta T$ $c_{\text{ice}} = 2{,}090$ J/(kg·°C)

Stage 2: Melting Ice (0°C, solid → liquid)

$Q_2 = m L_f$ $L_f = 334{,}000$ J/kg — temperature stays at 0°C

Stage 3: Heating Water ($0$°C → $100$°C)

$Q_3 = m c_{\text{water}} \Delta T$ $c_{\text{water}} = 4{,}186$ J/(kg·°C)

Stage 4: Boiling Water (100°C, liquid → gas)

$Q_4 = m L_v$ $L_v = 2{,}260{,}000$ J/kg — temperature stays at 100°C

Stage 5: Heating Steam ($100$°C → $120$°C)

$Q_5 = m c_{\text{steam}} \Delta T$ $c_{\text{steam}} = 2{,}010$ J/(kg·°C)

Total Energy

$Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5$

The flat sections (stages 2 and 4) on the heating curve represent phase changes — energy is absorbed but temperature does not change.

Heating Curve Concepts 📊

Worked Example: Full Heating Curve Calculation

Problem: Calculate the total energy needed to convert 0.200 kg of ice at $-20$°C to steam at $120$°C.

Stage 1: Heat ice from $-20$°C to 0°C

$Q_1 = mc_{\text{ice}}\Delta T = (0.200)(2{,}090)(20) = 8{,}360 \text{ J}$

Stage 2: Melt ice at 0°C

$Q_2 = mL_f = (0.200)(334{,}000) = 66{,}800 \text{ J}$

Stage 3: Heat water from 0°C to 100°C

$Q_3 = mc_{\text{water}}\Delta T = (0.200)(4{,}186)(100) = 83{,}720 \text{ J}$

Stage 4: Vaporize water at 100°C

$Q_4 = mL_v = (0.200)(2{,}260{,}000) = 452{,}000 \text{ J}$

Stage 5: Heat steam from 100°C to 120°C

$Q_5 = mc_{\text{steam}}\Delta T = (0.200)(2{,}010)(20) = 8{,}040 \text{ J}$

Total:

$Q_{\text{total}} = 8{,}360 + 66{,}800 + 83{,}720 + 452{,}000 + 8{,}040 = 618{,}920 \text{ J} \approx 619 \text{ kJ}$

Notice that Stage 4 (vaporization) accounts for $452/619 \approx 73\%$ of the total energy!

Multi-Step Heating Calculation 🔢

Calculate the total energy to convert 0.500 kg of ice at $-10$°C to steam at $110$°C.

Use: $c_{\text{ice}} = 2{,}090$ J/(kg·°C), $L_f = 334{,}000$ J/kg, $c_{\text{water}} = 4{,}186$ J/(kg·°C), $L_v = 2{,}260{,}000$ J/kg, $c_{\text{steam}} = 2{,}010$ J/(kg·°C).

1. $Q_1$: Energy to heat ice from $-10$°C to 0°C (in kJ)
2. $Q_2$: Energy to melt all the ice (in kJ)
3. $Q_3$: Energy to heat water from 0°C to 100°C (in kJ)
4. $Q_4$: Energy to vaporize all the water (in kJ)
5. $Q_5$: Energy to heat steam from 100°C to 110°C (in kJ)
6. $Q_{\text{total}}$ (in kJ)

Round all answers to 3 significant figures.

Exit Quiz

📐 Phase Diagrams, Critical Point & Vapor Pressure

Part 4 of 7 — Reading Phase Diagrams Like a Pro

Phase diagrams encode a wealth of information about how substances behave under different conditions of temperature and pressure. For AP Physics 2, you need to interpret these diagrams and understand the special points on them.

Reading a Phase Diagram

A phase diagram has:

• x-axis: Temperature ($T$)
• y-axis: Pressure ($P$)

The Three Boundary Lines

1. Solid-Liquid line (fusion curve): Separates solid and liquid regions. Crossing this line = melting or freezing.
2. Liquid-Gas line (vaporization curve): Separates liquid and gas regions. Crossing this line = boiling or condensation.
3. Solid-Gas line (sublimation curve): Separates solid and gas regions. Crossing this line = sublimation or deposition.

How to Trace a Process

To find what happens when you change conditions:

• Heating at constant pressure: Move right along a horizontal line
• Compressing at constant temperature: Move up along a vertical line
• Each time you cross a boundary line, a phase change occurs

Example: Heating at 1 atm

Starting from a low temperature and moving right at $P = 1$ atm: $\text{Solid} \xrightarrow{\text{melting}} \text{Liquid} \xrightarrow{\text{boiling}} \text{Gas}$

Special Points on the Phase Diagram

Triple Point

The triple point is where all three boundary lines meet — the unique $(T, P)$ where solid, liquid, and gas coexist in equilibrium.

• For water: $T_{\text{tp}} = 0.01$°C, $P_{\text{tp}} = 611$ Pa (well below atmospheric pressure)
• For CO₂: $T_{\text{tp}} = -56.6$°C, $P_{\text{tp}} = 5.18$ atm (above atmospheric pressure — this is why CO₂ sublimates at 1 atm!)

Critical Point

The critical point is the endpoint of the liquid-gas boundary line. Above this temperature and pressure:

• There is no distinction between liquid and gas
• The substance is a supercritical fluid
• No amount of pressure can liquify the gas above the critical temperature

For water: $T_c = 374$°C, $P_c = 218$ atm

Vapor Pressure

The vapor pressure of a liquid is the pressure exerted by its vapor when in equilibrium with the liquid phase. It:

• Increases with temperature (molecules escape faster at higher $T$)
• Equals atmospheric pressure at the boiling point
• Is described by the Clausius-Clapeyron equation (beyond AP scope, but the concept is tested)

A liquid boils when its vapor pressure equals the external pressure above it.

Phase Diagram Concepts 📊

Vapor Pressure & Boiling 🌡️

Phase Diagram Reading Drill 🎯

For each process on a typical phase diagram:

🧮 Calorimetry with Phase Changes

Part 5 of 7 — Problem-Solving Workshop

The most challenging (and most common) AP problems combine $Q = mc\Delta T$ with $Q = mL$ in calorimetry — mixing substances at different temperatures where one or more may undergo a phase change.

The Master Equation

In an insulated system, energy is conserved:

$Q_{\text{lost}} + Q_{\text{gained}} = 0$

or equivalently:

$\sum Q_i = 0$

Each $Q_i$ may include heating/cooling steps AND phase-change steps.

Problem-Solving Strategy

Step 1: Determine if a Phase Change Occurs

Before solving, check: is there enough energy to complete the phase change?

Example: Mixing 0.10 kg of ice at 0°C with 0.20 kg of water at 25°C.

Energy available from cooling water to 0°C: $Q_{\text{available}} = m_w c_w \Delta T = (0.20)(4{,}186)(25) = 20{,}930 \text{ J}$

Energy needed to melt all the ice: $Q_{\text{needed}} = m_i L_f = (0.10)(334{,}000) = 33{,}400 \text{ J}$

Since $Q_{\text{available}} < Q_{\text{needed}}$, not all ice melts! The final temperature is 0°C with some ice remaining.

Step 2: Set Up the Energy Equation

• If the phase change does complete: include both $Q = mL$ and $Q = mc\Delta T$ terms
• If the phase change does not complete: the final temperature is at the phase-change temperature, and you solve for how much substance changes phase

Step 3: Solve for the Unknown

Common unknowns: final temperature $T_f$, mass of ice melted, or time required.

Strategy Check 🧠

Worked Example: Ice in Hot Water

Problem: 0.100 kg of ice at $-15$°C is placed in 0.300 kg of water at $60$°C in an insulated container. Find the final temperature.

Step 1: Check if all ice melts

Energy needed to bring ice to 0°C and melt it: $Q_{\text{ice to 0°C}} = (0.100)(2{,}090)(15) = 3{,}135 \text{ J}$ $Q_{\text{melt}} = (0.100)(334{,}000) = 33{,}400 \text{ J}$ $Q_{\text{total needed}} = 3{,}135 + 33{,}400 = 36{,}535 \text{ J}$

Energy available from water cooling to 0°C: $Q_{\text{available}} = (0.300)(4{,}186)(60) = 75{,}348 \text{ J}$

Since $75{,}348 > 36{,}535$: ✅ All ice melts. Final $T_f > 0$°C.

Step 2: Energy equation

$\underbrace{(0.100)(2{,}090)(15)}_{\text{heat ice}} + \underbrace{(0.100)(334{,}000)}_{\text{melt ice}} + \underbrace{(0.100)(4{,}186)(T_f - 0)}_{\text{heat meltwater}} + \underbrace{(0.300)(4{,}186)(T_f - 60)}_{\text{cool hot water}} = 0$

Step 3: Solve

$3{,}135 + 33{,}400 + 418.6 T_f + 1{,}255.8 T_f - 75{,}348 = 0$ $1{,}674.4 T_f = 38{,}813$ $T_f = 23.2°\text{C}$

Calorimetry Workshop 🔢

Problem: 0.200 kg of ice at 0°C is added to 0.400 kg of water at 50°C in an insulated container.

Use: $c_w = 4{,}186$ J/(kg·°C), $L_f = 334{,}000$ J/kg.

1. Energy needed to melt all the ice (in J)
2. Energy available from water cooling to 0°C (in J)
3. Does all the ice melt? Enter "yes" or "no"
4. Final temperature of the mixture (in °C, round to 1 decimal)

Round all answers to 3 significant figures.

Exit Quiz

🌍 Real-World Applications of Phase Changes

Part 6 of 7 — Physics in Everyday Life

Phase changes and latent heat are not just textbook concepts — they explain a huge number of everyday phenomena. Understanding the physics behind these applications deepens your intuition and prepares you for AP conceptual questions.

Evaporative Cooling

Why You Feel Cold After Swimming

When you step out of a pool, water on your skin evaporates (vaporizes). The latent heat of vaporization ($L_v = 2{,}260{,}000$ J/kg) must come from somewhere — it comes from your body and the water itself, cooling you down.

Sweating

Your body's primary cooling mechanism: sweat glands produce water on the skin surface. As it evaporates, it absorbs $\sim 2.26$ MJ per kg of sweat — a remarkably efficient cooling system!

On a humid day, evaporation is slower (air is already saturated with water vapor), so cooling is less effective → you feel hotter.

Dogs Panting

Dogs don't sweat through skin. Instead, they pant — rapidly breathing over their wet tongue and respiratory tract. Evaporation from these moist surfaces provides cooling.

Desert Water Bags

Canvas water bags hung on the front of old cars: water seeps through the canvas and evaporates, cooling the remaining water inside. The same principle is behind clay pot coolers used in hot, dry climates.

Evaporative Cooling Check 💧

More Applications

Pressure Cookers

A sealed pot increases the pressure above the water → the boiling point rises above 100°C (to about 120°C at 2 atm). Food cooks faster because the water is hotter while still liquid.

$\text{Higher pressure} \rightarrow \text{Higher boiling point} \rightarrow \text{Faster cooking}$

Freeze-Drying (Lyophilization)

Food is frozen, then placed under very low pressure (below the triple point). Ice sublimates directly to vapor without passing through the liquid phase. This preserves the food's structure — the food doesn't get soggy.

Used for: astronaut food, instant coffee, pharmaceutical drugs, biological samples.

Ice Skating

The pressure under the narrow blade may slightly lower ice's melting point (water's unusual negative-slope fusion curve). A thin layer of liquid water forms under the blade, reducing friction. (The full explanation also involves surface melting and friction heating.)

Salt on Icy Roads

Salt lowers the freezing point of water (colligative property). Ice at $-5$°C in contact with salt forms a brine solution that has a freezing point below $-5$°C, so the ice melts. This is effective down to about $-18$°C for NaCl.

Applications Quiz 🔬

Application Identification 🎯

Match each phenomenon with its primary physics principle:

🎯 Synthesis & AP Review

Part 7 of 7 — Putting It All Together

This final part connects all the concepts from the topic and prepares you for AP-style questions. We'll review the concept map, address common mistakes, and work through mixed problems.

Concept Map: Phase Changes & Latent Heat

Core Equations

Key Values for Water

Connections to Other Topics

• Thermodynamics: Phase changes involve internal energy changes at constant temperature
• Kinetic Theory: Temperature measures average KE; phase changes alter PE
• Pressure: Affects boiling/melting points (phase diagrams)
• Ideal Gas Law: Applies to the gas phase; breaks down near phase transitions

Common Mistakes Quiz ⚠️

Identify the CORRECT statement in each question:

Mixed Problem Set 🔢

Use: $c_w = 4{,}186$ J/(kg·°C), $c_{\text{ice}} = 2{,}090$ J/(kg·°C), $L_f = 334{,}000$ J/kg, $L_v = 2{,}260{,}000$ J/kg.

1. How much energy (in kJ) is released when 0.500 kg of steam at 100°C condenses AND the resulting water cools to 40°C?

2. 0.050 kg of ice at $-10$°C is placed in 0.200 kg of water at 30°C. Find the final temperature (in °C, round to 1 decimal).

3. What mass of steam at 100°C must be condensed to provide enough energy to melt 2.00 kg of ice at 0°C? (in kg, round to 2 decimals)

Round all answers to 3 significant figures.

AP FRQ Preview

AP Physics 2 free-response questions on this topic typically ask you to:

Common FRQ Formats

1. Multi-step heating/cooling: "Calculate the total energy to convert X kg of ice at $T_1$ to steam at $T_2$." You must identify all stages, write the equation for each, and sum them.

2. Calorimetry with unknowns: "Ice at 0°C is mixed with water at $T$. Find the final temperature." You must check whether the phase change completes before setting up the energy equation.

3. Graphical analysis: Given a heating curve, identify phases, phase changes, and calculate specific heats or latent heats from the graph data.

4. Conceptual justification: "Explain why steam at 100°C causes worse burns than water at 100°C." Reference latent heat of vaporization being released during condensation.

Key Phrases for Full Credit

• "Energy is conserved in an isolated system"
• "During a phase change, temperature remains constant because energy changes intermolecular potential energy, not kinetic energy"
• "The latent heat of vaporization is much larger than the latent heat of fusion because vaporization requires completely separating molecules"

Final Mastery Quiz 🏆