Phase Changes and Latent Heat

States of matter, phase transitions, and latent heat

🧊💧💨 Phase Changes and Latent Heat

States of Matter

Solid

Fixed shape and volume
Particles vibrate in fixed positions
Strong intermolecular forces

Liquid

Fixed volume, takes shape of container
Particles can flow past each other
Moderate intermolecular forces

Gas

No fixed shape or volume
Particles move freely
Weak intermolecular forces

Phase Changes

Six Types of Phase Transitions:

Adding Energy:

Melting (Fusion): Solid → Liquid
Vaporization: Liquid → Gas
Sublimation: Solid → Gas (direct)

Removing Energy: 4. Freezing (Solidification): Liquid → Solid 5. Condensation: Gas → Liquid 6. Deposition: Gas → Solid (direct)

💡 Key Insight: During a phase change, temperature remains constant even as energy is added/removed!

Latent Heat

Latent heat is energy required to change phase without changing temperature.

Heat of Fusion (L_f)

Energy to melt (or freeze) 1 kg of substance:

$Q = mL_f$

For water: $L_f = 334,000$ J/kg = 334 kJ/kg

Heat of Vaporization (L_v)

Energy to vaporize (or condense) 1 kg of substance:

$Q = mL_v$

For water: $L_v = 2,260,000$ J/kg = 2260 kJ/kg

Note: $L_v > L_f$ because you must completely overcome intermolecular forces for vaporization.

Heating Curve for Water

When heating ice from -20°C to steam at 120°C:

Stage 1: Ice warms (-20°C → 0°C) $Q_1 = m c_{ice} \Delta T = m(2090)(20)$

Stage 2: Ice melts at 0°C (phase change) $Q_2 = m L_f = m(334,000)$ Temperature stays at 0°C!

Stage 3: Water warms (0°C → 100°C) $Q_3 = m c_{water} \Delta T = m(4186)(100)$

Stage 4: Water boils at 100°C (phase change) $Q_4 = m L_v = m(2,260,000)$ Temperature stays at 100°C!

Stage 5: Steam warms (100°C → 120°C) $Q_5 = m c_{steam} \Delta T = m(2010)(20)$

Total energy: $Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5$

Why Temperature Stays Constant

During phase changes:

Energy goes into breaking molecular bonds
Does NOT increase kinetic energy (temperature)
Potential energy increases, kinetic energy constant
Both phases coexist at transition temperature

Example: Ice-water mixture stays at 0°C until all ice melts.

Evaporation vs. Boiling

Evaporation

Occurs at surface
Happens at any temperature
Faster molecules escape
Causes cooling (loses high-energy molecules)

Boiling

Occurs throughout liquid
Happens at specific temperature (boiling point)
Vapor pressure equals atmospheric pressure
Bubbles form inside liquid

Important Values for Water

| Property | Value | |----------|-------| | Melting point | 0°C (273 K) | | Boiling point | 100°C (373 K) at 1 atm | | L_f (fusion) | 334 kJ/kg | | L_v (vaporization) | 2260 kJ/kg | | c_ice | 2090 J/kg·°C | | c_water | 4186 J/kg·°C | | c_steam | 2010 J/kg·°C |

Pressure Dependence

Higher pressure → higher boiling point
- Pressure cooker: cooks faster at higher T
- Mountain top: water boils below 100°C
Lower pressure → lower boiling point
- Vacuum chamber: water boils at room temp

Melting point is less affected by pressure (except for water/ice).

Problem-Solving Strategy

Identify all stages of heating/cooling
Separate into segments:
- Temperature changes: $Q = mc\Delta T$
- Phase changes: $Q = mL$
Add all energy contributions: $Q_{total} = \sum Q_i$
Watch for mixed-phase problems (some melts, some doesn't)
Check if enough energy for complete phase change

Common Mistakes

❌ Forgetting temperature is constant during phase change ❌ Using wrong specific heat (ice vs. water vs. steam) ❌ Adding ΔT during phase change (it's zero!) ❌ Confusing L_f and L_v ❌ Not checking if phase change is complete

📚 Practice Problems

1Problem 1easy

❓ Question:

How much energy is required to melt 2.0 kg of ice at 0°C? (L_f = 334 kJ/kg)

💡 Show Solution

Given:

Mass: $m = 2.0$ kg
Temperature: Already at 0°C (melting point)
Heat of fusion: $L_f = 334$ kJ/kg $= 334,000$ J/kg

Find: Energy required $Q$

Solution:

Since ice is already at melting point, only need energy for phase change:

$Q = mL_f = (2.0)(334,000) = 668,000 \text{ J} = 668 \text{ kJ}$

Answer: 668 kJ is required to melt the ice

Note: Temperature stays at 0°C during melting!

2Problem 2easy

❓ Question:

How much energy is required to melt 2.0 kg of ice at 0°C? (L_f = 334 kJ/kg)

💡 Show Solution

Given:

Mass: $m = 2.0$ kg
Temperature: Already at 0°C (melting point)
Heat of fusion: $L_f = 334$ kJ/kg $= 334,000$ J/kg

Find: Energy required $Q$

Solution:

Since ice is already at melting point, only need energy for phase change:

$Q = mL_f = (2.0)(334,000) = 668,000 \text{ J} = 668 \text{ kJ}$

Answer: 668 kJ is required to melt the ice

Note: Temperature stays at 0°C during melting!

3Problem 3medium

❓ Question:

How much energy is needed to convert 0.50 kg of ice at -10°C to water at 20°C? (c_ice = 2090 J/kg·°C, L_f = 334 kJ/kg, c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Mass: $m = 0.50$ kg
Initial: -10°C (ice)
Final: 20°C (water)

Find: Total energy $Q_{total}$

Solution:

Stage 1: Heat ice from -10°C to 0°C $Q_1 = mc_{ice}\Delta T = (0.50)(2090)(10) = 10,450 \text{ J}$

Stage 2: Melt ice at 0°C $Q_2 = mL_f = (0.50)(334,000) = 167,000 \text{ J}$

Stage 3: Heat water from 0°C to 20°C $Q_3 = mc_{water}\Delta T = (0.50)(4186)(20) = 41,860 \text{ J}$

Total energy: $Q_{total} = Q_1 + Q_2 + Q_3$ $Q_{total} = 10,450 + 167,000 + 41,860$ $Q_{total} = 219,310 \text{ J} = 219 \text{ kJ}$

Answer: 219 kJ is required

Note: Most energy goes to melting (76%), not warming!

4Problem 4hard

❓ Question:

How much energy is required to convert 1.0 kg of ice at -10°C to steam at 110°C? Use: c_ice = 2100 J/(kg·°C), c_water = 4186 J/(kg·°C), c_steam = 2010 J/(kg·°C), L_f = 3.34 × 10⁵ J/kg, L_v = 2.26 × 10⁶ J/kg.

💡 Show Solution

Solution:

Five stages:

1) Heat ice from -10°C to 0°C: Q₁ = mc_ice ΔT = (1.0)(2100)(10) = 21,000 J

2) Melt ice at 0°C: Q₂ = mL_f = (1.0)(3.34 × 10⁵) = 334,000 J

3) Heat water from 0°C to 100°C: Q₃ = mc_water ΔT = (1.0)(4186)(100) = 418,600 J

4) Vaporize water at 100°C: Q₄ = mL_v = (1.0)(2.26 × 10⁶) = 2,260,000 J

5) Heat steam from 100°C to 110°C: Q₅ = mc_steam ΔT = (1.0)(2010)(10) = 20,100 J

Total: Q_total = 21,000 + 334,000 + 418,600 + 2,260,000 + 20,100 Q_total = 3.05 × 10⁶ J or 3.05 MJ or 3050 kJ

Most energy goes into vaporization!

5Problem 5medium

❓ Question:

How much energy is needed to convert 0.50 kg of ice at -10°C to water at 20°C? (c_ice = 2090 J/kg·°C, L_f = 334 kJ/kg, c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Mass: $m = 0.50$ kg
Initial: -10°C (ice)
Final: 20°C (water)

Find: Total energy $Q_{total}$

Solution:

Stage 1: Heat ice from -10°C to 0°C $Q_1 = mc_{ice}\Delta T = (0.50)(2090)(10) = 10,450 \text{ J}$

Stage 2: Melt ice at 0°C $Q_2 = mL_f = (0.50)(334,000) = 167,000 \text{ J}$

Stage 3: Heat water from 0°C to 20°C $Q_3 = mc_{water}\Delta T = (0.50)(4186)(20) = 41,860 \text{ J}$

Total energy: $Q_{total} = Q_1 + Q_2 + Q_3$ $Q_{total} = 10,450 + 167,000 + 41,860$ $Q_{total} = 219,310 \text{ J} = 219 \text{ kJ}$

Answer: 219 kJ is required

Note: Most energy goes to melting (76%), not warming!

6Problem 6hard

❓ Question:

💡 Show Solution

Solution:

Five stages:

1) Heat ice from -10°C to 0°C: Q₁ = mc_ice ΔT = (1.0)(2100)(10) = 21,000 J

2) Melt ice at 0°C: Q₂ = mL_f = (1.0)(3.34 × 10⁵) = 334,000 J

3) Heat water from 0°C to 100°C: Q₃ = mc_water ΔT = (1.0)(4186)(100) = 418,600 J

4) Vaporize water at 100°C: Q₄ = mL_v = (1.0)(2.26 × 10⁶) = 2,260,000 J

5) Heat steam from 100°C to 110°C: Q₅ = mc_steam ΔT = (1.0)(2010)(10) = 20,100 J

Total: Q_total = 21,000 + 334,000 + 418,600 + 2,260,000 + 20,100 Q_total = 3.05 × 10⁶ J or 3.05 MJ or 3050 kJ

Most energy goes into vaporization!

7Problem 7hard

❓ Question:

A 0.10 kg piece of ice at 0°C is placed in 0.50 kg of water at 30°C in an insulated container. Does all the ice melt? If so, what is the final temperature? If not, how much ice melts? (L_f = 334 kJ/kg, c_water = 4186 J/kg·°C)

💡 Show Solution

Given:

Ice: $m_i = 0.10$ kg at 0°C
Water: $m_w = 0.50$ kg at 30°C
$L_f = 334,000$ J/kg, $c_w = 4186$ J/kg·°C

Solution:

Step 1: Find maximum energy available from cooling water to 0°C. $Q_{available} = m_w c_w \Delta T = (0.50)(4186)(30) = 62,790 \text{ J}$

Step 2: Find energy needed to melt all ice. $Q_{needed} = m_i L_f = (0.10)(334,000) = 33,400 \text{ J}$

Step 3: Compare. $Q_{available} = 62,790 \text{ J} > Q_{needed} = 33,400 \text{ J}$

All ice melts! ✓ Excess energy warms the resulting water.

Step 4: Find final temperature.

Energy conservation: $Q_{ice} + Q_{water} = 0$ $m_i L_f + m_i c_w(T_f - 0) + m_w c_w(T_f - 30) = 0$

$33,400 + (0.10)(4186)T_f + (0.50)(4186)(T_f - 30) = 0$ $33,400 + 418.6T_f + 2093T_f - 62,790 = 0$ $2511.6T_f = 29,390$ $T_f = 11.7°\text{C}$

Answer: All ice melts, final temperature is 11.7°C

8Problem 8hard

❓ Question:

💡 Show Solution

Given:

Ice: $m_i = 0.10$ kg at 0°C
Water: $m_w = 0.50$ kg at 30°C
$L_f = 334,000$ J/kg, $c_w = 4186$ J/kg·°C

Solution:

Step 1: Find maximum energy available from cooling water to 0°C. $Q_{available} = m_w c_w \Delta T = (0.50)(4186)(30) = 62,790 \text{ J}$

Step 2: Find energy needed to melt all ice. $Q_{needed} = m_i L_f = (0.10)(334,000) = 33,400 \text{ J}$

Step 3: Compare. $Q_{available} = 62,790 \text{ J} > Q_{needed} = 33,400 \text{ J}$

All ice melts! ✓ Excess energy warms the resulting water.

Step 4: Find final temperature.

Energy conservation: $Q_{ice} + Q_{water} = 0$ $m_i L_f + m_i c_w(T_f - 0) + m_w c_w(T_f - 30) = 0$

$33,400 + (0.10)(4186)T_f + (0.50)(4186)(T_f - 30) = 0$ $33,400 + 418.6T_f + 2093T_f - 62,790 = 0$ $2511.6T_f = 29,390$ $T_f = 11.7°\text{C}$

Answer: All ice melts, final temperature is 11.7°C

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