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Phase Changes and Latent Heat

States of matter, phase transitions, and latent heat

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🧊💧💨 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:

  1. Melting (Fusion): Solid → Liquid
  2. Vaporization: Liquid → Gas
  3. 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=mLfQ = mL_f

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

Heat of Vaporization (L_v)

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

Q=mLvQ = mL_v

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

Note: Lv>LfL_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) Q1=mciceΔT=m(2090)(20)Q_1 = m c_{ice} \Delta T = m(2090)(20)

Stage 2: Ice melts at 0°C (phase change) Q2=mLf=m(334,000)Q_2 = m L_f = m(334,000) Temperature stays at 0°C!

Stage 3: Water warms (0°C → 100°C) Q3=mcwaterΔT=m(4186)(100)Q_3 = m c_{water} \Delta T = m(4186)(100)

Stage 4: Water boils at 100°C (phase change) Q4=mLv=m(2,260,000)Q_4 = m L_v = m(2,260,000) Temperature stays at 100°C!

Stage 5: Steam warms (100°C → 120°C) Q5=mcsteamΔT=m(2010)(20)Q_5 = m c_{steam} \Delta T = m(2010)(20)

Total energy: Qtotal=Q1+Q2+Q3+Q4+Q5Q_{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

PropertyValue
Melting point0°C (273 K)
Boiling point100°C (373 K) at 1 atm
L_f (fusion)334 kJ/kg
L_v (vaporization)2260 kJ/kg
c_ice2090 J/kg·°C
c_water4186 J/kg·°C
c_steam2010 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

  1. Identify all stages of heating/cooling
  2. Separate into segments:
    • Temperature changes: Q=mcΔTQ = mc\Delta T
    • Phase changes: Q=mLQ = mL
  3. Add all energy contributions: Qtotal=∑QiQ_{total} = \sum Q_i
  4. Watch for mixed-phase problems (some melts, some doesn't)
  5. 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=2.0m = 2.0 kg
  • Temperature: Already at 0°C (melting point)
  • Heat of fusion: Lf=334L_f = 334 kJ/kg =334,000= 334,000 J/kg

Find: Energy required QQ

Solution:

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

Q=mLf=(2.0)(334,000)=668,000 J=668 kJQ = 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=2.0m = 2.0 kg
  • Temperature: Already at 0°C (melting point)
  • Heat of fusion: Lf=334L_f = 334 kJ/kg =334,000= 334,000 J/kg

Find: Energy required QQ

Solution:

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

Q=mLf=(2.0)(334,000)=668,000 J=668 kJQ = 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 3hard

❓ 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!

4Problem 4medium

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

Find: Total energy QtotalQ_{total}

Solution:

Stage 1: Heat ice from -10°C to 0°C Q1=mciceΔT=(0.50)(2090)(10)=10,450 JQ_1 = mc_{ice}\Delta T = (0.50)(2090)(10) = 10,450 \text{ J}

Stage 2: Melt ice at 0°C Q2=mLf=(0.50)(334,000)=167,000 JQ_2 = mL_f = (0.50)(334,000) = 167,000 \text{ J}

Stage 3: Heat water from 0°C to 20°C Q3=mcwaterΔT=(0.50)(4186)(20)=41,860 JQ_3 = mc_{water}\Delta T = (0.50)(4186)(20) = 41,860 \text{ J}

Total energy: Qtotal=Q1+Q2+Q3Q_{total} = Q_1 + Q_2 + Q_3 Qtotal=10,450+167,000+41,860Q_{total} = 10,450 + 167,000 + 41,860 Qtotal=219,310 J=219 kJQ_{total} = 219,310 \text{ J} = 219 \text{ kJ}

Answer: 219 kJ is required

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

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

Find: Total energy QtotalQ_{total}

Solution:

Stage 1: Heat ice from -10°C to 0°C Q1=mciceΔT=(0.50)(2090)(10)=10,450 JQ_1 = mc_{ice}\Delta T = (0.50)(2090)(10) = 10,450 \text{ J}

Stage 2: Melt ice at 0°C Q2=mLf=(0.50)(334,000)=167,000 JQ_2 = mL_f = (0.50)(334,000) = 167,000 \text{ J}

Stage 3: Heat water from 0°C to 20°C Q3=mcwaterΔT=(0.50)(4186)(20)=41,860 JQ_3 = mc_{water}\Delta T = (0.50)(4186)(20) = 41,860 \text{ J}

Total energy: Qtotal=Q1+Q2+Q3Q_{total} = Q_1 + Q_2 + Q_3 Qtotal=10,450+167,000+41,860Q_{total} = 10,450 + 167,000 + 41,860 Qtotal=219,310 J=219 kJQ_{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:

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: mi=0.10m_i = 0.10 kg at 0°C
  • Water: mw=0.50m_w = 0.50 kg at 30°C
  • Lf=334,000L_f = 334,000 J/kg, cw=4186c_w = 4186 J/kg·°C

Solution:

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

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

Step 3: Compare. Qavailable=62,790 J>Qneeded=33,400 JQ_{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: Qice+Qwater=0Q_{ice} + Q_{water} = 0 miLf+micw(Tf−0)+mwcw(Tf−30)=0m_i L_f + m_i c_w(T_f - 0) + m_w c_w(T_f - 30) = 0

33,400+(0.10)(4186)Tf+(0.50)(4186)(Tf−30)=033,400 + (0.10)(4186)T_f + (0.50)(4186)(T_f - 30) = 0 33,400+418.6Tf+2093Tf−62,790=033,400 + 418.6T_f + 2093T_f - 62,790 = 0 2511.6Tf=29,3902511.6T_f = 29,390 Tf=11.7°CT_f = 11.7°\text{C}

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

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: mi=0.10m_i = 0.10 kg at 0°C
  • Water: mw=0.50m_w = 0.50 kg at 30°C
  • Lf=334,000L_f = 334,000 J/kg, cw=4186c_w = 4186 J/kg·°C

Solution:

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

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

Step 3: Compare. Qavailable=62,790 J>Qneeded=33,400 JQ_{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: Qice+Qwater=0Q_{ice} + Q_{water} = 0 miLf+micw(Tf−0)+mwcw(Tf−30)=0m_i L_f + m_i c_w(T_f - 0) + m_w c_w(T_f - 30) = 0

33,400+(0.10)(4186)Tf+(0.50)(4186)(Tf−30)=033,400 + (0.10)(4186)T_f + (0.50)(4186)(T_f - 30) = 0 33,400+418.6Tf+2093Tf−62,790=033,400 + 418.6T_f + 2093T_f - 62,790 = 0 2511.6Tf=29,3902511.6T_f = 29,390 Tf=11.7°CT_f = 11.7°\text{C}

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

Explain using:
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