Unit 6 · Topic 6.5 Beta

Energy of Phase Changes

During a phase change, energy overcomes (or forms) attractions between particles, so the temperature stays constant.

Practice 4: Model AnalysisPractice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

Put a pot of water on the stove. The temperature climbs steadily to 100 °C, then stops, even though the burner is still on full. For several minutes the water just boils, sitting at 100 °C. All that energy is going somewhere, but not into making the water hotter.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. When water boils, what is overcome?

  1. Hydrogen bonds between water molecules
  2. O–H covalent bonds inside each molecule
  3. Ionic bonds
  4. Nothing
Show the answer

Phase changes overcome attractions between molecules; the molecules stay intact.

  • Correct: Hydrogen bonds between water molecules:
  • O–H covalent bonds inside each molecule:
  • Ionic bonds:
  • Nothing:

2. How much energy warms 10.0 g of water by 5.0 °C? (c = 4.18 J/(g·°C))

  1. 209 J
  2. 50 J
  3. 20.9 J
  4. 2.09 J
Show the answer

q = mcΔT = 10.0 × 4.18 × 5.0 = 209 J.

  • Correct: 209 J:
  • 50 J:
  • 20.9 J:
  • 2.09 J:

Part 4 · See it

See it first

Heating curve of water, temperature against energy added (not to scale): ice warms to 0 °C, a flat part while ice melts (q = n·ΔHfus), liquid warms to 100 °C (q = m·c·ΔT), a much longer flat part while water boils (q = n·ΔHvap), then steam warms. Ice and steam warm along steeper slopes than the liquid, since their specific heats are about half of liquid water's. On the flat parts the temperature is constant and the energy overcomes attractions between molecules.
On the sloped parts one phase warms (q = mcΔT). On the flat parts two phases coexist and the temperature stays constant (q = nΔH). LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Energy is added to a solidits particles vibrate faster and its temperature rises until it reaches the melting point
  2. At the melting point, added energy overcomes the attractions holding particles in placethe solid melts at constant temperature; potential energy rises, kinetic energy does not
  3. Once melted, added energy speeds the particles up againthe liquid warms until the boiling point
  4. Boiling separates the particles almost completelythe boiling plateau is longest: ΔHvap is larger than ΔHfus
  5. Running the curve backward (cooling)condensing and freezing release the same energies: q = −nΔH

Part 6 · Key ideas

Key ideas

  • The enthalpy of fusion (ΔHfus) is the energy to melt 1 mol of solid at its melting point; the enthalpy of vaporization (ΔHvap) is the energy to vaporize 1 mol of liquid.
  • Melting, vaporizing and subliming absorb energy (q = +nΔH); freezing and condensing release it (q = −nΔH).
  • ΔHvap > ΔHfus: vaporizing separates molecules almost completely.
  • On a heating curve, slopes use q = mcΔT and flat parts use q = nΔH; add the steps.
  • Stronger attractions between particles mean larger ΔHfus and ΔHvap.

Part 7 · Misconception

A common mistake

The wrong idea: While water boils, the energy from the burner makes the water molecules move faster.

What actually happens: The temperature stays at 100 °C, so the average speed of the molecules does not change. The energy goes into separating the molecules from each other.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

Heating curve of Compound Q

A 47.05 g sample of solid Compound Q (molar mass 94.10 g/mol) at 20.0 °C is heated by a heater that supplies 150. J of energy to the sample each second. Melting runs from 9.6 s to 47.2 s; boiling runs from 142.3 s to 295.3 s.

0501001502000100200300Time (s)Temperature (°C)
Data table
Time (s)Compound Q
020
9.641
47.241
142.3182
295.3182
303.8200

1. During which time interval is Compound Q entirely liquid?

  1. From 47.2 s to 142.3 s
  2. From 9.6 s to 47.2 s
  3. From 142.3 s to 295.3 s
  4. From 0 s to 9.6 s
Show the answer

Sloped parts show one phase warming; flat parts show two phases changing at constant temperature. The liquid alone is the slope between the two flat parts.

  • Correct: From 47.2 s to 142.3 s: Right: melting has finished at the end of the first flat part, and boiling has not yet begun.
  • From 9.6 s to 47.2 s: This flat part is melting: solid and liquid are both present.
  • From 142.3 s to 295.3 s: This flat part is boiling: liquid and gas are both present.
  • From 0 s to 9.6 s: Before the first flat part, the sample is a solid warming up.

2. Calculate the molar enthalpy of fusion of Compound Q, in kJ/mol.

Type a number and its unit.

Show the answer

Melting time = 47.2 s − 9.6 s = 37.6 s. Energy = 37.6 s × 150. J/s = 5640 J = 5.64 kJ. Moles = 47.05 g ÷ 94.10 g/mol = 0.500 mol. ΔHfus = 5.64 kJ ÷ 0.500 mol = 11.28 → 11.3 kJ/mol.

  • Answer: 11.3 kJ/mol

3. The boiling plateau is about four times as long as the melting plateau. Which explanation is correct?

  1. Boiling breaks the covalent bonds inside the molecules of Compound Q; melting leaves them intact.
  2. The temperature is higher during boiling, so each molecule needs more energy to move.
  3. Boiling pulls the molecules almost fully apart; melting just loosens their attractions.
  4. There is more liquid than solid in the sample, so boiling takes longer.
Show the answer

Melting turns a fixed arrangement into a liquid where molecules still touch; vaporizing pulls them apart into a gas. More attractions are overcome, so ΔHvap > ΔHfus.

  • Boiling breaks the covalent bonds inside the molecules of Compound Q; melting leaves them intact.: No covalent bonds break in either change; the molecules are still Compound Q in the gas.
  • The temperature is higher during boiling, so each molecule needs more energy to move.: The plateau length measures energy to overcome attractions, not the temperature.
  • Correct: Boiling pulls the molecules almost fully apart; melting just loosens their attractions.: Right: in the liquid, molecules still touch and attract; boiling overcomes nearly all those attractions.
  • There is more liquid than solid in the sample, so boiling takes longer.: The same 0.500 mol of Compound Q melts and then boils.

4. Between 142.3 s and 295.3 s, what happens to the average kinetic energy of the molecules of Compound Q?

  1. It increases, because the heater keeps adding 150. J of energy to the sample each second.
  2. It decreases, because the molecules spread far apart from each other as they become a gas.
  3. It increases, because covalent bonds break and release energy.
  4. It stays the same, because the energy goes into overcoming attractions between molecules.
Show the answer

Temperature tracks average kinetic energy. On a flat part, the added energy separates molecules (potential energy goes up) while their average speed stays constant.

  • It increases, because the heater keeps adding 150. J of energy to the sample each second.: The energy is added, but it does not speed the molecules up while the temperature is flat.
  • It decreases, because the molecules spread far apart from each other as they become a gas.: Spreading out raises their potential energy; their average speed stays the same at constant temperature.
  • It increases, because covalent bonds break and release energy.: No covalent bonds break, and breaking bonds would absorb energy, not release it.
  • Correct: It stays the same, because the energy goes into overcoming attractions between molecules.: Right: constant temperature means constant average kinetic energy; the energy raises the potential energy.

5. Why does the temperature of a melting ice-water mixture stay at 0 °C while energy is added?

  1. The added energy is lost to the room as fast as it is added.
  2. The thermometer stops reading temperature changes while ice is present.
  3. The added energy overcomes attractions between water molecules instead of speeding them up.
  4. The added energy breaks O–H bonds inside the water molecules.
Show the answer

During a phase change, energy goes into potential energy (separating molecules), not kinetic energy, so the temperature is constant until the change is complete.

  • The added energy is lost to the room as fast as it is added.: In a sealed, insulated container, the temperature still stays at 0 °C until the ice is gone.
  • The thermometer stops reading temperature changes while ice is present.: The thermometer works fine; the temperature really is constant.
  • Correct: The added energy overcomes attractions between water molecules instead of speeding them up.: Right: potential energy rises as the solid structure comes apart; average kinetic energy, and so temperature, stays the same.
  • The added energy breaks O–H bonds inside the water molecules.: The molecules stay intact as H₂O; only attractions between them are overcome.

6. How much energy, in kJ, is needed to turn 18.0 g of ice at −10.0 °C into liquid water at 25.0 °C? Use c(ice) = 2.09 J/(g·°C), c(water) = 4.18 J/(g·°C), ΔHfus = 6.01 kJ/mol and 18.02 g/mol.

Type a number and its unit.

Show the answer

Warm the ice: 18.0 × 2.09 × 10.0 = 376 J. Melt: (18.0 ÷ 18.02) mol × 6.01 kJ/mol = 6.003 kJ. Warm the liquid: 18.0 × 4.18 × 25.0 = 1881 J. Total = 0.376 + 6.003 + 1.881 = 8.261 → 8.26 kJ.

  • Answer: 8.26 kJ

7. A drink is cooled with either 50 g of ice at 0 °C or 50 g of liquid water at 0 °C. Which cools the drink more, and why?

  1. The cold water, because liquid water has a higher specific heat than ice.
  2. They cool it equally, because both start at 0 °C.
  3. The ice, because melting it absorbs energy from the drink before the meltwater starts to warm.
  4. The cold water, because ice releases energy as it melts.
Show the answer

Melting is endothermic. The ice takes energy from the drink to melt (6.01 kJ/mol, about 334 J/g) and then more to warm, so it cools the drink more.

  • The cold water, because liquid water has a higher specific heat than ice.: Both warm up from 0 °C, but only the ice also absorbs energy by melting.
  • They cool it equally, because both start at 0 °C.: Same temperature, but the ice must also melt, which absorbs energy.
  • Correct: The ice, because melting it absorbs energy from the drink before the meltwater starts to warm.: Right: each gram of ice absorbs about 334 J just to melt, on top of the energy it absorbs warming up.
  • The cold water, because ice releases energy as it melts.: Melting is endothermic: it absorbs energy.

Part 9 · Summary

Summary

During a phase change, energy overcomes (or forms) attractions between particles, so the temperature stays constant. The energy is n × ΔHfus for melting and n × ΔHvap for boiling, with ΔHvap the larger. A heating curve combines sloped parts (q = mcΔT) and flat parts (q = nΔH).

Part 10 · Up next

What comes next

Part 11 · Connections

Connections