Chemistry · Simulation and virtual lab Free
Changes of state: heating and cooling curves
Heat ice until it boils, or let melted stearic acid cool, record the temperature every 30 seconds and explain the flat parts of the graph with the particle model.
- You will learn to
- Plot heating and cooling curves from temperature readings taken every 30 seconds.
- Identify melting, freezing and boiling points and explain the flat parts with the particle model.
- Estimate the specific latent heat of vaporisation of water, and compare pure and impure substances.
- You will measure
- Temperature (°C) every 30 s; energy (kJ) and mass of water (g); then L_v in MJ/kg and the freezing point.
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: beaker, crushed ice, immersion heater with joulemeter, thermometer, stirrer, balance, stopwatch; boiling tube of stearic acid, clamp stand.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To plot heating and cooling curves, to identify melting, freezing and boiling points from the flat parts of the graphs, to explain them with the kinetic particle model, and to estimate the specific latent heat of vaporisation of water.
What to do
- Choose Heat ice until it boils: 200 g of crushed ice, 500 W heater, lagged beaker, clock × 10, record every 30 s.
- Switch the heater on. Watch the thermometer, the beaker and the particles. Note what happens at 0 °C and at 100 °C.
- Let the water boil for at least 3 minutes (sim time), then switch the heater off. Look at the graph and the joulemeter and balance readings in the table.
- Use two rows taken while boiling: Lv = (energy supplied) ÷ (mass boiled away). Compare with the box under the graph.
- Repeat with the beaker not lagged. Why is the value of Lv now different?
- Choose Cooling curve of stearic acid. Tap Start cooling, clock × 30. Find the freezing point from the flat part. Then repeat with the impure sample and compare.
Results
Questions
Why does the temperature stay at 0 °C while the ice melts, although the heater is on?
The energy supplied is used to break the bonds (forces of attraction) between the particles of the solid, not to make them move faster. The average kinetic energy, and so the temperature, stays the same until all the ice has melted.
Which part of the heating curve is the steepest, and why?
The ice part: ice has a smaller specific heat capacity (2 100 J/kg K) than water (4 200 J/kg K), so the same energy raises its temperature more.
Why is the flat part at 100 °C much longer than the flat part at 0 °C?
The specific latent heat of vaporisation (2.26 MJ/kg) is much larger than that of fusion (0.334 MJ/kg): to boil, the particles must be separated completely, which needs far more energy than loosening them in melting.
Describe the arrangement and motion of the particles in a solid, a liquid and a gas.
Solid: close together in a regular pattern, vibrating about fixed positions. Liquid: close together but irregular, sliding past each other. Gas: far apart, moving fast and randomly in all directions.
How does the cooling curve show that a substance is pure?
A pure substance freezes at one fixed temperature: the cooling curve has a flat part. An impure substance freezes over a range of temperatures, below the pure freezing point, with no flat part.
Why is the value of L_v too large when the beaker is not lagged?
Some energy from the heater is lost to the surroundings and does not boil water. The calculation assumes all the energy boiled the water, so E ÷ m comes out too large.
For teachers: expected results and how the model works
Water (lagged, 200 g ice at −10 °C, 500 W): ice reaches 0 °C after 8.4 s; melting lasts 0.2 × 334 000 ÷ 500 = 134 s; the water reaches 100 °C after a further 168 s (about 310 s in total); then 0.221 g boils away every second, so any two rows taken while boiling give L_v = 2.26 MJ/kg. Not lagged: heat flows in below 25 °C and out above it (1.2 W per K), so melting is shorter, heating is slower and L_v comes out too large. Stearic acid (20 g in a boiling tube, cooling in air): falls from 85 °C to 69 °C in about 3 minutes, stays at 69.0 °C for about 14 minutes while freezing, then cools slowly. The impure sample starts to freeze at 66 °C and finishes at 59 °C with no flat part. Use clock × 30 for the cooling curve.
Exam practice Exam set
National-exam style questions on this simulation (16 marks). Try each one on paper first, then open the marking scheme.
A 300 W immersion heater is placed in boiling water on a balance. In 5.0 minutes the mass falls by 40 g. (a) Calculate the energy supplied. (b) Calculate the specific latent heat of vaporisation. (c) Suggest why the accepted value is smaller. (5 marks)
Marking scheme: (a) E = Pt = 300 × 300 (1) = 9.0 × 10⁴ J (1). (b) L = E ÷ m = 9.0 × 10⁴ ÷ 0.040 (1) = 2.25 × 10⁶ J/kg (1). (c) Some energy is lost to the surroundings, so not all of it boils water (1).
Calculate the energy needed to change 0.50 kg of ice at −10 °C into water at 20 °C. (c of ice = 2 100 J/kg K, c of water = 4 200 J/kg K, L_f = 3.34 × 10⁵ J/kg) (4 marks)
Marking scheme: Warm the ice: 0.50 × 2 100 × 10 = 10 500 J (1). Melt: 0.50 × 3.34 × 10⁵ = 167 000 J (1). Warm the water: 0.50 × 4 200 × 20 = 42 000 J (1). Total = 2.20 × 10⁵ J (1).
Sketch the cooling curve of molten stearic acid from 85 °C to room temperature. Label the liquid, freezing and solid parts and mark the freezing point (69 °C). (4 marks)
Marking scheme: Temperature falls steeply at first (liquid) (1); a horizontal section at 69 °C labelled freezing / solid + liquid (1); then falls again, more slowly, towards room temperature (solid) (1); axes labelled temperature / °C and time / s (1).
Use the kinetic theory to explain why sweating cools the body. (3 marks)
Marking scheme: The fastest-moving water particles escape from the surface of the sweat (evaporation) (1); the average kinetic energy of the remaining particles falls, so the temperature falls (1); the latent heat of vaporisation is taken from the skin (1).
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Changes of state: heating and cooling curves
Aim: To plot heating and cooling curves, to identify melting, freezing and boiling points from the flat parts of the graphs, to explain them with the kinetic particle model, and to estimate the specific latent heat of vaporisation of water.
Apparatus (real laboratory): beaker, crushed ice, immersion heater with joulemeter, thermometer, stirrer, balance, stopwatch; boiling tube of stearic acid, clamp stand.
What to do (with the simulation on www.donatstem.co.rw)
- Choose Heat ice until it boils: 200 g of crushed ice, 500 W heater, lagged beaker, clock × 10, record every 30 s.
- Switch the heater on. Watch the thermometer, the beaker and the particles. Note what happens at 0 °C and at 100 °C.
- Let the water boil for at least 3 minutes (sim time), then switch the heater off. Look at the graph and the joulemeter and balance readings in the table.
- Use two rows taken while boiling: Lv = (energy supplied) ÷ (mass boiled away). Compare with the box under the graph.
- Repeat with the beaker not lagged. Why is the value of Lv now different?
- Choose Cooling curve of stearic acid. Tap Start cooling, clock × 30. Find the freezing point from the flat part. Then repeat with the impure sample and compare.
Results
| # | t (s) | Temperature (°C) | Energy (kJ) | Mass of water (g) | State |
|---|
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
| 7 | | | | | |
| 8 | | | | | |
| 9 | | | | | |
| 10 | | | | | |
| 11 | | | | | |
| 12 | | | | | |
| 13 | | | | | |
| 14 | | | | | |
Graph
Plot temperature (°C) on the vertical axis against time (s). Join the points with a smooth line and label each part with the state of the substance.
Questions
- Why does the temperature stay at 0 °C while the ice melts, although the heater is on? [2]
- Which part of the heating curve is the steepest, and why? [2]
- Why is the flat part at 100 °C much longer than the flat part at 0 °C? [2]
- Describe the arrangement and motion of the particles in a solid, a liquid and a gas. [2]
- How does the cooling curve show that a substance is pure? [2]
- Why is the value of L_v too large when the beaker is not lagged? [2]
- A 300 W immersion heater is placed in boiling water on a balance. In 5.0 minutes the mass falls by 40 g. (a) Calculate the energy supplied. (b) Calculate the specific latent heat of vaporisation. (c) Suggest why the accepted value is smaller. [5]
- Calculate the energy needed to change 0.50 kg of ice at −10 °C into water at 20 °C. (c of ice = 2 100 J/kg K, c of water = 4 200 J/kg K, L_f = 3.34 × 10⁵ J/kg) [4]
- Sketch the cooling curve of molten stearic acid from 85 °C to room temperature. Label the liquid, freezing and solid parts and mark the freezing point (69 °C). [4]
- Use the kinetic theory to explain why sweating cools the body. [3]
Donat Sciences and MathsChemistry · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Changes of state: heating and cooling curves
Expected results
Water (lagged, 200 g ice at −10 °C, 500 W): ice reaches 0 °C after 8.4 s; melting lasts 0.2 × 334 000 ÷ 500 = 134 s; the water reaches 100 °C after a further 168 s (about 310 s in total); then 0.221 g boils away every second, so any two rows taken while boiling give L_v = 2.26 MJ/kg. Not lagged: heat flows in below 25 °C and out above it (1.2 W per K), so melting is shorter, heating is slower and L_v comes out too large. Stearic acid (20 g in a boiling tube, cooling in air): falls from 85 °C to 69 °C in about 3 minutes, stays at 69.0 °C for about 14 minutes while freezing, then cools slowly. The impure sample starts to freeze at 66 °C and finishes at 59 °C with no flat part. Use clock × 30 for the cooling curve.
Answers
- Why does the temperature stay at 0 °C while the ice melts, although the heater is on? [2]
The energy supplied is used to break the bonds (forces of attraction) between the particles of the solid, not to make them move faster. The average kinetic energy, and so the temperature, stays the same until all the ice has melted.
- Which part of the heating curve is the steepest, and why? [2]
The ice part: ice has a smaller specific heat capacity (2 100 J/kg K) than water (4 200 J/kg K), so the same energy raises its temperature more.
- Why is the flat part at 100 °C much longer than the flat part at 0 °C? [2]
The specific latent heat of vaporisation (2.26 MJ/kg) is much larger than that of fusion (0.334 MJ/kg): to boil, the particles must be separated completely, which needs far more energy than loosening them in melting.
- Describe the arrangement and motion of the particles in a solid, a liquid and a gas. [2]
Solid: close together in a regular pattern, vibrating about fixed positions. Liquid: close together but irregular, sliding past each other. Gas: far apart, moving fast and randomly in all directions.
- How does the cooling curve show that a substance is pure? [2]
A pure substance freezes at one fixed temperature: the cooling curve has a flat part. An impure substance freezes over a range of temperatures, below the pure freezing point, with no flat part.
- Why is the value of L_v too large when the beaker is not lagged? [2]
Some energy from the heater is lost to the surroundings and does not boil water. The calculation assumes all the energy boiled the water, so E ÷ m comes out too large.
- A 300 W immersion heater is placed in boiling water on a balance. In 5.0 minutes the mass falls by 40 g. (a) Calculate the energy supplied. (b) Calculate the specific latent heat of vaporisation. (c) Suggest why the accepted value is smaller. [5]
(a) E = Pt = 300 × 300 (1) = 9.0 × 10⁴ J (1). (b) L = E ÷ m = 9.0 × 10⁴ ÷ 0.040 (1) = 2.25 × 10⁶ J/kg (1). (c) Some energy is lost to the surroundings, so not all of it boils water (1).
- Calculate the energy needed to change 0.50 kg of ice at −10 °C into water at 20 °C. (c of ice = 2 100 J/kg K, c of water = 4 200 J/kg K, L_f = 3.34 × 10⁵ J/kg) [4]
Warm the ice: 0.50 × 2 100 × 10 = 10 500 J (1). Melt: 0.50 × 3.34 × 10⁵ = 167 000 J (1). Warm the water: 0.50 × 4 200 × 20 = 42 000 J (1). Total = 2.20 × 10⁵ J (1).
- Sketch the cooling curve of molten stearic acid from 85 °C to room temperature. Label the liquid, freezing and solid parts and mark the freezing point (69 °C). [4]
Temperature falls steeply at first (liquid) (1); a horizontal section at 69 °C labelled freezing / solid + liquid (1); then falls again, more slowly, towards room temperature (solid) (1); axes labelled temperature / °C and time / s (1).
- Use the kinetic theory to explain why sweating cools the body. [3]
The fastest-moving water particles escape from the surface of the sweat (evaporation) (1); the average kinetic energy of the remaining particles falls, so the temperature falls (1); the latent heat of vaporisation is taken from the skin (1).