1.2.1 Conservation of mass during changes of state
The five changes of state
Change of state
A physical change in which a substance changes between the solid, liquid and gas states without becoming a different substance.
Melting
The change of state from a solid to a liquid.
Freezing
The change of state from a liquid to a solid.
Evaporation
The change of state from a liquid to a gas.
Condensation
The change of state from a gas to a liquid.
Sublimation
The change of state directly from a solid to a gas without passing through the liquid state.
- A change of state happens when a substance moves between the solid, liquid and gas states, for example when ice turns into liquid water.
- In every change of state the particles stay exactly the same particles, and only their arrangement and movement change.
- Melting turns a solid into a liquid, as when a knob of butter becomes a golden liquid in a hot frying pan.
- Freezing is the reverse of melting and turns a liquid into a solid, as when water in an ice-cube tray becomes ice in the freezer.
- Evaporation turns a liquid into a gas, as when a puddle on the pavement slowly disappears on a warm afternoon.
- Condensation is the reverse of evaporation and turns a gas into a liquid, as when water vapour from a hot shower forms droplets on a cold bathroom mirror.
- Sublimation turns a solid straight into a gas without the substance ever passing through the liquid state.
- Solid carbon dioxide, known as dry ice, sublimes at −78 ∘C-78\ ^\circ\text{C}−78 ∘C, so a lump of it shrinks away to nothing without ever leaving a puddle.
- Melting and freezing are opposite processes, and so are evaporation and condensation, so each of these changes can be undone by its partner.
Conservation of mass
Conservation of mass
The principle that the total mass of a closed system stays the same during any change, because particles are neither created nor destroyed.
- Mass is a measure of the amount of matter in an object, and it depends on how many particles the object contains and on the mass of each particle.
- A change of state does not create any new particles, and it does not destroy any of the particles already there.
- Each particle also keeps exactly the same mass, because a water molecule in ice is identical to a water molecule in liquid water or in steam.
- The number of particles and the mass of each particle are both unchanged, so the total mass of the substance stays the same.
- This is conservation of mass, and it holds for melting, freezing, evaporation, condensation and sublimation alike.
- For example, 200 g200\ \text{g}200 g of ice that melts completely gives exactly 200 g200\ \text{g}200 g of liquid water.
- If that water is frozen again, the result is still 200 g200\ \text{g}200 g of ice.
- What can change is the volume, because the spacing between the particles changes when the state changes.
- Water is an unusual case, since ice takes up slightly more space than the liquid it formed from, which is why a full glass bottle of water can crack in a freezer even though its mass has not changed.
Closed and open systems
System
An object or group of objects whose energy stores are being considered.
Closed system
A system that no matter can enter or leave, so its total mass stays constant.
Open system
A system that matter can enter or leave, so its measured mass can change.
- A system is the object or group of objects you choose to study, such as a flask and the substance inside it.
- In a closed system, no matter can get in or out, so a balance shows the total mass staying the same while a change of state happens inside.
- A sealed plastic bag containing ice cubes is a closed system, because the meltwater stays inside the bag.
- In an open system, matter can cross the boundary of the system, so the mass measured on a balance can go down or up.
- An uncovered dish of water left on a windowsill loses mass over a few days, because water particles evaporate and drift away into the air of the room.
- That mass has not been destroyed, since every one of those particles still exists as water vapour in the surrounding air.
- If the air in the room were included in the system, the total mass would be found to stay constant.
- A cold can of drink taken out of the fridge gains a little mass in the same way, because water vapour from the air condenses into droplets on its outside surface.
- That extra mass has come from the air around the can rather than being created.
- So whenever a measured mass seems to change during a change of state, the explanation is that particles have entered or left the part being weighed.
Weighing a change of state
- A sealed plastic bag holding some ice cubes is placed on a balance, and the reading is 56.2 g56.2\ \text{g}56.2 g.
- The ice melts completely, but none of the water can leave the sealed bag.
- The balance still reads 56.2 g56.2\ \text{g}56.2 g, because the same particles are present in a different state.
- An open beaker of water reads 148.0 g148.0\ \text{g}148.0 g on Monday and 139.5 g139.5\ \text{g}139.5 g on Thursday.
- This 8.5 g8.5\ \text{g}8.5 g is the mass of water that evaporated and left the open beaker as water vapour, so it has moved rather than been destroyed.
Mass and temperature while stearic acid freezes
- Aim: to use a data logger to record how the temperature and the mass of a substance change as it cools and freezes, and to show that the mass stays constant during the change of state.
- Apparatus: a boiling tube about one third full of solid stearic acid, a beaker of water just off the boil to act as a water bath, a temperature probe connected to a data logger and laptop, an electronic balance reading to 0.01 g0.01\ \text{g}0.01 g, a small empty beaker to stand the tube in, a test-tube holder, a loose plug of cotton wool, paper towels and eye protection.
- Method:
- Stand the boiling tube in the hot water bath until all of the waxy white stearic acid has melted into a clear, colourless liquid at about 80 ∘C80\ ^\circ\text{C}80 ∘C.
- Push the temperature probe into the centre of the liquid so that its tip is surrounded by stearic acid and does not touch the glass.
- Plug the mouth of the tube loosely with cotton wool around the probe lead, so that nothing can drop in or splash out.
- Lift the tube out with the test-tube holder, wipe the outside completely dry with a paper towel, and stand it upright in the small beaker on the balance.
- Start the data logger so that it records the temperature every 10 s10\ \text{s}10 s, and write down the balance reading at the start and then every minute.
- Leave the tube undisturbed for about 151515 minutes while it cools to about 50 ∘C50\ ^\circ\text{C}50 ∘C, keeping hands and draughts away from the balance.
- Note the time when cloudy white solid first appears at the edges and the time when the whole tube has set solid.
- Plot a graph of temperature against time and a graph of mass against time, using the same time axis for both.
- Variables: time is the independent variable, temperature and mass are the dependent variables, and the tube, its contents and its surroundings stay the same throughout.
- Results: the temperature falls quickly at first, stays almost level at about 69 ∘C69\ ^\circ\text{C}69 ∘C while the stearic acid freezes, and then falls again once it is fully solid.
- Conclusion: the mass reading stays the same from start to finish, to within the resolution of the balance, because no particles enter or leave the tube while the liquid turns into a solid.
- Watch out: water left on the outside of the tube evaporates during the run and makes the mass appear to fall, a draught or a knock makes the balance reading drift, and a balance that only reads to 1 g1\ \text{g}1 g is too coarse to show whether the mass is truly constant.
- Safety: wear eye protection, handle the hot tube only with the holder because molten stearic acid causes burns, and never seal the tube with a rubber bung while it is hot because the trapped air expands as it is heated.
- Do not say that mass is lost when water evaporates from an open container, because the particles still exist as water vapour in the air.
- Do not say that particles get bigger, heavier or lighter when a substance changes state, because each particle keeps the same size and mass and only the spacing changes.
- Do not describe the gaps between the particles of a gas as being filled with air, because the space between the particles is empty.
- Do not confuse mass with volume, because a change of state can change the volume of a substance while its mass stays the same.
Explaining conservation of mass
- An explain question about mass during a change of state usually carries two marks, one for the particles and one for the system.
- For the first mark, state that the number of particles stays the same because no particles are created or destroyed.
- For the second mark, state that no particles can enter or leave because the container is sealed, or that particles have escaped into the air if the mass of an open container has fallen.
- Writing only that mass is conserved repeats the question and does not earn the explanation marks.
- Name the change of state precisely, because calling a solid turning straight into a gas evaporation loses the mark for sublimation.
- Name the change of state in which a solid turns directly into a gas.
- Explain why the total mass of a substance stays the same when it melts.
- State what makes a system closed rather than open.
- An open dish of water loses 5 g5\ \text{g}5 g overnight, so explain where the 5 g5\ \text{g}5 g has gone.
- Describe how a data logger and a balance can be used to show that mass stays constant while stearic acid freezes.
1.2.2 Physical and chemical changes
Properties of materials
Property
A characteristic of a material, such as its density, melting point, colour or hardness, that can be observed or measured.
- Every material has a set of properties that let you identify it and decide what it can be used for.
- Properties include the density, the melting point, the boiling point, the colour, the hardness and whether the material conducts electricity.
- Pure water, for example, is colourless, freezes at 0 ∘C0\ ^\circ\text{C}0 ∘C, boils at 100 ∘C100\ ^\circ\text{C}100 ∘C and has a density of 1000 kg/m31000\ \text{kg/m}^31000 kg/m3.
- These properties come from the particles the material is made of, so a material keeps its properties for as long as its particles stay the same.
- Some properties look different in different states, since ice is hard while water flows, but those differences come from how the particles are arranged rather than from a change in the particles themselves.
Physical changes
Physical change
A change in which no new substance is formed, so the material recovers its original properties when the change is reversed.
- In a physical change, no new substance is made, and the particles of the material stay the same particles.
- Only the arrangement, spacing or movement of the particles changes.
- Because the particles are unchanged, reversing the change brings back the original material with all of its original properties.
- Every change of state is a physical change.
- Ice that melts into water can be frozen again, and the new ice has the same melting point, density and appearance as the ice you started with.
- Water that evaporates from a saucepan can be condensed on a cold lid, and the droplets that form are pure water again.
- Candle wax that melts near the flame runs down the side of the candle and sets again as solid wax with the same melting point.
- Dissolving is also a physical change, because salt stirred into water can be recovered as solid salt by evaporating the water away.
- Changing the shape of a material, such as bending a paper clip or squashing a drinks can, is a physical change too, because the metal is still the same metal.
Chemical changes
Chemical change
A change in which one or more new substances are formed, so the original material does not recover its properties when the conditions are reversed.
- In a chemical change, the atoms of the starting materials are rearranged to make one or more new substances.
- The new substances have different properties from the materials you started with.
- Reversing the conditions does not undo a chemical change, so cooling, freezing or drying the products does not bring back the original material.
- Bread that has been toasted stays brown and crisp after it cools, because the toasting has formed new substances on its surface.
- Frying an egg turns the runny, see-through egg white into a firm white solid, and letting the egg cool does not make it runny again.
- Burning a match turns the wood and the chemicals in the match head into ash, smoke and gases, none of which become a match again when they are cooled.
- Iron that rusts forms a flaky orange-brown solid that is weaker than iron and has quite different properties.
- Signs that a new substance has formed include a new colour, a gas being given off, a new smell, or a change that cannot be undone by reversing the conditions.
- Some chemical changes can be undone by carrying out a different chemical reaction, but that is not the same as reversing the conditions as you can for melting or freezing.
Telling the two changes apart
- The test is to imagine reversing the change and then ask whether the material gets back its original properties.
- If reversing the change gives back the original material with the same properties, the change is physical.
- If reversing the conditions leaves a different material with different properties, the change is chemical.
- How easy or quick the reversal is does not matter, because condensing all the steam from a boiling kettle back into water is awkward, yet boiling is still a physical change.
- A big change in appearance is not proof of a chemical change either, because ice, liquid water and steam look very different but are all the same substance.
Classifying changes
- Candle wax melts and sets again as solid wax with the same melting point, so melting wax is a physical change.
- Steam from a kettle condenses on a cold window into droplets of pure water, so condensation is a physical change.
- Salt dissolved in water is left behind as white crystals when the water evaporates, so dissolving is a physical change.
- Bread turns brown and crisp in a toaster and stays brown and crisp after it cools, so toasting is a chemical change.
- Magnesium ribbon burns with a bright white flame and leaves a white powder that is not a shiny metal, so burning is a chemical change.
- Do not call melting, freezing, evaporation, condensation or sublimation chemical changes, because the substance is the same before and after.
- Do not say that a physical change is one that is easy to reverse, because the test is whether the original properties come back when the change is reversed.
- Do not treat a change in appearance as proof of a chemical change, because water and ice look different but are the same substance.
Explaining whether a change is physical
- Questions on this idea often give one mark for naming the type of change and a second mark for the reason.
- Name the reverse process, for example that the liquid wax freezes back into solid wax when it is cooled.
- Then state that the material has the same properties as before, or that no new substance has been formed.
- For a chemical change, state that a new substance is formed and that reversing the conditions does not recover the original material.
- The single word reversible is too vague to earn the reason mark on its own.
- State what happens to the particles of a substance during a physical change.
- Explain why melting is described as a physical change.
- Give one reason why toasting bread is a chemical change.
- Describe how you could show that dissolving salt in water is a physical change.
- Explain why being easy to reverse is not a good test for a physical change.
1.2.3 Heating, temperature and changes of state
Energy stored in a system
System
An object or group of objects whose energy stores are being considered.
Internal energy
The total energy stored by the particles of a system in their kinetic and potential energy stores.
Heating
The transfer of energy from a hotter object to a cooler object because of the temperature difference between them.
- A system is the object or group of objects you are studying, such as a pan of water on a hob.
- The particles in a system store energy in two ways.
- The particles are always moving or vibrating, so they have energy in their kinetic energy stores.
- The particles attract one another and are held at certain distances apart, so they also have energy in their potential energy stores.
- The total kinetic and potential energy of all the particles is the internal energy of the system.
- Heating transfers energy into a system from something hotter, such as a hob, a flame or an electric heater.
- That energy goes into the particles, so heating always increases the internal energy of the system.
- The extra energy then has one of two effects: it either raises the temperature of the system or produces a change of state.
- Cooling works the other way round, because energy leaves the system, its internal energy decreases, and the substance either falls in temperature or changes state.
Raising the temperature
Temperature
A measure of how hot or cold an object is, which depends on the average kinetic energy of its particles.
- When a substance is heated and is not changing state, the energy transferred goes mainly into the kinetic energy stores of its particles.
- The particles then move faster, so the average kinetic energy of the particles goes up.
- Temperature depends on the average kinetic energy of the particles, so a higher average kinetic energy means a higher temperature.
- In a solid, heating makes the particles vibrate faster and further about their fixed positions.
- In a liquid, heating makes the particles slide past one another more quickly.
- In a gas, heating makes the particles travel faster in random directions and collide more often.
Producing a change of state
Melting point
The temperature at which a pure solid changes into a liquid.
Boiling point
The temperature at which a pure liquid changes into a gas throughout the whole of the liquid.
- When a solid reaches its melting point, further heating stops raising its temperature and makes the solid melt instead.
- During melting, the energy transferred goes into the potential energy stores of the particles rather than their kinetic energy stores.
- This energy is used to overcome the forces of attraction that hold the particles in fixed positions, so the particles break out of their regular pattern and can move past one another.
- The average kinetic energy of the particles does not change, so the temperature stays constant until all of the solid has melted.
- Pure ice melts at 0 ∘C0\ ^\circ\text{C}0 ∘C, and a glass of iced water stays at about 0 ∘C0\ ^\circ\text{C}0 ∘C until every piece of ice has gone, even though it is taking in energy from the warm room.
- When a liquid reaches its boiling point, further heating makes it boil, and the energy is used to pull the particles completely apart so that they escape as a gas.
- Water boils at 100 ∘C100\ ^\circ\text{C}100 ∘C at normal atmospheric pressure, and a pan of boiling pasta water stays at 100 ∘C100\ ^\circ\text{C}100 ∘C however high the hob is turned up, with the extra energy making it boil away faster.
- Once the change of state is complete, the energy goes back into the kinetic energy stores of the particles and the temperature starts to rise again.
- The same idea works in reverse, because a liquid that is freezing or a gas that is condensing gives out energy while its temperature stays constant.
- The internal energy of a system therefore goes up during melting and boiling, even though its temperature does not.
Heating curves
Heating curve
A graph of temperature against time for a substance heated at a steady rate, showing where its temperature rises and where it changes state.
- A heating curve shows how the temperature of a substance changes with time when it is heated at a steady rate.
- Temperature goes on the vertical axis in ∘C^\circ\text{C}∘C, and time goes on the horizontal axis in s\text{s}s or min\text{min}min.
- For ice taken from a freezer at −18 ∘C-18\ ^\circ\text{C}−18 ∘C and heated steadily, the graph has four main sections.
- The line first slopes upwards as the solid ice warms from −18 ∘C-18\ ^\circ\text{C}−18 ∘C to 0 ∘C0\ ^\circ\text{C}0 ∘C.
- The line then runs flat at 0 ∘C0\ ^\circ\text{C}0 ∘C while the ice melts, with solid and liquid present together.
- The line slopes upwards again as the liquid water warms from 0 ∘C0\ ^\circ\text{C}0 ∘C to 100 ∘C100\ ^\circ\text{C}100 ∘C.
- The line runs flat a second time at 100 ∘C100\ ^\circ\text{C}100 ∘C while the water boils and turns into steam.
- The flat sections show where the substance is changing state, and the sloping sections show where its temperature is rising in a single state.
- The flat section for boiling lasts much longer than the one for melting, because separating the particles completely to form a gas takes far more energy than loosening them to form a liquid.
- A cooling curve is the mirror image, with a flat section at the boiling point while the gas condenses and another at the melting point while the liquid freezes.
Reading a heating curve
- A beaker of crushed ice starts at −10 ∘C-10\ ^\circ\text{C}−10 ∘C and is heated at a steady rate.
- From 000 to 222 minutes the temperature rises to 0 ∘C0\ ^\circ\text{C}0 ∘C, so the ice is warming as a solid and its particles are gaining kinetic energy.
- From 222 to 999 minutes the temperature stays at 0 ∘C0\ ^\circ\text{C}0 ∘C, so the ice is melting and the energy is overcoming the forces between the particles.
- From 999 minutes onwards the temperature rises steadily, so all the ice has melted and the liquid water is warming.
- The internal energy of the contents of the beaker increases during all three sections, including the flat one.
Recording the temperature of ice as it is heated
- Aim: to record the temperature of crushed ice at regular intervals as it is heated until the water boils, and to plot a heating curve from the readings.
- Apparatus: a 250 cm3250\ \text{cm}^3250 cm3 beaker, crushed ice straight from a freezer, a thermometer with a range of −10 ∘C-10\ ^\circ\text{C}−10 ∘C to 110 ∘C110\ ^\circ\text{C}110 ∘C, a clamp stand with boss and clamp, a stirring rod, a Bunsen burner with tripod, gauze and heatproof mat, a stopwatch and eye protection.
- Method:
- Fill the beaker about half full with crushed ice, because small pieces pack closely around the thermometer bulb and warm evenly.
- Clamp the thermometer so that its bulb sits in the middle of the ice and does not touch the bottom or sides of the beaker, where the glass is hotter than the contents.
- Read the starting temperature with your eye level with the top of the liquid column, and expect it to be a little below 0 ∘C0\ ^\circ\text{C}0 ∘C.
- Light the Bunsen burner on a small blue flame under the gauze and start the stopwatch at the same moment.
- Stir the contents gently with the stirring rod and record the temperature every 30 s30\ \text{s}30 s, writing beside each reading whether you see solid only, solid and liquid, liquid only, or bubbling liquid.
- Keep the flame the same size all the way through, so that energy is supplied at a steady rate.
- Carry on until the water has been boiling steadily for about 333 minutes, then turn off the gas.
- Plot temperature on the vertical axis against time on the horizontal axis, and join the points with a smooth line.
- Variables: time is the independent variable, temperature is the dependent variable, and the mass of ice and the size of the flame are kept the same.
- Results: the temperature rises to 0 ∘C0\ ^\circ\text{C}0 ∘C, stays level while the ice melts, rises steadily through the liquid, and then levels off close to 100 ∘C100\ ^\circ\text{C}100 ∘C while the water boils.
- Watch out: a bulb touching the glass reads too high, an unstirred beaker leaves warm water at the bottom and ice at the top so the flat section looks sloped, and reading the scale from above or below gives a parallax error.
- Safety: wear eye protection, stand up while heating, keep clear of the steam because it scalds, never stir with the thermometer because it can break, and leave the beaker to cool before moving it.
- Do not say that the temperature rises while a substance melts or boils, because the energy is going into the potential energy stores of the particles.
- Do not say that no energy is being transferred during a flat section of a heating curve, because heating carries on and the internal energy keeps increasing.
- Do not confuse temperature with internal energy, because a bath of warm water has far more internal energy than a mug of boiling water even though its temperature is lower.
- Do not use the phrase heat energy, because heating is a way of transferring energy rather than a store of energy.
Explaining the flat sections
- A question about a flat section of a heating curve usually carries two or three marks.
- State that energy is still being transferred to the substance, so its internal energy is increasing.
- State that this energy is used to overcome the forces of attraction between the particles, so that they can change their arrangement.
- State that the kinetic energy of the particles is not increasing, so the temperature stays the same.
- For a sloping section, link the rising temperature to the particles gaining kinetic energy and moving faster.
- Name the two energy stores that together make up the internal energy of a system.
- Describe the two effects that heating can have on a substance.
- Explain why the temperature of melting ice stays at 0 ∘C0\ ^\circ\text{C}0 ∘C even though energy is being transferred to it.
- Sketch the shape of a heating curve for ice warmed from −10 ∘C-10\ ^\circ\text{C}−10 ∘C until the water boils.
- State the range of thermometer you would choose to record the temperature of ice as it is heated to boiling.
1.2.4 Specific heat capacity and specific latent heat
Specific heat capacity
Specific heat capacity
The energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of a substance by 1 ∘C1\ ^\circ\text{C}1 ∘C.
- Different materials need different amounts of energy to warm up by the same number of degrees.
- The specific heat capacity of a material is the energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of it by 1 ∘C1\ ^\circ\text{C}1 ∘C.
- Its symbol is ccc, and its unit is the joule per kilogram per degree Celsius, J/kg ∘C\text{J/kg}\,^\circ\text{C}J/kg∘C.
- Water has a specific heat capacity of about 4200 J/kg ∘C4200\ \text{J/kg}\,^\circ\text{C}4200 J/kg∘C, so 4200 J4200\ \text{J}4200 J is needed to warm 1 kg1\ \text{kg}1 kg of water by just 1 ∘C1\ ^\circ\text{C}1 ∘C.
- Aluminium has a specific heat capacity of about 900 J/kg ∘C900\ \text{J/kg}\,^\circ\text{C}900 J/kg∘C and copper about 390 J/kg ∘C390\ \text{J/kg}\,^\circ\text{C}390 J/kg∘C, so the same energy warms 1 kg1\ \text{kg}1 kg of copper by more than ten times as much as 1 kg1\ \text{kg}1 kg of water.
- A material with a high specific heat capacity warms up slowly and cools down slowly, because it must gain or lose a lot of energy for each degree of temperature change.
- A material with a low specific heat capacity warms up and cools down quickly.
- On a sunny day at the beach, the dry sand becomes too hot to walk on barefoot while the sea stays cool, partly because sand has a much lower specific heat capacity than water.
- Central heating systems pump hot water round the radiators because each kilogram of water carries a large amount of energy from the boiler to the rooms.
- Specific heat capacity applies only while a substance stays in the same state, because the temperature does not change at all while the state is changing.
Specific latent heat
Specific latent heat
The energy needed to change the state of 1 kg1\ \text{kg}1 kg of a substance without any change in its temperature.
Specific latent heat of fusion
The energy needed to change 1 kg1\ \text{kg}1 kg of a substance from solid to liquid without any change in its temperature.
Specific latent heat of vaporisation
The energy needed to change 1 kg1\ \text{kg}1 kg of a substance from liquid to gas without any change in its temperature.
- During a change of state, energy is transferred to or from a substance while its temperature stays constant.
- That energy goes into or comes out of the potential energy stores of the particles as they move apart or closer together, rather than changing their kinetic energy.
- The specific latent heat of a substance is the energy needed to change the state of 1 kg1\ \text{kg}1 kg of it without changing its temperature.
- Its symbol is LLL, and its unit is the joule per kilogram, J/kg\text{J/kg}J/kg.
- There is no degree Celsius in the unit, because the temperature does not change during a change of state.
- The specific latent heat of fusion applies when a substance melts or freezes.
- The specific latent heat of vaporisation applies when a substance boils or condenses.
- For water, the specific latent heat of fusion is about 3.34×105 J/kg3.34 \times 10^{5}\ \text{J/kg}3.34×105 J/kg and the specific latent heat of vaporisation is about 2.26×106 J/kg2.26 \times 10^{6}\ \text{J/kg}2.26×106 J/kg.
- The value for vaporisation is almost seven times larger, because turning a liquid into a gas means pulling the particles completely apart, whereas melting only frees them from their fixed positions.
- The energy taken in when 1 kg1\ \text{kg}1 kg of a substance melts is given out again when 1 kg1\ \text{kg}1 kg of it freezes, and the same is true for boiling and condensing.
- This is why a scald from steam at 100 ∘C100\ ^\circ\text{C}100 ∘C is much worse than one from water at the same temperature, since the steam releases its latent heat of vaporisation as it condenses on the skin.
Comparing the two quantities
- Both quantities describe the energy needed per kilogram of a substance, but they describe different things happening to it.
- Specific heat capacity is about a change in temperature with no change of state.
- Specific latent heat is about a change of state with no change in temperature.
- During a temperature change, the energy transferred mainly increases the kinetic energy of the particles.
- During a change of state, the energy transferred increases the potential energy of the particles.
- The unit of specific heat capacity is J/kg ∘C\text{J/kg}\,^\circ\text{C}J/kg∘C, while the unit of specific latent heat is J/kg\text{J/kg}J/kg.
- On a heating curve, specific heat capacity decides how steep the sloping sections are, and specific latent heat decides how long the flat sections last.
Specific heat capacity of a metal
- Aim: to measure the energy supplied to a metal block and the rise in its temperature, and use them to find the specific heat capacity of the metal.
- Apparatus: a 1 kg1\ \text{kg}1 kg aluminium block with two holes drilled in its top, a 12 V12\ \text{V}12 V electric immersion heater, a low-voltage power supply, a joulemeter, a thermometer reading to 0.5 ∘C0.5\ ^\circ\text{C}0.5 ∘C or a temperature probe and data logger, a pipette with a little water, insulating material such as foam lagging, a heatproof mat, a stopwatch and an electronic balance.
- Method:
- Measure the mass of the block on the electronic balance and write it in kilograms.
- Stand the block on the heatproof mat and wrap its sides and top in insulation, leaving the two holes uncovered.
- Slide the heater into the larger hole and the thermometer into the smaller one, adding a few drops of water to the thermometer hole so the bulb makes good thermal contact with the metal.
- Connect the heater to the power supply through the joulemeter, and reset the joulemeter to zero.
- Record the starting temperature of the block once the reading is steady.
- Switch on and let the heater run for about 101010 minutes, recording the joulemeter reading and the temperature every minute.
- Switch off and keep watching the thermometer, recording the highest temperature reached, because energy carries on spreading from the hot heater into the block for a short time.
- Work out the temperature rise by subtracting the starting temperature from the highest temperature.
- Variables: the energy supplied is the independent variable, the temperature rise is the dependent variable, and the block, its mass and its insulation stay the same.
- Maths: the specific heat capacity is c=ΔEm Δθc = \dfrac{\Delta E}{m\,\Delta\theta}c=mΔθΔE, where ΔE\Delta EΔE is the joulemeter reading, mmm is the mass of the block and Δθ\Delta\thetaΔθ is the temperature rise.
- Graph: plotting temperature against energy supplied gives a straight line, and its gradient equals 1m c\dfrac{1}{m\,c}mc1, so the result uses every reading rather than just the first and last.
- Results: a typical class value for aluminium is about 1000 J/kg ∘C1000\ \text{J/kg}\,^\circ\text{C}1000 J/kg∘C, a little above the accepted value of about 900 J/kg ∘C900\ \text{J/kg}\,^\circ\text{C}900 J/kg∘C.
- Watch out: some of the energy recorded by the joulemeter warms the heater, the thermometer and the surrounding air instead of the block, so the calculated value comes out too high, and a dry thermometer hole makes the reading lag behind the true temperature.
- Safety: the heater and block become hot enough to burn, so leave them on the heatproof mat to cool before handling, and keep water away from the power supply.
- Do not mix up the two units, because specific heat capacity has a degree Celsius in its unit and specific latent heat does not.
- Do not use specific latent heat for a temperature change, or specific heat capacity for a change of state.
- Do not swap fusion and vaporisation, because fusion goes with melting and freezing while vaporisation goes with boiling and condensing.
- Do not leave 1 kg1\ \text{kg}1 kg out of either definition, because both quantities are measured per kilogram.
Writing the definitions
- A definition of specific heat capacity needs three ideas: the energy needed, 1 kg1\ \text{kg}1 kg of the substance, and a temperature rise of 1 ∘C1\ ^\circ\text{C}1 ∘C.
- A definition of specific latent heat needs the energy needed, 1 kg1\ \text{kg}1 kg of the substance, a change of state, and no change in temperature.
- An answer such as the heat needed to make something hotter scores nothing, because it gives no mass and no size of temperature change.
- When explaining the insulation in the practical, say that it reduces energy transfer to the surroundings so that more of the measured energy heats the block, rather than writing that it makes the result more accurate.
- When explaining why the calculated value is too high, say that some energy was transferred to the surroundings, so the energy that heated the block was less than the joulemeter reading.
- State the definition of specific heat capacity.
- State the unit of specific latent heat and explain why it has no temperature in it.
- Name the specific latent heat used when ice melts and the one used when steam condenses.
- Explain why dry sand on a beach heats up much faster than the sea on a sunny day.
- Explain why the specific heat capacity found for a metal block in the practical is usually higher than the accepted value.
1.2.5 Internal energy and temperature change
What decides the energy needed
Internal energy
The total energy stored by the particles of a system in their kinetic and potential energy stores.
- When a material is heated and stays in the same state, the energy transferred increases its internal energy and raises its temperature.
- The energy needed for this depends on three things: the mass of the material, its specific heat capacity and the change in temperature.
- Doubling the mass doubles the energy needed, because there are twice as many particles to speed up.
- Doubling the temperature rise doubles the energy needed, because each particle has to gain twice as much kinetic energy.
- A material with double the specific heat capacity needs double the energy for the same mass and the same temperature rise.
- A kettle filled to the top takes several times longer to boil than one holding just enough water for a single mug, because the energy needed is proportional to the mass of water.
The equation
Specific heat capacity
The energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of a substance by 1 ∘C1\ ^\circ\text{C}1 ∘C.
- The three factors combine into one equation: change in thermal energy=mass×specific heat capacity×change in temperature\text{change in thermal energy} = \text{mass} \times \text{specific heat capacity} \times \text{change in temperature}change in thermal energy=mass×specific heat capacity×change in temperature.
- In symbols, this is ΔE=m c Δθ\Delta E = m\,c\,\Delta\thetaΔE=mcΔθ.
- ΔE\Delta EΔE is the change in thermal energy in joules, J\text{J}J.
- mmm is the mass in kilograms, kg\text{kg}kg.
- ccc is the specific heat capacity in J/kg ∘C\text{J/kg}\,^\circ\text{C}J/kg∘C.
- Δθ\Delta\thetaΔθ is the change in temperature in ∘C^\circ\text{C}∘C, where θ\thetaθ is the Greek letter theta.
- The symbol Δ\DeltaΔ means change in, so Δθ\Delta\thetaΔθ is always worked out as the final temperature minus the starting temperature.
- You do not have to recall this equation from memory, but you must be able to choose it from a list of equations and apply it correctly.
- The equation applies only when there is no change of state, because the temperature does not change while a substance is melting or boiling.
Calculating the energy change
- Start by writing the equation, then work out Δθ\Delta\thetaΔθ if the question gives a starting and a final temperature.
- Convert the mass into kilograms, remembering that 1 kg=1000 g1\ \text{kg} = 1000\ \text{g}1 kg=1000 g.
- Substitute the values, calculate, and finish with the unit, which for energy is the joule, J\text{J}J.
- Large answers can be written in standard form or in kilojoules, where 1 kJ=1000 J1\ \text{kJ} = 1000\ \text{J}1 kJ=1000 J.
- When a material cools down, Δθ\Delta\thetaΔθ is negative and so ΔE\Delta EΔE comes out negative, which shows that the internal energy has decreased as energy is transferred to the surroundings.
Heating water in a kettle
- A kettle heats 1.5 kg1.5\ \text{kg}1.5 kg of water from 15 ∘C15\ ^\circ\text{C}15 ∘C to 100 ∘C100\ ^\circ\text{C}100 ∘C, and the specific heat capacity of water is 4200 J/kg ∘C4200\ \text{J/kg}\,^\circ\text{C}4200 J/kg∘C.
- Write the equation.
- Work out the temperature change.
- Substitute the values and calculate.
- The internal energy of the water increases by 5.4×105 J5.4 \times 10^{5}\ \text{J}5.4×105 J, which is about 540 kJ540\ \text{kJ}540 kJ.
Mass in grams and a falling temperature
- A 250 g250\ \text{g}250 g aluminium baking tray cools from 180 ∘C180\ ^\circ\text{C}180 ∘C to 20 ∘C20\ ^\circ\text{C}20 ∘C, and the specific heat capacity of aluminium is 900 J/kg ∘C900\ \text{J/kg}\,^\circ\text{C}900 J/kg∘C.
- Convert the mass into kilograms.
- Work out the temperature change, which is negative because the tray is cooling.
- Substitute the values and calculate.
- The negative sign shows that the internal energy of the tray falls by 36 000 J36\,000\ \text{J}36000 J as energy is transferred to the kitchen.
Rearranging the equation
- Questions can give you the energy and ask for one of the other quantities instead.
- To find the specific heat capacity, divide both sides by m Δθm\,\Delta\thetamΔθ to give c=ΔEm Δθc = \dfrac{\Delta E}{m\,\Delta\theta}c=mΔθΔE.
- To find the mass, divide both sides by c Δθc\,\Delta\thetacΔθ to give m=ΔEc Δθm = \dfrac{\Delta E}{c\,\Delta\theta}m=cΔθΔE.
- To find the temperature change, divide both sides by m cm\,cmc to give Δθ=ΔEm c\Delta\theta = \dfrac{\Delta E}{m\,c}Δθ=mcΔE.
- If the question asks for a final temperature, find Δθ\Delta\thetaΔθ first and then add it to the starting temperature, or subtract it if the material is cooling.
- In a calculation with several steps, keep the unrounded value in your calculator and round only the final answer, usually to the same number of significant figures as the data.
Finding a final temperature
- An immersion heater transfers 63 000 J63\,000\ \text{J}63000 J to 0.50 kg0.50\ \text{kg}0.50 kg of water at 18 ∘C18\ ^\circ\text{C}18 ∘C, and the specific heat capacity of water is 4200 J/kg ∘C4200\ \text{J/kg}\,^\circ\text{C}4200 J/kg∘C.
- Rearrange the equation to make Δθ\Delta\thetaΔθ the subject.
- Substitute the values and calculate.
- The water is being heated, so the temperature change is added to the starting temperature.
Finding a specific heat capacity
- A 0.80 kg0.80\ \text{kg}0.80 kg block of an unknown metal gains 15 600 J15\,600\ \text{J}15600 J, and its temperature rises from 20 ∘C20\ ^\circ\text{C}20 ∘C to 70 ∘C70\ ^\circ\text{C}70 ∘C.
- Work out the temperature change.
- Rearrange the equation, substitute the values and calculate.
- This value matches the specific heat capacity of copper, so the block is probably made of copper.
- Do not substitute the final temperature in place of Δθ\Delta\thetaΔθ, because the equation needs the change in temperature.
- Do not use a mass in grams, because the unit of ccc is per kilogram and the answer would come out 100010001000 times too big.
- Do not use this equation for melting or boiling, because the temperature is not changing during a change of state.
- Do not write the unit of specific heat capacity as J/kg\text{J/kg}J/kg, because that is the unit of specific latent heat.
Showing your working
- Calculations on this equation are usually worth two to four marks, and most of those marks are for the method rather than the final number.
- Write the equation first, then show the substitution with every value in the correct unit, because a correct substitution earns credit even if the arithmetic then slips.
- In a two-step question, a wrong value from the first step is carried forward, so finishing the method still scores the later marks.
- Check that the size of your answer makes sense, since warming a mug of tea needs tens of thousands of joules rather than millions or just a handful.
- Give the unit with the final answer, using the unit printed on the answer line where there is one.
- Write the equation that links change in thermal energy, mass, specific heat capacity and change in temperature.
- State the three factors that decide how much energy is needed to warm a material.
- Calculate the energy needed to warm 2.0 kg2.0\ \text{kg}2.0 kg of water from 20 ∘C20\ ^\circ\text{C}20 ∘C to 50 ∘C50\ ^\circ\text{C}50 ∘C, using c=4200 J/kg ∘Cc = 4200\ \text{J/kg}\,^\circ\text{C}c=4200 J/kg∘C.
- Rearrange ΔE=m c Δθ\Delta E = m\,c\,\Delta\thetaΔE=mcΔθ to make Δθ\Delta\thetaΔθ the subject.
- Explain why ΔE\Delta EΔE comes out negative when a material cools.
1.2.6 Energy transfer during a change of state
Energy and changes of state
Latent heat
The energy transferred to or from a substance during a change of state while its temperature stays constant.
Specific latent heat
The energy needed to change the state of 1 kg1\ \text{kg}1 kg of a substance without any change in its temperature.
- While a substance changes state, energy is transferred to or from it but its temperature stays the same.
- This energy is called latent heat, from a Latin word meaning hidden, because a thermometer shows no sign of it.
- When a substance melts or boils, it takes in energy, and that energy goes into the potential energy stores of the particles as they move further apart.
- When a substance freezes or condenses, it gives out the same amount of energy as the particles move closer together again.
- The energy involved depends on the mass that changes state and on the specific latent heat of the substance.
- Twice the mass of ice needs twice the energy to melt, which is why a large bag of ice keeps a cool box cold for longer than a handful of cubes.
The equation
Specific latent heat of fusion
The energy needed to change 1 kg1\ \text{kg}1 kg of a substance from solid to liquid without any change in its temperature.
Specific latent heat of vaporisation
The energy needed to change 1 kg1\ \text{kg}1 kg of a substance from liquid to gas without any change in its temperature.
- The equation is thermal energy for a change in state=mass×specific latent heat\text{thermal energy for a change in state} = \text{mass} \times \text{specific latent heat}thermal energy for a change in state=mass×specific latent heat.
- In symbols, this is E=m LE = m\,LE=mL.
- EEE is the energy transferred in joules, J\text{J}J.
- mmm is the mass that changes state in kilograms, kg\text{kg}kg.
- LLL is the specific latent heat in joules per kilogram, J/kg\text{J/kg}J/kg.
- Use the specific latent heat of fusion for melting or freezing, and the specific latent heat of vaporisation for boiling or condensing.
- There is no temperature in the equation, because the temperature does not change while the state is changing.
- The mass must be the mass that changes state, so if only part of a block of ice melts, you use only the mass that melts.
- You do not have to recall this equation from memory, but you must be able to choose it from a list of equations and apply it correctly.
- Rearranging gives L=EmL = \dfrac{E}{m}L=mE for the specific latent heat and m=ELm = \dfrac{E}{L}m=LE for the mass.
Calculating the energy
- Write the equation, convert any mass in grams into kilograms, substitute the values, and give the answer with its unit.
- Values of specific latent heat are large, so they are usually given in standard form, such as 3.34×105 J/kg3.34 \times 10^{5}\ \text{J/kg}3.34×105 J/kg.
- Enter standard form into your calculator with the ×10x\times 10^{x}×10x key rather than typing out the zeros, which avoids slipping a digit.
- Some questions combine both kinds of energy change, with a substance first warming up to its melting or boiling point and then changing state.
- For these, work out each stage separately, using ΔE=m c Δθ\Delta E = m\,c\,\Delta\thetaΔE=mcΔθ for warming and E=m LE = m\,LE=mL for changing state, and then add the energies together.
Melting ice
- A 0.25 kg0.25\ \text{kg}0.25 kg block of ice at 0 ∘C0\ ^\circ\text{C}0 ∘C melts into water at 0 ∘C0\ ^\circ\text{C}0 ∘C, and the specific latent heat of fusion of water is 3.34×105 J/kg3.34 \times 10^{5}\ \text{J/kg}3.34×105 J/kg.
- Melting the ice takes 83 500 J83\,500\ \text{J}83500 J, and the temperature stays at 0 ∘C0\ ^\circ\text{C}0 ∘C throughout.
Finding a specific latent heat
- A heater transfers 45 000 J45\,000\ \text{J}45000 J to water that is already boiling, and during that time 20 g20\ \text{g}20 g of the water turns into steam.
- Convert the mass into kilograms.
- Rearrange the equation, substitute the values and calculate.
- This is the specific latent heat of vaporisation of water, close to the accepted value of 2.26×106 J/kg2.26 \times 10^{6}\ \text{J/kg}2.26×106 J/kg.
Warming ice and then melting it
- A 0.50 kg0.50\ \text{kg}0.50 kg block of ice from a freezer is at −18 ∘C-18\ ^\circ\text{C}−18 ∘C and is warmed until it has all melted at 0 ∘C0\ ^\circ\text{C}0 ∘C.
- The specific heat capacity of ice is 2100 J/kg ∘C2100\ \text{J/kg}\,^\circ\text{C}2100 J/kg∘C, and its specific latent heat of fusion is 3.34×105 J/kg3.34 \times 10^{5}\ \text{J/kg}3.34×105 J/kg.
- Stage 1 warms the solid ice from −18 ∘C-18\ ^\circ\text{C}−18 ∘C to 0 ∘C0\ ^\circ\text{C}0 ∘C, a temperature change of 18 ∘C18\ ^\circ\text{C}18 ∘C.
- Stage 2 melts the ice at a constant 0 ∘C0\ ^\circ\text{C}0 ∘C.
- The total energy is the sum of the two stages.
- Melting needs almost nine times as much energy as warming the ice, which is why ice sits at 0 ∘C0\ ^\circ\text{C}0 ∘C for so long once it starts to melt.
Specific latent heat of vaporisation of water
- Aim: to measure the energy needed to boil away a measured mass of water, and use it to find the specific latent heat of vaporisation of water.
- Apparatus: a 12 V12\ \text{V}12 V electric immersion heater, a low-voltage power supply, a joulemeter, a 250 cm3250\ \text{cm}^3250 cm3 glass beaker wrapped in insulating foam, an electronic balance reading to 0.1 g0.1\ \text{g}0.1 g, a clamp stand with boss and clamp, a heatproof mat, water and eye protection.
- Method:
- Stand the insulated beaker on the electronic balance and pour in about 150 cm3150\ \text{cm}^3150 cm3 of water.
- Clamp the heater so that its coil is fully under the water but does not touch the beaker, so that its weight does not rest on the balance.
- Connect the heater to the power supply through the joulemeter, switch on, and heat the water until it is boiling steadily.
- Once the water is boiling, reset the joulemeter to zero and record the balance reading at the same moment.
- Let the water boil for about 555 minutes, checking that the coil stays covered the whole time.
- Record the joulemeter reading and the new balance reading at the same moment, then switch off.
- Subtract the final balance reading from the first one to find the mass of water that has turned into steam.
- Why wait for boiling: the readings start only once the water is boiling, so that all of the energy recorded goes into changing state rather than into warming the water.
- Maths: the specific latent heat is L=EmL = \dfrac{E}{m}L=mE, where EEE is the joulemeter reading in J\text{J}J and mmm is the mass of water boiled away in kg\text{kg}kg.
- Results: a typical class result is about 2.4×106 J/kg2.4 \times 10^{6}\ \text{J/kg}2.4×106 J/kg, a little above the accepted value of 2.26×106 J/kg2.26 \times 10^{6}\ \text{J/kg}2.26×106 J/kg.
- Alternative: the specific latent heat of fusion of stearic acid can be found in the same way, by recording the energy supplied only while its temperature stays level at about 69 ∘C69\ ^\circ\text{C}69 ∘C and measuring the mass of solid that melts in that time.
- Watch out: energy lost to the surroundings makes the calculated value too high, splashing water out of the beaker makes it too low, and the bubbling makes the balance reading flicker, so take the reading quickly and at the same moment as the joulemeter reading.
- Safety: wear eye protection, keep hands away from the steam because it scalds, never let the heater coil run dry because it becomes dangerously hot, and keep the power supply away from the water.
- Do not include a temperature change in E=m LE = m\,LE=mL, because the temperature does not change while the state is changing.
- Do not use the total mass of a substance when only part of it changes state.
- Do not use the latent heat of vaporisation in a melting question, or the latent heat of fusion in a boiling question.
- Do not say that energy disappears during a change of state, because it is stored in the potential energy stores of the particles and is given out again when the substance changes back.
Multi-step energy questions
- In a question with several stages, label each stage, such as warming the ice and melting the ice, so the examiner can follow your method.
- Each stage usually carries its own marks, so a slip in one stage does not cost you the marks for the other.
- Check that you have chosen the right specific latent heat by matching it to the change of state named in the question.
- Give large answers in standard form or in kilojoules, and make sure the unit matches the number you write.
- Write the equation that links thermal energy for a change in state, mass and specific latent heat.
- Explain why the equation for a change of state does not include a change in temperature.
- Calculate the energy needed to melt 0.40 kg0.40\ \text{kg}0.40 kg of ice at 0 ∘C0\ ^\circ\text{C}0 ∘C, using L=3.34×105 J/kgL = 3.34 \times 10^{5}\ \text{J/kg}L=3.34×105 J/kg.
- State why the readings in the latent heat practical start only once the water is boiling.
- Describe how to find the total energy needed to warm ice from below 0 ∘C0\ ^\circ\text{C}0 ∘C and then melt it.
