Physical Properties & Physical Changes · Grade 6 Science
Chapter: Matter, Mixtures & Water PropertiesPhysical Properties & Physical Changes
Learning goals
- I can measure density and use it to compare two materials.
- I can decide whether a change is physical by asking what substance is present afterwards.
- I can explain why a single property cannot identify an unknown material.
Readings for this unit
Before this unit
Read the chapter opening and the named observation or diagram. Clarify one word or unit that blocks understanding. Recall the relevant earlier measurement, matter, or energy idea.
Study the unit across three 35–45-minute class sessions. Use the worked examples after an attempt and the supplied records when actual equipment is not available.
Gold has a density of about 19.3 g/cm3. Pyrite, the mineral miners called fool's gold, has a density of about 5.0 g/cm3. Both are metallic and yellow, and a great many people in the 1849 California rush carried home sacks of the wrong one.
Colour alone could not separate them, and neither could lustre. What separated them was any second property: pyrite scratches glass and gold does not, pyrite leaves a greenish-black streak on a white tile while gold leaves a yellow one, and a 100 g lump of gold occupies only about 5.2 cm3 while 100 g of pyrite occupies about 20 cm3.
That is the whole lesson in one example. Properties you can measure without destroying the sample are powerful, no single one of them identifies anything, and a change that alters shape or state has not altered what the substance is.
Build your understanding
Where this fits
The last lesson established that particles arrange themselves differently in solids, liquids and gases. This lesson is about the properties that follow from what a substance is rather than from how much of it you have, and about telling a change of form from a change of substance. The next lesson covers changes that do create a new substance.
Use three flexible 40-45-minute sessions: measuring properties without destroying the sample; physical change and what reversibility does not prove; density as a ratio and identifying an unknown, then the practice set, revision and the exit check. Workshop, review, quiz and exams have separate time.
Part 1 · Measuring without destroying
A physical property is anything you can determine while leaving the substance chemically unchanged: colour, hardness, melting point, density, whether it conducts electricity, whether it is attracted to a magnet. Some of these depend on how much you have and some do not, and the distinction matters. Mass and volume change when you take a bigger piece, so neither identifies a material. Density, melting point and hardness stay the same whatever the size of the sample, which is exactly why those are the ones used for identification. A 1 g crumb of aluminium and a 50 kg block both have a density of 2.70 g/cm3 and both melt at 660 degrees Celsius.
Hardness is measured on the Mohs scale, which ranks minerals by what scratches what: talc is 1, gypsum 2, calcite 3, quartz 7 and diamond 10. The test is genuinely practical, because a fingernail is about 2.5, a copper coin about 3.5 and a steel knife about 5.5, so three everyday objects divide the whole range into useful bands. Gold at 2.5 to 3 can be marked with a coin; pyrite at 6 to 6.5 cannot be scratched by a knife and will scratch glass instead. None of these tests consumes the sample, which is the point of calling them physical. Magnetism is worth adding to the list because it is so sharply decisive. Iron, nickel and cobalt are attracted to a magnet and almost nothing else in an ordinary classroom is, so a magnet separates iron filings from sand in seconds and tells you instantly that a shiny grey metal is not aluminium, which a magnet ignores completely.
Pause and explain
Why is density useful for identifying a material when mass is not?
Check your reasoning after answering
Because density does not depend on how much you have. A larger piece of aluminium has more mass and more volume, but the ratio between them stays at 2.70 g/cm3, so the number describes the material itself. Mass changes with every sample size, so it tells you about the particular lump in front of you and nothing about what it is made of.
Part 2 · Changing the form, keeping the substance
A physical change alters state, shape or arrangement while the substance stays what it was. Melting ice gives you water, which is the same substance in a different state. Tearing paper gives you smaller pieces of paper. Dissolving 10 g of salt in 100 mL of water gives you salt water, and the salt is still salt, which you can prove by boiling the water away and weighing what is left: close to 10 g of salt, unchanged. In each case the particles have been rearranged or separated, and nothing new has been made.
The common shortcut, that physical changes are reversible and chemical ones are not, is unreliable in both directions and you should drop it. Tearing paper is physical and essentially impossible to undo. Smashing a plate is physical and permanent. Meanwhile some chemical changes reverse readily under the right conditions. The question that actually works is simpler: after the change, is the same substance present? Ask that rather than asking whether you could get back to where you started, and the awkward cases stop being awkward. Dissolving is the case that causes most argument, so it is worth being explicit. When salt dissolves it seems to vanish, which feels like it must have become something else, but the mass is conserved exactly: 100 g of water plus 10 g of salt gives 110 g of solution, and boiling the water away returns the salt. Nothing was created or destroyed, the particles were simply separated and mixed among the water particles.
Pause and explain
Why is 'can it be reversed?' a poor test for whether a change is physical?
Check your reasoning after answering
Because it gives wrong answers both ways. Tearing paper and smashing a plate are physical changes that cannot practically be undone, while some chemical changes reverse easily in the right conditions. The reliable question is whether the same substance is present afterwards, which is about identity rather than about how difficult the change is to reverse.
Part 3 · Density as a ratio
Density is mass divided by volume, and the units say so: grams per cubic centimetre, written g/cm3, or grams per millilitre for liquids, where 1 mL and 1 cm3 are the same volume. A piece of iron with a mass of 78.7 g and a volume of 10 cm3 has a density of 7.87 g/cm3. Take a piece twice as large and both numbers double, to 157.4 g and 20 cm3, and the ratio is unchanged. That is what it means to say density is a property of the material rather than of the object.
Measuring the volume of an awkward shape is done by displacement, and the method is worth knowing precisely. Fill a measuring cylinder to a marked level, say 50 mL, lower the object in gently so nothing splashes out, and read the new level, say 68 mL. The object displaced 18 mL, so its volume is 18 cm3. Read the cylinder at eye level and at the bottom of the curved surface the water makes, because reading from above can easily shift the value by 1 mL or more, and on a volume of 18 cm3 that is an error of around 6 per cent in the density. For a regular shape you can skip displacement and measure directly instead. A rectangular block 4.0 cm by 3.0 cm by 2.0 cm has a volume of 24 cm3, found by multiplying the three lengths, and that will usually be more accurate than displacement because a ruler can be read more finely than a measuring cylinder. Use displacement when the shape makes the arithmetic impossible, not as a default.
Pause and explain
A stone raises the water in a cylinder from 50 mL to 68 mL and has a mass of 47 g. What is its density, and what could go wrong in the measurement?
Check your reasoning after answering
The stone displaced 18 mL, so its volume is 18 cm3 and the density is 47 divided by 18, about 2.6 g/cm3. The main risks are splashing water out when lowering the stone, which makes the volume look too small, and reading the cylinder from above rather than at eye level, which on 18 cm3 can shift the result by several per cent.
Part 4 · One property is never enough
Return to gold and pyrite. Both are metallic yellow solids, so colour and lustre put them in the same box. Density separates them decisively at 19.3 against 5.0 g/cm3, and so does hardness, 2.5 to 3 against 6 to 6.5, and so does the streak test, where rubbing the mineral on unglazed white tile leaves yellow for gold and greenish-black for pyrite. Any one of those three would have saved a prospector a journey, and the colour that fooled him is a perfectly real property that simply happens to be shared.
This is why identification is always a matter of combining properties and stating how confident you are. A classroom test can usually narrow an unknown to a small group of candidates rather than proving a unique answer, and saying so is part of the result. There is also a safety rule that follows directly: never identify an unknown substance by tasting it, smelling it closely, heating it or breaking it open, because those tests work by exposing you to whatever it turns out to be, which is the one thing you do not yet know. Professional laboratories face the same problem and solve it with instruments rather than senses. A spectrometer identifies a substance from the particular wavelengths of light it absorbs or emits, which is effectively a fingerprint and needs no contact at all. The logic is identical to the classroom version: gather several independent properties, and trust the identification in proportion to how many of them agree.
Pause and explain
Gold and pyrite are both metallic yellow. Name two measurements that separate them and explain why colour is still a real property.
Check your reasoning after answering
Density separates them at 19.3 against 5.0 g/cm3, and hardness at 2.5 to 3 against 6 to 6.5; the streak test also gives yellow against greenish-black. Colour remains a genuine physical property that can be measured and reported, it simply happens to be shared by these two, which shows why identification needs several properties rather than one.
Apply the whole unit · Identify the unknown
A grey metallic block has a mass of 108 g. Lowered into a measuring cylinder, it raises the water level from 50.0 mL to 90.0 mL. A steel knife does not scratch it. Candidate materials: aluminium at 2.70 g/cm3, iron at 7.87, lead at 11.3.
Your task
Calculate the density, choose between the candidates, and state how confident you can be.
What your response should show
- Calculates the volume by displacement and the density with correct units
- Matches the value to a candidate and rules the others out with figures
- Brings the second property in rather than relying on density alone
- States the limit of the identification instead of claiming certainty
Response guidance — after your attempt
Find the volume from the displacement and calculate the density with its unit.
Compare your value with the three candidates and say which it matches.
Use the hardness observation and state what your identification does and does not establish.
The block displaced 90.0 minus 50.0, which is 40.0 mL, so its volume is 40.0 cm3. The density is 108 divided by 40.0, which is 2.70 g/cm3. That matches aluminium exactly and is nowhere near iron at 7.87 or lead at 11.3, so the density alone rules both of those out: a 40 cm3 piece of lead would have a mass of about 452 g, more than four times what was measured. The hardness observation is consistent, since aluminium is harder than a fingernail but is not especially soft. What this establishes is that the block behaves like aluminium on the two properties tested, which is strong evidence given how far apart the three candidates are. What it does not establish is that the block is pure aluminium rather than an aluminium alloy, since many alloys have densities within a few per cent of the pure metal, and separating those would need a test this investigation did not do.
Scientific references for this unit
- US Geological Survey: mineral identification, streak and the Mohs hardness scale
- US Geological Survey Water Science School: density and measuring volume
- National Institute of Standards and Technology: densities of common materials
- Royal Society of Chemistry: physical and chemical change, and safe laboratory practice
Parent / teacher teaching notes
Begin with the literal observation, quantity, or diagram relationship. Ask the learner to explain why the evidence supports the claim, check a relevant unit or boundary, then discuss an alternative. The supplied numbers are illustrative, not real observations. A parent can read aloud and allow a spoken rehearsal before writing. Judge science and evidence rather than political, faith, or family agreement. Use the complete data option whenever an actual procedure is unavailable or unsuitable.
The key idea
A physical property can be measured without turning the substance into something else, and a physical change alters form while leaving identity intact. Because different substances share individual properties, identification always needs more than one.
Worked examples
Example 1 · Mass, volume, and shape
- Divide mass by volume and include units.
- Distinguish shape from chemical identity.
- State the model’s condition about no material exchange.
Example 2 · Transfer to a different case
- Identify the distinction and conditions in this new case.
- Use the relevant mechanism or quantity; preserve units and the chosen boundary.
- Explain why the answer fits and what further evidence would be needed.
Example 3 · Recognize a change in form · a focused example
- State the literal condition or observation in the example.
- Apply the distinction explained in this teaching part.
- Check whether the claim stays within the given evidence.
Avoid these mistakes
- Misleading claim: Every change that is hard to reverse must be chemical.
- Misleading claim: The largest sample must have the greatest density.
- Omitting units, measurement timing, or a model boundary makes the evidence hard to interpret.
- Treating illustrative classroom data as real field measurements or copying a model is not an independent explanation.
Check your understanding
Why is transparency insufficient for identification?
Discuss the answer
Different substances can share that appearance. Additional appropriate measurements and known source information are needed.
Does irreversibility alone prove chemical change?
Discuss the answer
No. A physical change can be difficult to undo; evidence about substance identity is more relevant.
What is the larger piece’s density if it is 60 g and 20 cm³?
Discuss the answer
60 ÷ 20 = 3 g/cm³; size changed but the ratio stayed the same.
How can a second property strengthen identification?
Discuss the answer
It distinguishes possibilities that the first property cannot separate; the conclusion should still match the available evidence.
Sustained scientific explanation and revision
These activities are scheduled in the parent plan. Save a first response, a correction, and the reason for the change in the science notebook.
Activity 1 · Physical Properties & Physical Changes · Model and explain
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Physical Properties & Physical Changes, study Parts 3–4 and attempt the worked case: Calculate density and explain why the shape change is physical under the stated model. Explain the steps before opening the model.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
48 ÷ 24 = 2 g/cm³. Reshaping changes outline while the modeled material and measured mass/volume remain. It is described as physical because no new substance is formed in this model, not simply because the shape might be restored.
Activity 2 · Physical Properties & Physical Changes · Practice, revise and exit
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Physical Properties & Physical Changes, complete practice questions 1–6 and both choice checks, then answer independently: A material is blue and heavy. Can those words uniquely identify it? Save a correction explaining one changed claim or method.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
No. Blue depends on the observation conditions, and heavy depends on sample size and comparison. Measured properties and additional evidence are needed.
Words to know
- physical property
- A characteristic measured or observed without changing chemical identity.
- density
- Mass per unit volume under stated conditions.
- evidence
- A relevant observation, measurement, or record used to evaluate a claim.
- system boundary
- The chosen limit specifying what is included and which exchanges cross it.
- uncertainty
- A stated limit in measurement or knowledge, distinct from a careless mistake.
In the real world
Jewelers and scientists use density like a fingerprint: because pure gold always has a density of 19.3 g/cm³, measuring a coin's mass and volume can reveal whether it is real gold or a cheaper fake.
Exit check
A material is blue and heavy. Can those words uniquely identify it?
Check your explanation after answering
No. Blue depends on the observation conditions, and heavy depends on sample size and comparison. Measured properties and additional evidence are needed.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.