Density & Floating — Print · Grade 8 Science

Chapter: Properties of Matter

Density & Floating

Learning goals

Readings for this unit

Before this unit

Divide a measured mass by a measured volume and write units.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions. Study the four visual parts across the scheduled sessions, then use the illustrated worked case and independent transfer questions. Extend or revisit a session when the learner cannot explain the figure; completion time is a proposed allocation, not a mastery guarantee.

Start with something to explain

What would explain the observations in density & floating?

Illustrative evidence case. Illustrative block A has mass 24 g and volume 8 cm³; B has mass 48 g and volume 16 cm³. A separate hollow object has mass 48 g and enclosed volume 80 cm³.

Same density, different mass and volumeSame density, different mass and volumeA: 24 g / 8 cm³B: 48 g / 16 cm³A: 3 g/cm³B: 3 g/cm³Box sizes are schematic; use the measured volumes.Illustrative model • use the labels and stated assumptions
Doubling both mass and volume leaves their ratio unchanged. The schematic boxes are not geometrically scaled to measured volumes, so calculate from their labels. Density describes mass per volume, whereas mass alone describes the amount in each sample.

Notice and wonder

Why does B not have twice A’s density? Write two observations from the supplied record without guessing their cause. Include a quantity and its unit when given. An observation is what was recorded; an inference is what you think explains it.

Write one question about why the pattern occurs and one question about a comparison that could distinguish competing explanations. A testable question identifies what could be observed or measured.

Your first model

Before reading the model answer, sketch the named objects or particles and label what you think changes. Calculate the three average densities and use modeled water density 1 g/cm³ to predict floating. Explain why mass alone does not rank floating behavior. Keep this first version so you can compare it with your later explanation.

Build the language from the idea

density
Mass divided by volume.
displacement
The fluid volume moved aside by a submerged object.
average density
Total mass divided by the total volume enclosed by an object.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

Mass tells how much matter a sample contains. Volume tells how much space it occupies. Density compares them: density = mass ÷ volume. A small metal sample can be denser than a large wooden block even though the wooden block has more total mass. Include units, such as grams per cubic centimeter (g/cm³).

Same density, different mass and volumeSame density, different mass and volumeA: 24 g / 8 cm³B: 48 g / 16 cm³A: 3 g/cm³B: 3 g/cm³Box sizes are schematic; use the measured volumes.Illustrative model • use the labels and stated assumptions
Doubling both mass and volume leaves their ratio unchanged. The schematic boxes are not geometrically scaled to measured volumes, so calculate from their labels. Density describes mass per volume, whereas mass alone describes the amount in each sample.

Samples of the same uniform substance have approximately the same density at the same temperature and pressure. If one sample has twice the volume, it also has twice the mass. For an irregular solid that does not dissolve or absorb water, volume can be measured by the rise in water level when it is submerged: 1 mL is the same volume as 1 cm³.

Measure an irregular object by displacementMeasure an irregular object by displacementBefore: 40 mLAfter: 58 mLObject volume = displaced liquid volume = 58 − 40 = 18 mL.1 mL = 1 cm³; read the scale and use complete submersion.
The two graduated-cylinder records show a rise from 40 to 58 mL when an object is fully submerged without trapped bubbles or spilled liquid. Its displaced volume is 18 mL, equivalent to 18 cm³. The cylinders are schematic: use the written readings rather than measuring the artwork, and do not confuse final total volume with object volume.

An object with average density below that of its surrounding fluid can float; one with greater average density tends to sink. Average density includes enclosed air. A hollow steel ship can float even though solid steel is denser than water. The upward buoyant force balances the ship’s weight once it floats steadily.

Density compares equal volumes: a cubic centimeter of a denser material has more mass. For a regular box, find volume by multiplying length, width and height. For an irregular solid that does not dissolve, measure the rise in water level after complete immersion.

Floating depends on an object's average density compared with the fluid. A hollow steel boat can float because its enclosed air greatly increases volume without much added mass. The material's density and the whole object's average density are different quantities.

Make a fair density test with a balance, a graduated container and recorded units. Read water level at eye level, remove trapped bubbles and repeat the measurement. If densities are very close, uncertainty can prevent a confident prediction; collect better measurements instead of forcing a conclusion.

Build your understanding

Where this fits

This unit helps answer the chapter question: How can the properties of a material help us explain it and choose a useful application? Keep a notebook table with the question, evidence, explanation and one unresolved issue. Revise a previous model only when you can name the new evidence or idea that changes it.

Use three teaching sessions of about 40–45 minutes: first observe the phenomenon, draft a model and study Parts 1–2; next explain Parts 3–4 with the figures and evidence; then work the application, examples, practice, model revision and exit check. Divide any difficult part into smaller steps and revisit after feedback. Investigations, review and quizzes have separate sessions.

Part 1 · Compare equal volumes rather than total size

Mass measures the amount of matter; volume measures occupied space. Density is mass divided by volume. To compare materials fairly, imagine equal-sized portions. A large low-density block can have more total mass than a tiny high-density piece, so heavier does not always mean denser.

Same density, different mass and volumeSame density, different mass and volumeA: 24 g / 8 cm³B: 48 g / 16 cm³A: 3 g/cm³B: 3 g/cm³Box sizes are schematic; use the measured volumes.Illustrative model • use the labels and stated assumptions
Doubling both mass and volume leaves their ratio unchanged. The schematic boxes are not geometrically scaled to measured volumes, so calculate from their labels. Density describes mass per volume, whereas mass alone describes the amount in each sample.

Read the visual model

Which comparison decides density: equal masses or equal volumes?

Visual reasoning — after your attempt

Comparing the masses of equal volumes directly shows which is denser. Calculating mass divided by volume also works when the samples have different sizes.

A 20 cm³ sample of material A has mass 40 g; its density is 2 g/cm³. A 5 cm³ sample of B has mass 15 g; its density is 3 g/cm³. B is denser even though the sample of A has greater mass.

Pause and explain

Which comparison decides density: equal masses or equal volumes?

Check your reasoning after answering

Comparing the masses of equal volumes directly shows which is denser. Calculating mass divided by volume also works when the samples have different sizes.

Part 2 · Measure volume and carry the units

For a rectangular block, multiply length, width, and height using the same length unit. For a suitable irregular object, measure the increase in water level after complete immersion. This displacement method assumes the sample does not dissolve, absorb appreciable water, or trap bubbles that distort the reading.

Measure an irregular object by displacementMeasure an irregular object by displacementBefore: 40 mLAfter: 58 mLObject volume = displaced liquid volume = 58 − 40 = 18 mL.1 mL = 1 cm³; read the scale and use complete submersion.
The two graduated-cylinder records show a rise from 40 to 58 mL when an object is fully submerged without trapped bubbles or spilled liquid. Its displaced volume is 18 mL, equivalent to 18 cm³. The cylinders are schematic: use the written readings rather than measuring the artwork, and do not confuse final total volume with object volume.

Read the visual model

First find the change in liquid reading; it is the volume of the submerged object. Then combine that volume with a separately measured mass. A water displacement of 18 mL does not mean the object itself has a mass of 18 g.

If the object has mass 54 g, what density follows from these readings?

Visual reasoning — after your attempt

Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.

A cylinder rises from 30 mL to 42 mL. The object volume is 12 mL, equivalent to 12 cm³. If mass is 24 g, density is 24/12=2 g/cm³. Use the change in reading, not the final total water volume.

Pause and explain

What error comes from using 42 mL as the object volume?

Check your reasoning after answering

It includes the original water volume. The object displaced only 12 mL; using 42 mL would underestimate its density.

Part 3 · Connect floating to the whole object

Floating depends on the average density of the whole object compared with the surrounding fluid, under the simple introductory model. Enclosed air contributes volume with little mass. A hollow boat can therefore have a lower average density than water even when its hull material is denser than water.

Three density calculations distinguish solid material from a sealed hollow object.Solid sample A54 g / 20 cm³ = 2.7Density in g/cm³Sample B16 g / 20 cm³ = 0.8Density in g/cm³Sealed hollow A54 g / 90 cm³ = 0.6Density in g/cm³Compare each whole-object average with water: 1.0 g/cm³
Supplied-data comparison. The hollow object has the same mass as sample A but a larger sealed outer volume. Its material density stays the same; its whole-object average changes.

Read the visual model

Why can a solid steel piece sink while a steel boat floats?

Visual reasoning — after your attempt

The boat includes enclosed air and displaces enough water to balance its weight. Its whole-object average density can be lower than water, unlike the solid piece.

If water enters a hollow boat, its total mass increases and its effective average density can increase. A floating object settles until its upward buoyant force balances its weight; the fraction submerged depends on the amount of fluid displaced.

Pause and explain

Why can a solid steel piece sink while a steel boat floats?

Check your reasoning after answering

The boat includes enclosed air and displaces enough water to balance its weight. Its whole-object average density can be lower than water, unlike the solid piece.

Part 4 · Decide how confident a prediction should be

Measurements are estimates. Read a graduated cylinder at eye level and repeat mass and volume measurements. Compare densities using the fluid temperature and the same units. A value very close to the fluid density may not support a confident prediction because small measurement differences can change the comparison.

Repeat volume readings before judging a boundary caseRepeat volume readings before judging a boundary caseRepeated mass 24 g; volume readings vary slightly.Volume (cm³)7.588.5123Trial numberDensity range: 24/8.1 to 24/7.9 ≈ 2.96 to 3.04 g/cm³.
With a supplied mass of 24 g, repeated volume readings of 7.9, 8.0 and 8.1 cm³ lead to density estimates of about 3.04, 3.00 and 2.96 g/cm³. The dots are repeat readings, not a trend over time. A prediction close to a comparison-fluid density needs measurement uncertainty and the whole object’s average density considered.

Read the visual model

A repeat is another measurement under intended comparable conditions. Variation can come from the method or object. Report a reasonable range and ask whether it changes the prediction instead of selecting only the most convenient reading.

Would these results justify claiming an exact density of 3.0000 g/cm³? Explain how the graph helps.

Visual reasoning — after your attempt

No. The repeated volume readings vary, giving a density range of roughly 2.96–3.04 g/cm³. The graph makes that variation visible and discourages unsupported precision.

A measured density of 0.99 g/mL with uncertainty about 0.03 g/mL overlaps water near 1.00 g/mL. Record this uncertainty and improve the measurement rather than declaring certain floating from the central value alone.

Pause and explain

Why should near-equal measured densities be treated cautiously?

Check your reasoning after answering

The plausible density range can cross the fluid density. More precise measurements or a controlled observation are needed for a confident prediction.

Density as identification evidence

An unknown sample can be compared with reference densities, but one matching ratio is not always a unique identity.

Density is mass divided by volume. To investigate a sample, measure its mass with a suitable balance and its volume by dimensions or safe displacement. Displacement is the volume of liquid moved aside by an immersed object. Subtract the initial from the final vessel reading, use compatible units and make sure the object is fully immersed. A soluble or absorbent object can change during the test and needs another method.

Density as identification evidenceDensity as identification evidenceQuantity (g/cm³)Sample U: density2.7Reference A2.7Reference B7.9Reference C2.7Illustrative model • read the labels and caption
The supplied sample and reference bars show matching values for more than one reference. Bar lengths compare stated densities, not object masses. The matching A/C values demonstrate why density can support or rule out a candidate without uniquely identify every sample; actual reference conditions and uncertainty must also be checked.

A measurement has uncertainty from scale divisions, reading angle, trapped air, method and sample variation. Compare a result with references under appropriate temperature and composition conditions. A small numerical difference may not distinguish two materials if the uncertainty is larger. A porous or hollow object’s average density is not automatically its solid material density. Repeated matched measurements can show consistency but do not repair an unsuitable volume method.

Use density as one piece of identification evidence, alongside characteristic properties such as solubility, melting behavior or conductivity. Solubility describes how much of a substance can dissolve under stated conditions; it is not the speed of dissolving. Two different materials can have overlapping densities. State whether the evidence is consistent with an identity, rejects a proposed identity, or remains insufficient. Do not label a sample certain merely because one table number matches.

Explain and apply

An illustrative unknown has mass 54 g and displacement 20 cm³. References A/C are both 2.7 g/cm³ and B is 7.9. Argue which candidates remain and plan one additional property comparison.

  • Calculate with correct units.
  • Distinguish support from unique identification.
  • Choose relevant further evidence and controls.
Reasoning and marking guidance — after your attempt

Density is 54/20=2.7 g/cm³. A and C remain consistent while B is inconsistent under matched conditions and adequate measurement precision. Density alone cannot distinguish A from C; choose a safe characteristic-property comparison with supplied reference data, such as solubility for approved samples. Name uncertainty and whether the object is homogeneous.

Apply the whole unit · Two samples and a hollow design

Illustrative measurements: sample A has mass 54 g and displaces 20 mL; sample B has mass 16 g and volume 20 cm³. A hollow object made using A has total mass 54 g and sealed outer volume 90 cm³. Use water density 1.0 g/cm³.

Your task

Calculate all three densities, predict their behavior in water under the stated assumptions, and explain how changing shape can change whole-object density without changing material density. Name a measurement that needs checking.

What your response should show

  • Use the supplied observations and measurements accurately; include units where needed.
  • Explain the science behind your answer, rather than giving only a number or label.
  • Show a labeled model, calculation, or evidence comparison as requested.
  • State an assumption or limitation and suggest a useful next measurement.
Response guidance — after your attempt

A is 54/20=2.7 g/cm³ and sinks in the simple model. B is 16/20=0.8 g/cm³ and can float. The sealed hollow object averages 54/90=0.6 g/cm³ and can float. Its material remains 2.7 g/cm³; added enclosed volume changes the average. Check sealing, actual outer volume, bubbles, and mass.

Scientific references and adapted teaching resources

Parent / teacher teaching notes

Use a mass/volume/density table before discussing boats. Rehearse 1 mL=1 cm³ and the displacement subtraction. Let the learner explain the hollow case with a labeled cross-section, then check with a different set of numbers. Adaptation credit: OpenSciEd middle-school teaching routines and public task structures by OpenSciEd, CC BY 4.0. This independent version adds original explanatory text, data and figures, terminology support and a home-study schedule. It does not reproduce the complete teacher editions or their classroom validation. The illustrated study “Density as identification evidence” is required curriculum content. Teach it before the associated application or distribute it over the assigned unit sessions; it is not an optional enrichment. Its practice/review/quiz response provides evidence of understanding. Model/data work must not be recorded as an actual physical experiment.

The key idea

Density is mass per volume. Floating depends on the object’s average density compared with the fluid.

Worked examples

Example 1

A sample has mass 36 g and volume 12 cm³. Will it sink in water of density 1 g/cm³?
  1. Density = 36 ÷ 12 = 3 g/cm³.
  2. 3 g/cm³ is greater than 1 g/cm³.
Answer: It sinks, assuming it does not dissolve or trap enough additional air to change its average density.

Example 2 · Worked application

Two blocks have masses 24 g and 72 g and volumes 8 cm³ and 24 cm³. Are their densities different?
  1. Calculate density for each, rather than comparing masses alone.
  2. 24/8=3 g/cm³; 72/24=3 g/cm³.
  3. Both have the same mass per equal volume, though the larger block has three times the mass.
Answer: Both densities are 3 g/cm³.
Check the reasoning

Multiply density by volume: 3×8=24 and 3×24=72, recovering the masses.

Example 3

A solid has mass 54 g and raises water from 20 mL to 40 mL. Find density and predict its behavior in water of density 1 g/mL.
  1. Displaced volume is 40−20=20 mL.
  2. Density is 54/20=2.7 g/mL, greater than water's density.
Answer: 2.7 g/mL; the solid sinks if fully wetted, without trapped air or other support.

Example 4 · Find and fix the mistake

Context: Two blocks have masses 24 g and 72 g and volumes 8 cm³ and 24 cm³. Are their densities different? A student claims: “The 72 g block must be denser because it is heavier.” Evaluate the claim and write a corrected explanation.
  1. Test the claim against the stated conditions: Multiply density by volume: 3×8=24 and 3×24=72, recovering the masses.
  2. Density compares mass per volume. A larger sample can be heavier while having the same density.
Answer: Density compares mass per volume. A larger sample can be heavier while having the same density.
Check the reasoning

Multiply density by volume: 3×8=24 and 3×24=72, recovering the masses.

Example 5 · Interpret the visual model

Measure an irregular object by displacementMeasure an irregular object by displacementBefore: 40 mLAfter: 58 mLObject volume = displaced liquid volume = 58 − 40 = 18 mL.1 mL = 1 cm³; read the scale and use complete submersion.
The two graduated-cylinder records show a rise from 40 to 58 mL when an object is fully submerged without trapped bubbles or spilled liquid. Its displaced volume is 18 mL, equivalent to 18 cm³. The cylinders are schematic: use the written readings rather than measuring the artwork, and do not confuse final total volume with object volume.
If the object has mass 54 g, what density follows from these readings?
  1. Identify the represented objects, axes, labeled quantities and stated conditions before choosing a rule.
  2. First find the change in liquid reading; it is the volume of the submerged object. Then combine that volume with a separately measured mass. A water displacement of 18 mL does not mean the object itself has a mass of 18 g.
  3. Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.
Answer: Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.
Check the reasoning

Keep the model’s stated limits explicit: The two graduated-cylinder records show a rise from 40 to 58 mL when an object is fully submerged without trapped bubbles or spilled liquid. Its displaced volume is 18 mL, equivalent to 18 cm³. The cylinders are schematic: use the written readings rather than measuring the artwork, and do not confuse final total volume with object volume.

Avoid these mistakes

Check your understanding

A solid has mass 54 g and raises water from 20 mL to 40 mL. Find density and predict its behavior in water of density 1 g/mL.

Discuss the answer

2.7 g/mL; the solid sinks if fully wetted, without trapped air or other support.

Why can a steel ship float while a small solid steel piece sinks?

Discuss the answer

Density compares equal volumes: a cubic centimeter of a denser material has more mass. For a regular box, find volume by multiplying length, width and height. For an irregular solid that does not dissolve, measure the rise in water level after complete immersion. Floating depends on an object's average density compared with the fluid. A hollow steel boat can float because its enclosed air greatly increases volume without much added mass. The material's density and the whole object's average density are different quantities.

Return to the phenomenon

Illustrative block A has mass 24 g and volume 8 cm³; B has mass 48 g and volume 16 cm³. A separate hollow object has mass 48 g and enclosed volume 80 cm³. Calculate the three average densities and use modeled water density 1 g/cm³ to predict floating. Explain why mass alone does not rank floating behavior. Revise your first explanation using a named idea from the lesson. Mark the changed part and the evidence that supports it.

Compare explanations after your attempt

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

What should we investigate next?

Which condition could be changed to test a remaining uncertainty in this explanation? State the predicted observation and what result would challenge your model.

Investigate, explain and revise

Use these tasks after the three teaching sessions as needed. Discuss an attempted explanation before opening response guidance; a matching sentence is less important than a supported mechanism.

Activity 1 · Explain the new evidence case

Assigned reading and preparation

Use the visible scenario above, Parts 1–4 and your first notebook model. No outside reading or equipment is required.

Your task

Illustrative block A has mass 24 g and volume 8 cm³; B has mass 48 g and volume 16 cm³. A separate hollow object has mass 48 g and enclosed volume 80 cm³. Calculate the three average densities and use modeled water density 1 g/cm³ to predict floating. Explain why mass alone does not rank floating behavior. Write 100–150 words or give a labeled drawing with an equivalent oral explanation.

Success criteria

  • Use two relevant stated records or conditions.
  • Explain the mechanism rather than only repeat a definition.
  • Label observation and inference.
  • Name a changed model feature or a remaining uncertainty.
Hint

Read the stated boundary and conditions before selecting a relationship. The unit’s worked examples show intermediate steps.

Response guidance — discuss after your attempt

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

Activity 2 · Read and extend the figure

Assigned reading and preparation

Study the new labeled figure and its caption, then compare it with the associated teaching part.

Your task

Why does B not have twice A’s density? Explain what each important label represents. Add one missing feature that would be needed for a different question, and explain why the original figure leaves it out.

Success criteria

  • Answer the figure question with its stated labels.
  • Connect the visual representation to the scientific idea.
  • Identify a specific limitation from the caption.
  • Propose a useful change without misrepresenting supplied evidence.
Hint

Doubling both mass and volume leaves their ratio unchanged. The schematic boxes are not geometrically scaled to measured volumes, so calculate from their labels. Density describes mass per volume, whereas mass alone describes the amount in each sample.

Response guidance — discuss after your attempt

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³. Doubling both mass and volume leaves their ratio unchanged. The schematic boxes are not geometrically scaled to measured volumes, so calculate from their labels. Density describes mass per volume, whereas mass alone describes the amount in each sample.

Words to know

density
Mass divided by volume.
displacement
The fluid volume moved aside by a submerged object.
average density
Total mass divided by the total volume enclosed by an object.
model
A representation that explains selected features and makes predictions; it leaves other details out.
evidence
Relevant observations or records used to support or challenge a claim.
system boundary
The chosen limit separating the objects being studied from their surroundings.
assumption
A condition taken as true for a model or calculation and needing to be checked against the case.

In the real world

Exit check

Why does B not have twice A’s density? Explain the relevant mechanism and one model limitation.

Check your explanation after answering

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³. Doubling both mass and volume leaves their ratio unchanged. The schematic boxes are not geometrically scaled to measured volumes, so calculate from their labels. Density describes mass per volume, whereas mass alone describes the amount in each sample.

Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.

Problem 1

A block has mass 24 g and volume 8 cm³. Find its density.
Answer and explanation

3 g/cm³

24 ÷ 8 = 3.

Problem 2

Which measurement is needed with mass to calculate density?
  1. Time
  2. Volume
  3. Speed
  4. Force
Answer and explanation

Volume

Density = mass ÷ volume.

Problem 3

Why can a steel ship float?
Answer and explanation

Its hull encloses air, reducing its average density below that of water.

Problem 4

A solid has mass 54 g and raises water from 20 mL to 40 mL. Find density and predict its behavior in water of density 1 g/mL. Explain the reasoning in your own words.
Answer and explanation

2.7 g/mL; the solid sinks if fully wetted, without trapped air or other support.

Displaced volume is 40−20=20 mL.

Density is 54/20=2.7 g/mL, greater than water's density.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 5

An insoluble solid has mass 35 g and displaces 14 mL of water. Find its density and show the volume method. Show the reasoning, not just the conclusion.
Hints

Multiply density by volume: 3×8=24 and 3×24=72, recovering the masses.

Answer and explanation

The solid volume is 14 mL; density=35/14=2.5 g/mL.

Problem 6

A hollow object has total mass 90 g and external volume 120 mL. Predict floating in water at 1 g/mL using its average density.
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

90/120=0.75 g/mL, below water's density. It can float under the model, provided it retains its enclosed air and does not take in water.

Problem 7

Why does B not have twice A’s density?
Answer and explanation

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³.

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 8

Illustrative block A has mass 24 g and volume 8 cm³; B has mass 48 g and volume 16 cm³. A separate hollow object has mass 48 g and enclosed volume 80 cm³. Calculate the three average densities and use modeled water density 1 g/cm³ to predict floating. Explain why mass alone does not rank floating behavior.
Answer and explanation

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 9

An illustrative unknown has mass 54 g and displacement 20 cm³. References A/C are both 2.7 g/cm³ and B is 7.9. Argue which candidates remain and plan one additional property comparison.
Hints

Use the illustrated study titled Density as identification evidence. Keep the observations, model and conditions separate.

Answer and explanation

Density is 54/20=2.7 g/cm³. A and C remain consistent while B is inconsistent under matched conditions and adequate measurement precision. Density alone cannot distinguish A from C; choose a safe characteristic-property comparison with supplied reference data, such as solubility for approved samples. Name uncertainty and whether the object is homogeneous.

Density is 54/20=2.7 g/cm³. A and C remain consistent while B is inconsistent under matched conditions and adequate measurement precision. Density alone cannot distinguish A from C; choose a safe characteristic-property comparison with supplied reference data, such as solubility for approved samples. Name uncertainty and whether the object is homogeneous.

Response criteria

  • Calculate with correct units.
  • Distinguish support from unique identification.
  • Choose relevant further evidence and controls.

Problem 10

Measure an irregular object by displacementMeasure an irregular object by displacementBefore: 40 mLAfter: 58 mLObject volume = displaced liquid volume = 58 − 40 = 18 mL.1 mL = 1 cm³; read the scale and use complete submersion.
The two graduated-cylinder records show a rise from 40 to 58 mL when an object is fully submerged without trapped bubbles or spilled liquid. Its displaced volume is 18 mL, equivalent to 18 cm³. The cylinders are schematic: use the written readings rather than measuring the artwork, and do not confuse final total volume with object volume.
For this unit’s illustrated case, if the object has mass 54 g, what density follows from these readings?
Answer and explanation

Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.

First find the change in liquid reading; it is the volume of the submerged object. Then combine that volume with a separately measured mass. A water displacement of 18 mL does not mean the object itself has a mass of 18 g.

Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 11

A completely submerged object raises a cylinder reading from 35 to 47 mL and has mass 30 g. Calculate density, state the volume assumptions and distinguish this from claiming an exact substance identity.
Answer and explanation

Volume 12 cm³; density 30/12 = 2.5 g/cm³. Assume no trapped bubbles, spill or incomplete submersion. Density alone does not uniquely identify the substance.

Use the change 47−35, not the final reading.

Divide the independently measured mass by that volume.

Volume 12 cm³; density 30/12 = 2.5 g/cm³. Assume no trapped bubbles, spill or incomplete submersion. Density alone does not uniquely identify the substance.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

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