Density: $m = \rho V$
Why can two metal cubes of the same size feel completely different in weight in your hand, and how can you work out what an unknown object is made of using only a balance and a measuring cylinder?
Prior knowledge
- In direct proportionality the measurement points lie, within the measurement uncertainty, on a straight line through the origin, and its slope is the ratio $y : x$chapter Graphs and fitting a straight line
- Measurement results always contain error, and a result is given with a suitable number of significant figureschapter Measuring and the reliability of a measurement
- Quantities are calculated in SI units, and units are converted using powers of tenchapter Quantities, units and the SI system
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 3.
Think about it
At a recycling point there are two grey metal cubes on the table. They are exactly the same size, but one feels much heavier in your hand than the other. One is aluminium and the other is iron, but the cubes have no markings. On the table there are a balance and a ruler.
Which quantities of the cubes should you measure so that you can work out what they are made of?
Optional Recap: direct proportionality
Predict
Make your predictions before investigating 2 questions
Prediction 1: a piece three times as big
Prediction 2: two substances in the same coordinate system
Investigate
Aluminium pieces: is mass proportional to volume?
Simulated example data
The table shows example data for five aluminium pieces of different sizes. The volume has been calculated from the dimensions and the mass measured with a balance. The numbers are example values, not real measurements, but they include the usual measurement error.
| Piece | $V$ ($\text{cm}^3$) | $m$ (g) |
|---|---|---|
| 1 | 10.0 | 26.8 |
| 2 | 20.0 | 54.4 |
| 3 | 30.0 | 80.6 |
| 4 | 40.0 | 108.4 |
| 5 | 50.0 | 134.6 |
Calculate the ratio $m : V$ for each row.
Unknown metal
Simulated example data
The table shows example data for pieces made of an unknown metal. The numbers are example values. You can calculate density in two ways: as the ratio $m : V$ of a single row, or as the ratio of the differences $\Delta m : \Delta V$ of two rows far apart. The latter uses the whole measuring range.
| Piece | $V$ ($\text{cm}^3$) | $m$ (g) |
|---|---|---|
| 1 | 5.0 | 45.1 |
| 2 | 10.0 | 89.2 |
| 3 | 15.0 | 135.0 |
| 4 | 20.0 | 178.9 |
| 5 | 25.0 | 224.3 |
Calculate the density of the metal using the first and last rows and the formula $\rho = \Delta m : \Delta V$. Finally, compare your result with the table in the book (the theory part of this chapter): aluminium $2.70$, iron $7.87$, copper $8.96$, lead $11.3$ and gold $19.3\,\text{g/cm}^3$. Give your answer in $\text{g/cm}^3$.
Hint 1: a guiding question
What is the change in mass and what is the change in volume between the first and the last row?
Hint 2: what is the connection
$\Delta m = 224.3\,\text{g} - 45.1\,\text{g}$ and $\Delta V = 25.0\,\text{cm}^3 - 5.0\,\text{cm}^3$.
Hint 3: an intermediate step
Show the solution
Given
first row $V_1 = 5.0\,\text{cm}^3$, $m_1 = 45.1\,\text{g}$ and last row $V_5 = 25.0\,\text{cm}^3$, $m_5 = 224.3\,\text{g}$
Find
the density $\rho$ of the metal in $\text{g/cm}^3$
Principle
Mass is directly proportional to volume, so the measurement points lie on the line $m = \rho V$, whose slope is the density. The slope is obtained as the ratio of the differences of two points far apart.
Formula
Conversions
Not needed: the masses are in grams and the volumes in cubic centimetres, and the answer is asked in $\text{g/cm}^3$.
Substitution
Answer
$8.96\,\tfrac{\text{g}}{\text{cm}^3}$
Three significant figures like the measurement results.
Reasonableness
The value matches copper in the table, $8.96\,\tfrac{\text{g}}{\text{cm}^3}$, so the metal is probably copper. Check with a single row:
which is almost the same.
Liquid in a container: what does the mass consist of?
Simulated example data
Liquid is poured into a measuring flask, and the balance shows the total mass $m_\text{tot}$ of the flask and the liquid. The table has example data. The numbers are example values.
| $V$ ($\text{cm}^3$) | $m_\text{tot}$ (g) |
|---|---|
| 0 | 42.1 |
| 20 | 57.9 |
| 40 | 73.5 |
| 60 | 89.6 |
| 80 | 105.1 |
Investigate how much the mass increases for every $20\,\text{cm}^3$.
Conclusion: how are mass and volume related?
Write based on the three tables: How do you know that the mass of the aluminium pieces is directly proportional to volume? How was the density of the unknown metal obtained? Which line, that of aluminium or that of the unknown metal, would be steeper in the coordinate system, and why? What must you do if a liquid is measured in a container?
Model answer
For the aluminium pieces the ratio $m : V$ was about $2.7\,\text{g/cm}^3$ in every row, so mass is directly proportional to volume and the ratio is the density of the substance. The density of the unknown metal was obtained from the differences of two points far apart, $\rho = \Delta m : \Delta V = 179.2\,\text{g} : 20.0\,\text{cm}^3 = 8.96\,\text{g/cm}^3$, and it matched copper. The line of copper is steeper than that of aluminium, because its density, that is its slope, is greater. The total mass of a liquid with its container is not proportional to volume, because the mass of the empty container is included. So the mass of the container must be subtracted from the total mass, or the density must be read from the slope of the line, in which case the mass of the container is the intercept. The data were example values with small measurement error. So you can work out what the cubes are made of by measuring their mass and volume and calculating the density from them.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
The mass $m$ of an object tells you how much matter the object contains, and its unit is the kilogram. The volume $V$ tells you how much space the object takes up; its SI unit is the cubic metre ($\text{m}^3$). Two objects of the same size may have different weights if they are made of different substances: a ten-cubic-centimetre pine cube is light, an aluminium cube is clearly heavier and an iron cube is the heaviest.
The difference comes from how tightly the matter is packed inside the volume. This is described by density $\rho$: it tells you how much mass a substance has per unit of volume. Iron is about sixteen times as dense as pine, so the same volume contains about sixteen times as much matter.
Mass is directly proportional to volume
When you measure pieces of different sizes made of the same substance, you notice a regularity: twice the volume gives twice the mass, and half the volume gives half the mass. So mass is directly proportional to volume, which means the ratio $m : V$ stays the same. That constant ratio is the density of the substance.
If the measurement results are plotted in a coordinate system with volume on the horizontal axis and mass on the vertical axis, the points lie, within the measurement uncertainty, on a straight line through the origin (chapter Graphs and fitting a straight line). The slope of the line is the density.
Density tells you what the substance is
Every pure substance has its own density in a given state of matter and at a given temperature, so density is like a fingerprint of the substance: once the mass and volume of an object have been measured, the calculated density can be compared with table values. The conclusion is not certain. Measurements always contain error (chapter Measuring and the reliability of a measurement), and some substances have almost the same density, so the result only tells you which substances the object may be made of.
Density
The density of a substance is the ratio of mass to volume. Solving this ratio for mass and for volume gives two other forms:
$$\begin{aligned} \rho &= \frac{m}{V} \\ m &= \rho V \\ V &= \frac{m}{\rho} \end{aligned}$$
where $\rho$ is the density (SI unit $\text{kg/m}^3$), $m$ the mass (kg) and $V$ the volume ($\text{m}^3$). The form $m = \rho V$ is a model of direct proportionality $y = kx$: the mass $m$ is $y$, the volume $V$ is $x$ and the density $\rho$ is the slope $k$.
In the figure, the points of each substance lie on their own line. The steeper the line, the greater the density. You can get the density from a graph by reading two points far apart on the line and calculating $\rho = \Delta m : \Delta V$. The result obtained this way is more accurate than the ratio $m : V$ of a single point, because it uses the whole line.
Units
The SI unit of density is $\text{kg/m}^3$, but for small objects the unit $\text{g/cm}^3$ is often used. The relationship between the units is
$$1\,\tfrac{\text{g}}{\text{cm}^3} = \frac{10^{-3}\,\text{kg}}{10^{-6}\,\text{m}^3} = 1000\,\tfrac{\text{kg}}{\text{m}^3}$$
In a calculation, keep all quantities in the same system of units. The volume conversions are $1\,\text{cm}^3 = 10^{-6}\,\text{m}^3$ and $1\,\text{dm}^3 = 10^{-3}\,\text{m}^3$. In addition, $1\,\text{dm}^3$ is one litre, and $1\,\text{ml} = 1\,\text{cm}^3$.
| Substance | $\rho$ ($\text{kg/m}^3$) | $\rho$ ($\text{g/cm}^3$) |
|---|---|---|
| pine | 500 | 0.50 |
| ethanol | 790 | 0.79 |
| cooking oil | 920 | 0.92 |
| ice | 920 | 0.92 |
| water | 1000 | 1.00 |
| aluminium | 2700 | 2.70 |
| iron | 7870 | 7.87 |
| copper | 8960 | 8.96 |
| lead | 11 300 | 11.3 |
| gold | 19 300 | 19.3 |
Measuring volume
The volume of a regular object is calculated from its dimensions: the volume of a cube is $a^3$ and the volume of a rectangular block is the area of the base times the height. The volume of an irregular object is measured by displacement of water: the object is lowered into a measuring cylinder, and the volume is the rise of the water level, $V = V_2 - V_1$.
A reading of a measuring cylinder contains uncertainty, for example from reading the scale and the curvature of the liquid surface. The difference of two readings is a smaller number than the readings themselves, but it includes the uncertainty of both readings, so the difference may have fewer significant figures than either reading. This limits the accuracy of the density.
The mass of a container in a measurement
If a liquid is measured in a container and the balance shows the total mass of the container and the liquid, the measurement points do not pass through the origin: the mass $m_0$ of the empty container is included. The total mass follows a linear model $m_\text{tot} = \rho V + m_0$. The density is still the slope of the line, and the constant $m_0$ is the point where the line crosses the vertical axis, that is, the mass of the empty container. The total mass is not directly proportional to volume, but the mass of the liquid itself is.
Common misconception
Many people think that a bigger object is denser because it is heavier. A bigger piece of iron is indeed heavier than a small one, but their density is the same: mass and volume grow together so that the ratio $m : V$ does not change. Density is a property of the substance, not of the size of the object.
Another mistake is to mix up units. If the mass is in grams and the volume in cubic metres, the result is a thousand times too large or too small. First convert everything to one system of units: either $\text{kg}$ and $\text{m}^3$ or $\text{g}$ and $\text{cm}^3$.
A third mistake is to think that the mass of an empty container belongs to the liquid. It must be subtracted from the total mass before the density is calculated.
Summary
- Density is mass per unit of volume: $\rho = m : V$, SI unit $\text{kg/m}^3$. Water 1000 $\text{kg/m}^3$, aluminium 2700 $\text{kg/m}^3$, iron 7870 $\text{kg/m}^3$.
- The mass of a given substance is directly proportional to volume: $m = \rho V$. In a graph, $m(V)$ is a straight line through the origin whose slope is the density.
- Solve the formula for the quantity you need: $\rho = m : V$, $m = \rho V$ or $V = m : \rho$. Keep the units in one system: $1\,\text{g/cm}^3 = 1000\,\text{kg/m}^3$.
- The volume of an irregular object is measured by displacement of water, $V = V_2 - V_1$. The mass of a container must be subtracted, or it shows up as the constant term $m_0$ of the line.
Apply
Worked example
Problem
A solid iron beam is $1.20\,\text{m}$ long, and its cross-section is a square with a side of 4.0 cm. What is the mass of the beam?
Step 1/7
Given and find. The length is $L = 1.20\,\text{m}$ and the side of the cross-section is $a = 4.0\,\text{cm}$. From the table, the density of iron is $\rho = 7870\,\text{kg/m}^3$. We are asked for the mass $m$ of the beam.
Step 2/7
Principle. The beam is iron all the way through, so its mass is directly proportional to volume, and the constant of proportionality is the density of iron. The volume is obtained as the product of the area of the cross-section, $a^2$, and the length.
Step 3/7
Rearranged formula. $$\begin{aligned} V &= a^2 L \\ m &= \rho V = \rho a^2 L \end{aligned}$$
Step 4/7
Unit conversions. The density is in $\text{kg/m}^3$, so the lengths are converted to metres: $a = 4.0\,\text{cm} = 0.040\,\text{m}$. The length is already in metres, and the mass comes out directly in kilograms.
Step 5/7
Substitution. $$\begin{aligned} m &= \rho a^2 L \\ &= 7870\,\tfrac{\text{kg}}{\text{m}^3} \cdot (0.040\,\text{m})^2 \cdot 1.20\,\text{m} \\ &= 7870\,\tfrac{\text{kg}}{\text{m}^3} \cdot 1.92 \cdot 10^{-3}\,\text{m}^3 \\ &= 15.11\,\text{kg} \end{aligned}$$
Step 6/7
Result. The mass of the beam is about $15\,\text{kg}$. The result is given with two significant figures, because the side $4.0\,\text{cm}$ is known to two figures; the table density $7870\,\text{kg/m}^3$ would be more precise.
Step 7/7
Reasonableness. The volume of the beam is $1.92\,\text{dm}^3$, about two litres, and the same volume of water would have a mass of about $1.9\,\text{kg}$. Iron is about 7.9 times as dense as water, so the estimate is $7.9 \cdot 1.9\,\text{kg} \approx 15\,\text{kg}$. The result matches.
Optional Find the error
Apply to a new situation
A stone found in a river has a mass of $187\,\text{g}$. A measuring cylinder contains $120\,\text{ml}$ of water, and when the stone is lowered in, the reading is $190\,\text{ml}$.
What is the density of the substance of the stone? Give your answer in $\text{kg/m}^3$.
Hint 1: a guiding question
What is the volume of the stone? The stone displaces its own volume of water.
Hint 2: what is the connection
The volume is
Then convert the mass to kilograms and the volume to cubic metres:
Hint 3: an intermediate step
Show the solution
Given
mass of the stone $m = 187\,\text{g}$; water reading before $V_1 = 120\,\text{ml}$ and after $V_2 = 190\,\text{ml}$
Find
the density $\rho$ of the substance of the stone in $\text{kg/m}^3$
Principle
A submerged stone displaces its own volume of water, so the volume of the stone is the difference of the readings. Density is mass divided by volume.
Formula
Conversions
$m = 187\,\text{g} = 0.187\,\text{kg}$.
Substitution
Answer
$2.7 \cdot 10^3\,\tfrac{\text{kg}}{\text{m}^3}$
Two significant figures, because the difference of the readings, $70\,\text{ml}$, is known to two figures.
Reasonableness
Rocks are usually about 2.5 to 3 times as dense as water, and $2.7 \cdot 10^3\,\text{kg/m}^3$ is in this range. Estimate: $190\,\text{g}$ and $70\,\text{cm}^3$ would give about $2.7\,\text{g/cm}^3$.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The volume of an aluminium object is $40\,\text{cm}^3$ and its mass is $108\,\text{g}$. What is the density of aluminium in $\text{g/cm}^3$?
Answer
$\rho$$= \dfrac{m}{V}$$= \dfrac{108\,\text{g}}{40\,\text{cm}^3}$$= 2.7\,\tfrac{\text{g}}{\text{cm}^3}$ - The volume of a water bottle is $1.5\,\text{dm}^3$. What is the mass of the water ($\rho = 1000\,\text{kg/m}^3$)? Which conversion is needed?
Answer
The volume is converted to cubic metres: $1.5\,\text{dm}^3 = 1.5 \cdot 10^{-3}\,\text{m}^3$. Then
$m$$= \rho V$$= 1000\,\tfrac{\text{kg}}{\text{m}^3} \cdot 1.5 \cdot 10^{-3}\,\text{m}^3$$= 1.5\,\text{kg}$
1. Concept
2. Density is a property of a substance
3. Calculation exercise
The edge of a pine cube is $20.0\,\text{cm}$ and the density of pine is $500\,\text{kg/m}^3$. What is the mass of the cube? Give your answer in kilograms.
Hint 1: a guiding question
First calculate the volume of the cube. The volume of a cube is the edge to the power of three.
Hint 2: what is the connection
Convert the edge to metres: $20.0\,\text{cm} = 0.200\,\text{m}$. The volume is $V = a^3$.
Hint 3: an intermediate step
Show the solution
Given
edge of the cube $a = 20.0\,\text{cm}$; density of pine $\rho = 500\,\tfrac{\text{kg}}{\text{m}^3}$
Find
the mass $m$ of the cube in kilograms
Principle
The cube is pine all the way through, so the mass is density times volume. The volume of a cube is the edge to the power of three.
Formula
$V = a^3$ and $m = \rho V = \rho a^3$
Conversions
The density is in $\text{kg/m}^3$, so the edge is converted to metres: $a = 20.0\,\text{cm} = 0.200\,\text{m}$.
Substitution
Answer
$4.00\,\text{kg}$
Three significant figures like the edge $20.0\,\text{cm}$.
Reasonableness
The volume of the cube is $8.00\,\text{dm}^3$, that is, eight litres. The same volume of water would be $8\,\text{kg}$, and pine is about half as dense as water, so the mass is about $4\,\text{kg}$.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
Optional 4. Explain and justify
A student measures the mass and volume of an unknown piece of metal and calculates a density of $8.9\,\text{g/cm}^3$. The student says: "The piece is copper." Explain why density is a good way to identify a substance and why the conclusion is still not completely certain.
Model answer
Density $\rho = m : V$ is a property of the substance: a large and a small piece of the same substance give the same result. That is why a measured density can be compared with table values. The result $8.9\,\text{g/cm}^3$ is very close to the value for copper, $8.96\,\text{g/cm}^3$, so copper is likely. The conclusion is not certain, because measurements always contain error, especially in the volume, and because other substances or alloys may have a density close to the same value. Other tests, such as colour or magnetism, add certainty.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Density
- The mass of a substance per unit volume.$$\rho = \dfrac{m}{V}$$
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 7-1Core
A grey metal ingot is weighed at a recycling point. Its mass is $340\,\text{g}$ and its volume is $30.0\,\text{cm}^3$.
What is the density of the substance of the ingot? Give your answer in $\text{g/cm}^3$.
Answer
$11.3\,\tfrac{\text{g}}{\text{cm}^3}$
Show the solution
Given
mass of the metal ingot $m = 340\,\text{g}$ and volume $V = 30.0\,\text{cm}^3$
Find
the density $\rho$ of the substance of the ingot in $\text{g/cm}^3$
Principle
Density is mass per unit of volume. The ingot is made of one substance, so the ratio of its mass to its volume is the density of the substance.
Formula
$\rho = \dfrac{m}{V}$
Conversions
Not needed: the mass is in grams and the volume in cubic centimetres, and the answer is asked in $\text{g/cm}^3$.
Substitution
$\rho$$= \dfrac{340\,\text{g}}{30.0\,\text{cm}^3}$$= 11.33\,\tfrac{\text{g}}{\text{cm}^3}$Answer
$11.3\,\tfrac{\text{g}}{\text{cm}^3}$
Three significant figures like the volume $30.0\,\text{cm}^3$.
Reasonableness
The value is close to the density of lead, $11.3\,\tfrac{\text{g}}{\text{cm}^3}$, so the ingot is probably lead. Estimate: $30\,\text{cm}^3$ and about $11\,\tfrac{\text{g}}{\text{cm}^3}$ would give about $330\,\text{g}$, which is close to the measured mass.
- Exercise 7-2Core
The volume of an aluminium plate is $35.0\,\text{cm}^3$, and the density of aluminium is $2.70\,\text{g/cm}^3$. What is the mass of the plate? Give your answer in grams.
Answer
$94.5\,\text{g}$
Show the solution
Given
volume of the aluminium plate $V = 35.0\,\text{cm}^3$ and density $\rho = 2.70\,\tfrac{\text{g}}{\text{cm}^3}$
Find
the mass $m$ of the plate in grams
Principle
The mass of a given substance is directly proportional to volume, and the constant of proportionality is the density. The plate is aluminium all the way through.
Formula
Solve for the mass from the definition of density $\rho = \dfrac{m}{V}$: $m = \rho V$
Conversions
Not needed: the density is in $\text{g/cm}^3$ and the volume in $\text{cm}^3$, so the mass comes out directly in grams.
Substitution
$m$$= 2.70\,\tfrac{\text{g}}{\text{cm}^3} \cdot 35.0\,\text{cm}^3$$= 94.5\,\text{g}$Answer
$94.5\,\text{g}$
Three significant figures like the given values.
Reasonableness
Water with a volume of $35\,\text{cm}^3$ would have a mass of $35\,\text{g}$. Aluminium is about 2.7 times as dense, so $2.7 \cdot 35\,\text{g} \approx 95\,\text{g}$. The result matches.
- Exercise 7-3
In the chemistry classroom, $316\,\text{g}$ of ethanol is measured on a balance. Its density is $0.79\,\text{g/cm}^3$.
What is the volume of the liquid? Give your answer in cubic centimetres.
Answer
$400\,\text{cm}^3$
Show the solution
Given
mass of the ethanol $m = 316\,\text{g}$ and density $\rho = 0.79\,\tfrac{\text{g}}{\text{cm}^3}$
Find
the volume $V$ of the liquid in cubic centimetres
Principle
Density is mass per unit of volume, so when the mass and the density are known, the volume is obtained by dividing the mass by the density.
Formula
Solve for the volume from $\rho = \dfrac{m}{V}$: $V = \dfrac{m}{\rho}$
Conversions
Not needed: the mass is in grams and the density in $\text{g/cm}^3$, so the gram cancels and $\text{cm}^3$ remains.
Substitution
$V$$= \dfrac{316\,\text{g}}{0.79\,\tfrac{\text{g}}{\text{cm}^3}}$$= 400\,\text{cm}^3$Answer
$400\,\text{cm}^3$
Two significant figures, because the density $0.79\,\tfrac{\text{g}}{\text{cm}^3}$ is known to two figures.
Reasonableness
Ethanol is less dense than water, so $316\,\text{g}$ takes up more space than $316\,\text{g}$ of water, which is $316\,\text{cm}^3$. The result $400\,\text{cm}^3$ is greater, as it should be.
- Exercise 7-4
The volume of a concrete slab is $0.75\,\text{m}^3$. The density of concrete is $2400\,\text{kg/m}^3$. What is the mass of the slab? Give your answer in kilograms.
Answer
$1800\,\text{kg}$
Show the solution
Given
volume of the slab $V = 0.75\,\text{m}^3$ and density of concrete $\rho = 2400\,\tfrac{\text{kg}}{\text{m}^3}$
Find
the mass $m$ of the slab in kilograms
Principle
The slab is concrete all the way through, so the mass is directly proportional to volume and the constant is the density.
Formula
$\rho = \dfrac{m}{V}$, so $m = \rho V$
Conversions
Not needed: the density is in $\text{kg/m}^3$ and the volume in $\text{m}^3$, so the mass comes out in kilograms.
Substitution
$m$$= 2400\,\tfrac{\text{kg}}{\text{m}^3} \cdot 0.75\,\text{m}^3$$= 1800\,\text{kg}$Answer
$1800\,\text{kg}$
Two significant figures like the volume $0.75\,\text{m}^3$.
Reasonableness
A cubic metre of water has a mass of $1000\,\text{kg}$ and concrete is about 2.4 times as dense, so $0.75\,\text{m}^3$ has a mass of about $0.75 \cdot 2400\,\text{kg} \approx 1800\,\text{kg}$: over a tonne, as you would expect of a concrete slab.
- Exercise 7-5
The mass of a block of ice is $36.8\,\text{kg}$ and its volume is $0.0400\,\text{m}^3$. What is the density of ice? Give your answer in $\text{kg/m}^3$.
Answer
$920\,\tfrac{\text{kg}}{\text{m}^3}$
Show the solution
Given
mass of the block of ice $m = 36.8\,\text{kg}$ and volume $V = 0.0400\,\text{m}^3$
Find
the density $\rho$ of ice in $\text{kg/m}^3$
Principle
Density is mass per unit of volume, and the block is made of one substance.
Formula
$\rho = \dfrac{m}{V}$
Conversions
Not needed: the mass is in kilograms and the volume in cubic metres, and the answer is asked in $\text{kg/m}^3$.
Substitution
$\rho$$= \dfrac{36.8\,\text{kg}}{0.0400\,\text{m}^3}$$= 920\,\tfrac{\text{kg}}{\text{m}^3}$Answer
$920\,\tfrac{\text{kg}}{\text{m}^3}$
Three significant figures like the given values.
Reasonableness
The result is slightly below the density of water, $1000\,\tfrac{\text{kg}}{\text{m}^3}$, which fits the fact that ice floats on water. Rough estimate: $37\,\text{kg} : 0.04\,\text{m}^3 \approx 925\,\tfrac{\text{kg}}{\text{m}^3}$.
- Exercise 7-6
A uniform aluminium block is sawn into two equal parts. What happens to the density of the substance?
Answer
The density is the same in both parts as in the original block.
Show the solution
Find
how the density changes when a uniform aluminium block is cut in half.
Principle
Density $\rho = \dfrac{m}{V}$ is the ratio of mass to volume, and it is a property of the substance. The block is aluminium all the way through, so each half is the same substance as the whole block.
Model answer
The density stays the same. When the block is cut in half, the mass $m$ and the volume $V$ both halve, so the ratio $m : V$ does not change: the density of each part is still $2.70\,\text{g/cm}^3$. Density does not depend on the size or shape of the object, only on the substance.
- Exercise 7-7
Cooking oil is poured carefully into a glass of water. The density of the oil is $920\,\text{kg/m}^3$ and that of water is $1000\,\text{kg/m}^3$. The oil does not mix with the water.
What happens in the glass?
Answer
The oil stays on top of the water, because the density of the oil is smaller than that of water.
Show the solution
Find
what happens when cooking oil is poured into water, when the density of the oil is $920\,\text{kg/m}^3$ and that of water is $1000\,\text{kg/m}^3$.
Principle
Density tells how much mass there is per unit of volume. A liquid with a smaller density is lighter in the same volume, and it settles on top of a liquid with a greater density.
Model answer
The density of the oil, $920\,\text{kg/m}^3$, is smaller than the density of water, $1000\,\text{kg/m}^3$, so the oil stays on top of the water as a layer of its own. The same is shown by the fact that the density of the oil is $0.92$ times the density of water: a cubic metre of oil has a mass of $920\,\text{kg}$ and a cubic metre of water $1000\,\text{kg}$. The oil does not mix with the water, so the density decides which liquid is on top; the thickness or the amount of the liquid does not matter.
- Exercise 7-8
The volume of a lead sinker for a fishing hook is $5.00\,\text{cm}^3$. The density of lead is $11.3\,\text{g/cm}^3$.
What is the mass of the sinker? Give your answer in grams.
Answer
$56.5\,\text{g}$
Show the solution
Given
volume of the lead sinker $V = 5.00\,\text{cm}^3$ and density of lead $\rho = 11.3\,\tfrac{\text{g}}{\text{cm}^3}$
Find
the mass $m$ of the sinker in grams
Principle
The sinker is lead all the way through, so the mass is directly proportional to volume and the constant of proportionality is the density of lead.
Formula
Solve for the mass from $\rho = \dfrac{m}{V}$: $m = \rho V$
Conversions
Not needed: the volume is in cubic centimetres and the density in $\text{g/cm}^3$, so $\text{cm}^3$ cancels and grams remain.
Substitution
$m$$= 11.3\,\tfrac{\text{g}}{\text{cm}^3} \cdot 5.00\,\text{cm}^3$$= 56.5\,\text{g}$Answer
$56.5\,\text{g}$
Three significant figures like the given values.
Reasonableness
Five cubic centimetres of water would have a mass of about $5\,\text{g}$. The density of lead is about $11$ times the density of water, so the mass is about $11 \cdot 5\,\text{g} = 55\,\text{g}$, which fits the result $56.5\,\text{g}$.
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 7-9Core
The dimensions of a gold ingot are $10.0\,\text{cm} \cdot 5.0\,\text{cm} \cdot 2.0\,\text{cm}$. The density of gold is $19\,300\,\text{kg/m}^3$. What is the mass of the ingot? Give your answer in kilograms.
Answer
$1.9\,\text{kg}$
Show the solution
Given
dimensions of the ingot $10.0\,\text{cm}$, $5.0\,\text{cm}$ and $2.0\,\text{cm}$; density of gold $\rho = 19\,300\,\tfrac{\text{kg}}{\text{m}^3}$
Find
the mass $m$ of the ingot in kilograms
Principle
The ingot is gold all the way through, so the mass is density times volume. The volume of a rectangular block is the product of the three dimensions.
Formula
$V = a\,b\,c$ and $m = \rho V = \rho\,a\,b\,c$
Conversions
The density is in $\text{kg/m}^3$, so the dimensions are converted to metres: $10.0\,\text{cm} = 0.100\,\text{m}$, $5.0\,\text{cm} = 0.050\,\text{m}$ and $2.0\,\text{cm} = 0.020\,\text{m}$.
Substitution
$m$$= 19\,300\,\tfrac{\text{kg}}{\text{m}^3} \cdot 0.100\,\text{m} \cdot 0.050\,\text{m} \cdot 0.020\,\text{m}$$= 19\,300\,\tfrac{\text{kg}}{\text{m}^3} \cdot 1.0 \cdot 10^{-4}\,\text{m}^3$$= 1.93\,\text{kg}$Answer
$1.9\,\text{kg}$
Two significant figures, because the dimensions $5.0\,\text{cm}$ and $2.0\,\text{cm}$ are known to two figures.
Reasonableness
The volume of the ingot is $100\,\text{cm}^3$, and the same volume of water would be $100\,\text{g}$. Gold is about 19 times as dense, so $19 \cdot 100\,\text{g} \approx 1.9\,\text{kg}$.
- Exercise 7-10Core
In a laboratory there is a bottle containing $2.5\,\text{dm}^3$ of ethanol. The density of ethanol is $790\,\text{kg/m}^3$.
What is the mass of the liquid? Give your answer in kilograms.
Answer
$2.0\,\text{kg}$
Show the solution
Given
volume of the ethanol $V = 2.5\,\text{dm}^3$ and density $\rho = 790\,\tfrac{\text{kg}}{\text{m}^3}$
Find
the mass $m$ of the liquid in kilograms
Principle
The liquid is one substance, so the mass is directly proportional to volume and the constant is the density.
Formula
$\rho = \dfrac{m}{V}$, so $m = \rho V$
Conversions
The density is per cubic metre, so
$2.5\,\text{dm}^3$$= 2.5 \cdot 10^{-3}\,\text{m}^3$$= 0.0025\,\text{m}^3$because $1\,\text{dm}^3 = 10^{-3}\,\text{m}^3$.
Substitution
$m$$= 790\,\tfrac{\text{kg}}{\text{m}^3} \cdot 0.0025\,\text{m}^3$$= 1.975\,\text{kg}$Answer
$2.0\,\text{kg}$
Two significant figures like the volume $2.5\,\text{dm}^3$ and the density $790\,\tfrac{\text{kg}}{\text{m}^3}$.
Reasonableness
Two and a half litres of water has a mass of $2.5\,\text{kg}$. Ethanol is less dense than water, so its mass is slightly smaller, about $0.79 \cdot 2.5\,\text{kg} \approx 2.0\,\text{kg}$.
- Exercise 7-11
The mass of a metal object is $89.0\,\text{g}$. The figure shows a measuring cylinder before and after the object is lowered into it.
What is the density of the substance of the object? Give your answer in $\text{kg/m}^3$.
Answer
$8900\,\tfrac{\text{kg}}{\text{m}^3}$
Show the solution
Given
mass of the object $m = 89.0\,\text{g}$; a figure of the measuring cylinder before and after the object is lowered in
Find
the density $\rho$ of the substance of the object in $\text{kg/m}^3$
Principle
A submerged object displaces its own volume of water, so the volume of the object is the difference of the water readings. Density is mass per unit of volume.
Reading the graph
The numbers on the scale are at $20\,\text{ml}$ intervals, the long marks are at $10\,\text{ml}$ intervals and the small marks at $2\,\text{ml}$ intervals. Before the object is lowered in, the water reaches the mark $V_1 = 40\,\text{ml}$. After that, the water reaches the long mark halfway between the numbers $40$ and $60$, $V_2 = 50\,\text{ml}$. An answer is accepted if it deviates by at most $3\,\%$.
Formula
$V = V_2 - V_1$ and $\rho = \dfrac{m}{V}$
Conversions
$V$$= 50\,\text{ml} - 40\,\text{ml}$$= 10\,\text{ml}$$= 10\,\text{cm}^3$$= 1.0 \cdot 10^{-5}\,\text{m}^3$$89.0\,\text{g} = 0.0890\,\text{kg}$.
Substitution
$\rho$$= \dfrac{0.0890\,\text{kg}}{1.0 \cdot 10^{-5}\,\text{m}^3}$$= 8900\,\tfrac{\text{kg}}{\text{m}^3}$Answer
$8900\,\tfrac{\text{kg}}{\text{m}^3}$
Two significant figures, because the volume $10\,\text{ml}$ is read from the scale to about $1\,\text{ml}$.
Reasonableness
Estimate: $89\,\text{g} : 10\,\text{cm}^3 \approx 9\,\tfrac{\text{g}}{\text{cm}^3}$, which is $9000\,\tfrac{\text{kg}}{\text{m}^3}$. The result is close to the density of copper, $8960\,\tfrac{\text{kg}}{\text{m}^3}$, so the object may be copper.
- Exercise 7-12
The mass of an empty measuring flask is $215\,\text{g}$. When $250\,\text{cm}^3$ of an unknown liquid is poured into the flask, the balance shows $412\,\text{g}$.
What is the density of the liquid? Give your answer in $\text{kg/m}^3$.
Answer
$788\,\tfrac{\text{kg}}{\text{m}^3}$
Show the solution
Given
mass of the empty flask $m_0 = 215\,\text{g}$, volume of the liquid $V = 250\,\text{cm}^3$ and total mass of the flask and the liquid $m_\text{tot} = 412\,\text{g}$
Find
the density $\rho$ of the liquid in $\text{kg/m}^3$
Principle
The balance shows the total mass of the flask and the liquid, but the density is calculated from the mass of the liquid alone. The mass of the flask is subtracted, and the rest is the mass of the liquid.
Formula
$m$$= m_\text{tot} - m_0$$\rho$$= \dfrac{m}{V}$$= \dfrac{m_\text{tot} - m_0}{V}$Conversions
$m$$= 412\,\text{g} - 215\,\text{g}$$= 197\,\text{g}$$= 0.197\,\text{kg}$$250\,\text{cm}^3$$= 250 \cdot 10^{-6}\,\text{m}^3$$= 2.50 \cdot 10^{-4}\,\text{m}^3$Substitution
$\rho$$= \dfrac{0.197\,\text{kg}}{2.50 \cdot 10^{-4}\,\text{m}^3}$$= 788\,\tfrac{\text{kg}}{\text{m}^3}$Answer
$788\,\tfrac{\text{kg}}{\text{m}^3}$
Three significant figures, because the difference of the masses, $197\,\text{g}$, and the volume, $250\,\text{cm}^3$, have been measured to three figures.
Reasonableness
The result is close to the density of ethanol, $790\,\tfrac{\text{kg}}{\text{m}^3}$, so the liquid is probably ethanol. If the mass of the flask had been forgotten, the wrong result would be $1650\,\tfrac{\text{kg}}{\text{m}^3}$, which would be more than water.
- Exercise 7-13
Objects of different sizes were made of a uniform metal, and the volume and mass of each were measured. Draw a $(V,m)$ graph of the measurement points and fit a straight line through the origin to the points.
What is the density of the substance in $\text{g/cm}^3$? Which substance is it? Use the table from the lesson.
Mass of the metal objects as a function of volume $V\,(\text{cm}^3)$ $m\,(\text{g})$ 10.0 27.3 20.0 53.5 30.0 81.6 40.0 107.9 50.0 135.4 Answer
$2.7\,\tfrac{\text{g}}{\text{cm}^3}$
Show the solution
Given
the measurement points $(V$; $m)$ in the table: $(10.0\,\text{cm}^3$; $27.3\,\text{g})$, $(20.0\,\text{cm}^3$; $53.5\,\text{g})$, $(30.0\,\text{cm}^3$; $81.6\,\text{g})$, $(40.0\,\text{cm}^3$; $107.9\,\text{g})$ and $(50.0\,\text{cm}^3$; $135.4\,\text{g})$
Find
the density $\rho$ of the substance in $\text{g/cm}^3$ and the name of the substance
Principle
The mass of a given substance is directly proportional to volume, $m = \rho V$, so the points lie on a straight line through the origin, and the slope of the line is the density. The line is drawn through the middle of the set of points, and the slope is calculated from points on the line, not from a single measurement point.
Reading the graph
The points lie on a rising straight line with points on both sides of it. The fitted line passes approximately through the points $(0\,\text{cm}^3$; $0\,\text{g})$ and $(50.0\,\text{cm}^3$; $135\,\text{g})$. The points are read from the line and chosen far apart.
Formula
$\rho$$= \dfrac{\Delta m}{\Delta V}$$= \dfrac{m_2 - m_1}{V_2 - V_1}$Conversions
Not needed: the mass is in grams and the volume in cubic centimetres, like the requested unit $\text{g/cm}^3$.
Substitution
$\rho$$= \dfrac{135\,\text{g} - 0\,\text{g}}{50.0\,\text{cm}^3 - 0\,\text{cm}^3}$$= 2.70\,\tfrac{\text{g}}{\text{cm}^3}$Answer
$2.7\,\tfrac{\text{g}}{\text{cm}^3}$
Two significant figures, because the line is fitted by eye. A fit by eye varies, so about $2.62\text{–}2.78\,\text{g/cm}^3$ is a good answer. In the table the density of aluminium is $2.70\,\text{g/cm}^3$, so the substance is aluminium.
Reasonableness
The line passes through the origin, as it should, because an object with a volume of zero has a mass of zero. Check: $27.3\,\text{g} : 10.0\,\text{cm}^3 = 2.73\,\tfrac{\text{g}}{\text{cm}^3}$ and $135.4\,\text{g} : 50.0\,\text{cm}^3 = 2.71\,\tfrac{\text{g}}{\text{cm}^3}$, so every point gives almost the same value.
- Exercise 7-14
Liquid was poured into a measuring cylinder in steps, and a balance showed the total mass $m_\text{tot}$ of the cylinder and the liquid as a function of the volume $V$ of the liquid. Draw a graph of the measurement points and fit a straight line to the points.
What is the density of the liquid in $\text{kg/m}^3$? Also determine the mass of the empty measuring cylinder and explain why the line does not pass through the origin.
Total mass of the measuring cylinder and the liquid as a function of the liquid volume $V\,(\text{cm}^3)$ $m_\text{tot}\,(\text{g})$ 20.0 103.6 40.0 121.2 60.0 140.5 80.0 158.9 100.0 176.8 Answer
$920\,\tfrac{\text{kg}}{\text{m}^3}$
Show the solution
Given
the measurement points $(V$; $m_\text{tot})$ in the table: $(20.0\,\text{cm}^3$; $103.6\,\text{g})$, $(40.0\,\text{cm}^3$; $121.2\,\text{g})$, $(60.0\,\text{cm}^3$; $140.5\,\text{g})$, $(80.0\,\text{cm}^3$; $158.9\,\text{g})$ and $(100.0\,\text{cm}^3$; $176.8\,\text{g})$
Find
the density $\rho$ of the liquid in $\text{kg/m}^3$, the mass $m_0$ of the empty measuring cylinder and the reason why the line does not pass through the origin
Principle
The balance shows the total mass of the measuring cylinder and the liquid, which follows the linear model $m_\text{tot} = \rho V + m_0$. The density of the liquid is the slope of the line, and the mass $m_0$ of the empty cylinder is the point where the line crosses the vertical axis.
Reading the graph
The points lie on a rising straight line with points on both sides of it. The fitted line passes approximately through the points $(0\,\text{cm}^3$; $85\,\text{g})$ and $(100\,\text{cm}^3$; $177\,\text{g})$. The point $(0\,\text{cm}^3$; $85\,\text{g})$ is where the line crosses the vertical axis.
Formula
$\rho$$= \dfrac{\Delta m_\text{tot}}{\Delta V}$$= \dfrac{m_2 - m_1}{V_2 - V_1}$$m_0 = m_1 - \rho V_1$
Conversions
$1\,\tfrac{\text{g}}{\text{cm}^3} = 1000\,\tfrac{\text{kg}}{\text{m}^3}$.
Substitution
$\rho$$= \dfrac{177\,\text{g} - 85\,\text{g}}{100\,\text{cm}^3 - 0\,\text{cm}^3}$$= \dfrac{92\,\text{g}}{100\,\text{cm}^3}$$= 0.92\,\tfrac{\text{g}}{\text{cm}^3}$$= 920\,\tfrac{\text{kg}}{\text{m}^3}$$m_0 = 85\,\text{g}$ (where the line crosses the vertical axis)
Answer
the density of the liquid is $920\,\tfrac{\text{kg}}{\text{m}^3}$, two significant figures, because the line is fitted by eye. The mass of the empty measuring cylinder is about $85\,\text{g}$. The line does not pass through the origin, because the balance also shows the mass of the empty measuring cylinder: when there is no liquid, $V = 0$, the balance still shows $m_0$.
Reasonableness
The density $920\,\tfrac{\text{kg}}{\text{m}^3}$ is the same as the density of cooking oil or ice in the table, so the liquid may be cooking oil. Check at the point $V = 60\,\text{cm}^3$: $85\,\text{g}\,{+}\,0.92 \cdot 60\,\text{g} = 140\,\text{g}$, measured $140.5\,\text{g}$.
- Exercise 7-15
A student determined the density of a metal object by the water displacement method. The mass of the object is $107\,\text{g}$. The reading of the measuring cylinder was $60\,\text{ml}$ before the object was lowered in and $72\,\text{ml}$ afterwards. The student calculated $\rho = 107\,\text{g} : 72\,\text{cm}^3 = 1.5\,\text{g/cm}^3$ and concluded that the object was light plastic. Find the student’s error and calculate the correct density in $\text{g/cm}^3$.
Which substance could the object be?
Answer
$8.9\,\tfrac{\text{g}}{\text{cm}^3}$
Show the solution
Given
mass of the object $m = 107\,\text{g}$, readings of the measuring cylinder $V_1 = 60\,\text{ml}$ and $V_2 = 72\,\text{ml}$; the student’s result $\rho = 1.5\,\tfrac{\text{g}}{\text{cm}^3}$
Find
the student’s error, the correct density $\rho$ in $\text{g/cm}^3$ and the substance
Principle
In the water displacement method the volume of the object is the rise of the water level, that is, the difference of the readings. The reading $V_2$ of the measuring cylinder also includes the water that was in the cylinder before the object was lowered in. Density is the mass of the object divided by the volume of the object.
Formula
$V = V_2 - V_1$ and $\rho = \dfrac{m}{V}$
Conversions
$V$$= 72\,\text{ml} - 60\,\text{ml}$$= 12\,\text{ml}$$= 12\,\text{cm}^3$because $1\,\text{ml} = 1\,\text{cm}^3$.
Substitution
$\rho$$= \dfrac{107\,\text{g}}{12\,\text{cm}^3}$$= 8.92\,\tfrac{\text{g}}{\text{cm}^3}$Answer
The student divided the mass by the reading $72\,\text{ml}$ after the object was lowered in, not by the difference of the readings, so the volume was six times too large and the density too small. The correct density is $8.9\,\tfrac{\text{g}}{\text{cm}^3}$, two significant figures, because the volume $12\,\text{ml}$ is a two-figure number. In the table the density of copper is $8.96\,\tfrac{\text{g}}{\text{cm}^3}$, so the object may be copper, not plastic.
Reasonableness
The result $8.9\,\tfrac{\text{g}}{\text{cm}^3}$ is over eight times the density of water, which fits a metal. The difference of the readings is smaller than either reading, so the difference has relatively more uncertainty: if the error of each reading is $1\,\text{ml}$, the volume is $12 \pm 2\,\text{ml}$ and the density can be about $7.6\text{–}10.7\,\tfrac{\text{g}}{\text{cm}^3}$.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 7-16
A copper object and an aluminium object have the same mass.
How many times as large is the volume of the aluminium object compared with the volume of the copper object? Use the densities in the table.
Answer
$3.32$
Show the solution
Given
the copper object and the aluminium object have equal masses, $m_\text{Cu} = m_\text{Al} = m$; from the table $\rho_\text{Cu} = 8960\,\tfrac{\text{kg}}{\text{m}^3}$ and $\rho_\text{Al} = 2700\,\tfrac{\text{kg}}{\text{m}^3}$
Find
the ratio of the volumes $V_\text{Al} : V_\text{Cu}$
Principle
The mass of each object is density times volume. Because the masses are equal, the products of density and volume are equal, so a greater density means a smaller volume.
Formula
$\rho_\text{Al} V_\text{Al} = \rho_\text{Cu} V_\text{Cu}$, from which
$$\frac{V_\text{Al}}{V_\text{Cu}} = \frac{\rho_\text{Cu}}{\rho_\text{Al}}$$Conversions
Not needed: the densities are in the same unit and the mass $m$ cancels.
Substitution
$\dfrac{V_\text{Al}}{V_\text{Cu}}$$= \dfrac{8960\,\tfrac{\text{kg}}{\text{m}^3}}{2700\,\tfrac{\text{kg}}{\text{m}^3}}$$= 3.32$Answer
$3.32$
Times, three significant figures like the densities $8960\,\tfrac{\text{kg}}{\text{m}^3}$ and $2.70\cdot10^{3}\,\tfrac{\text{kg}}{\text{m}^3}$.
Reasonableness
Aluminium is less dense, so more volume of it is needed to get the same mass. With rounded numbers $3 \cdot 2700 = 8100$ and $4 \cdot 2700 = 10\,800$, so the ratio $8960 : 2700$ is between three and four, closer to three, like $3.32$.
- Exercise 7-17
An object is made by casting together $100\,\text{cm}^3$ of aluminium and $50\,\text{cm}^3$ of iron. The volumes are assumed to add up.
What is the average density of the object? Give your answer in $\text{g/cm}^3$.
Answer
$4.4\,\tfrac{\text{g}}{\text{cm}^3}$
Show the solution
Given
aluminium $V_1 = 100\,\text{cm}^3$, $\rho_1 = 2.70\,\tfrac{\text{g}}{\text{cm}^3}$ and iron $V_2 = 50\,\text{cm}^3$, $\rho_2 = 7.87\,\tfrac{\text{g}}{\text{cm}^3}$; the volumes add up
Find
the average density $\bar{\rho}$ of the object in $\text{g/cm}^3$
Principle
Densities are not added or averaged directly. The masses and the volumes are added, and the density of the whole object is the total mass divided by the total volume.
Formula
$m_i = \rho_i V_i$, and
$$\bar{\rho} = \frac{m_1 + m_2}{V_1 + V_2} = \frac{\rho_1 V_1 + \rho_2 V_2}{V_1 + V_2}$$Conversions
Not needed: all quantities are in the units $\text{g}$ and $\text{cm}^3$.
Substitution
The masses are
$m_1$$= 2.70\,\tfrac{\text{g}}{\text{cm}^3} \cdot 100\,\text{cm}^3$$= 270\,\text{g}$$m_2$$= 7.87\,\tfrac{\text{g}}{\text{cm}^3} \cdot 50\,\text{cm}^3$$= 393.5\,\text{g}$so
$\bar{\rho}$$= \dfrac{270\,\text{g} + 393.5\,\text{g}}{150\,\text{cm}^3}$$= 4.42\,\tfrac{\text{g}}{\text{cm}^3}$Answer
$4.4\,\tfrac{\text{g}}{\text{cm}^3}$
Two significant figures like the volumes $50\,\text{cm}^3$.
Reasonableness
The result is between the densities of aluminium and iron, and closer to aluminium, because there is twice as much aluminium by volume. The ordinary mean of the two densities, $5.3\,\tfrac{\text{g}}{\text{cm}^3}$, would be too large.
- Exercise 7-18
A round copper wire has a diameter of $1.0\,\text{mm}$ and a mass of $0.50\,\text{kg}$.
How long is the wire? The cross-section of the wire is a circle with area $\pi r^2$, and the volume of the wire is the area times the length.
Answer
$71\,\text{m}$
Show the solution
Given
mass of the wire $m = 0.50\,\text{kg}$, diameter $d = 1.0\,\text{mm}$, so radius $r = 0.50\,\text{mm}$; from the table $\rho = 8960\,\tfrac{\text{kg}}{\text{m}^3}$
Find
the length $L$ of the wire in metres
Principle
The volume is obtained from the mass and the density. The wire is a long cylinder whose volume is the area of the cross-section times the length, so the length is obtained by dividing the volume by the area.
Formula
$V = \dfrac{m}{\rho}$, $A = \pi r^2$ and $V = A L$, so
$$L = \frac{V}{A} = \frac{m}{\rho\,\pi r^2}$$Conversions
The radius is
$r$$= 0.50\,\text{mm}$$= 0.50 \cdot 10^{-3}\,\text{m}$so $r^2 = 2.5 \cdot 10^{-7}\,\text{m}^2$.
Substitution
$V$$= \dfrac{0.50\,\text{kg}}{8960\,\tfrac{\text{kg}}{\text{m}^3}}$$= 5.58 \cdot 10^{-5}\,\text{m}^3$$A$$= \pi \cdot 2.5 \cdot 10^{-7}\,\text{m}^2$$= 7.85 \cdot 10^{-7}\,\text{m}^2$so
$L$$= \dfrac{5.58 \cdot 10^{-5}\,\text{m}^3}{7.85 \cdot 10^{-7}\,\text{m}^2}$$= 71.1\,\text{m}$Answer
$71\,\text{m}$
Two significant figures, because the diameter $1.0\,\text{mm}$ is known to two figures.
Reasonableness
Half a kilogram of copper has a volume of only about $56\,\text{cm}^3$, that is, half a decilitre. Drawn into a thin wire one thousandth of a metre thick, it is enough for tens of metres of length, so $71\,\text{m}$ is believable.
- Exercise 7-19
A student determines the density of a metal object by the water displacement method. The mass of the object is $113\,\text{g}$. The reading of the measuring cylinder is $50.0\,\text{ml}$ before the object is lowered in and $62.0\,\text{ml}$ afterwards.
Afterwards the student notices that an air bubble with a volume of $2.0\,\text{ml}$ had stuck to the surface of the object.
What is the true density of the object in $\text{g/cm}^3$? Also calculate the density the student originally obtained and its percentage error, deduce from the table which metal each density would suggest the object to be, and say how the bubble could have been removed or noticed.
Answer
$11.3\,\tfrac{\text{g}}{\text{cm}^3}$
Show the solution
Given
mass of the object $m = 113\,\text{g}$, readings of the measuring cylinder $V_1 = 50.0\,\text{ml}$ and $V_2 = 62.0\,\text{ml}$, volume of the air bubble $V_\text{b} = 2.0\,\text{ml}$; from the table the density of copper $8.96\,\tfrac{\text{g}}{\text{cm}^3}$ and of lead $11.3\,\tfrac{\text{g}}{\text{cm}^3}$
Find
the true density $\rho$ of the object in $\text{g/cm}^3$, the density $\rho'$ obtained by the student, its percentage error and the metals that the densities suggest
Principle
In the water displacement method the volume is the difference of the readings $V_2 - V_1$. The bubble displaces water along with the object, so the difference includes both the volume of the object and the volume of the bubble. The mass is weighed separately, so the error comes only from the volume.
Formula
$V' = V_2 - V_1$, $V = V' - V_\text{b}$, $\rho = \dfrac{m}{V}$ and $\rho' = \dfrac{m}{V'}$; the percentage error is $\dfrac{\rho' - \rho}{\rho} \cdot 100\,\%$
Conversions
$1\,\text{ml} = 1\,\text{cm}^3$, so $V' = 12.0\,\text{cm}^3$ and $V_\text{b} = 2.0\,\text{cm}^3$.
Substitution
$V'$$= 62.0\,\text{ml} - 50.0\,\text{ml}$$= 12.0\,\text{ml}$$V$$= 12.0\,\text{cm}^3 - 2.0\,\text{cm}^3$$= 10.0\,\text{cm}^3$The true density is
$\rho$$= \dfrac{113\,\text{g}}{10.0\,\text{cm}^3}$$= 11.3\,\tfrac{\text{g}}{\text{cm}^3}$and the student’s result is
$\rho'$$= \dfrac{113\,\text{g}}{12.0\,\text{cm}^3}$$= 9.42\,\tfrac{\text{g}}{\text{cm}^3}$The error is
$\dfrac{9.42 - 11.3}{11.3} \cdot 100\,\%$$= -16.7\,\%$Answer
The true density is $11.3\,\tfrac{\text{g}}{\text{cm}^3}$, three significant figures, because the mass and the readings are given to three figures. The student’s result $9.42\,\tfrac{\text{g}}{\text{cm}^3}$ is about $17\,\%$ too small, because the measured volume was too large. The true density matches lead ($11.3\,\tfrac{\text{g}}{\text{cm}^3}$), but the erroneous result is closer to copper ($8.96\,\tfrac{\text{g}}{\text{cm}^3}$), so the bubble can lead to the wrong metal. The bubble can be removed by tapping the object, and it can be noticed by looking at the cylinder from the side.
Reasonableness
The mass is correct and the volume too large, so the density has to come out too small, and it did: $9.42 < 11.3$. Check: $0.83 \cdot 11.3 \approx 9.4$.
- Exercise 7-20
An alloy object is cast from copper and aluminium. Its volume is $100\,\text{cm}^3$ and its mass is $520\,\text{g}$. The volumes are assumed to add up.
How many cubic centimetres of copper does the object contain? Use the densities in the table from the lesson. Also justify which metal there is more of by volume and which by mass.
Answer
$39.9\,\text{cm}^3$
Show the solution
Given
volume of the alloy object $V = 100\,\text{cm}^3$ and mass $m = 520\,\text{g}$; from the table $\rho_\text{Cu} = 8.96\,\tfrac{\text{g}}{\text{cm}^3}$ and $\rho_\text{Al} = 2.70\,\tfrac{\text{g}}{\text{cm}^3}$; the volumes add up
Find
the volume $V_\text{Cu}$ of copper in cubic centimetres
Principle
The total volume is the sum of the volumes of the parts and the total mass is the sum of the masses of the parts. The mass of each part is density times volume. There are two unknowns, $V_\text{Cu}$ and $V_\text{Al}$, so two equations are needed: the sum of the volumes and the sum of the masses.
Formula
For the volumes $V_\text{Cu} + V_\text{Al} = V$ and for the masses $\rho_\text{Cu} V_\text{Cu} + \rho_\text{Al} V_\text{Al} = m$. Substituting $V_\text{Al} = V - V_\text{Cu}$ into the second equation gives
$V_\text{Cu}$$= \dfrac{m - \rho_\text{Al} V}{\rho_\text{Cu} - \rho_\text{Al}}$Conversions
Not needed: the masses are in grams, the volumes in cubic centimetres and the densities in $\text{g/cm}^3$.
Substitution
$V_\text{Cu}$$= \dfrac{520\,\text{g} - 2.70\,\tfrac{\text{g}}{\text{cm}^3} \cdot 100\,\text{cm}^3}{8.96\,\tfrac{\text{g}}{\text{cm}^3} - 2.70\,\tfrac{\text{g}}{\text{cm}^3}}$$= \dfrac{250\,\text{g}}{6.26\,\tfrac{\text{g}}{\text{cm}^3}}$$= 39.9\,\text{cm}^3$Answer
$39.9\,\text{cm}^3$
Three significant figures, because the densities in the table have three figures and the data $520\,\text{g}$ and $100\,\text{cm}^3$ are read to three figures. The volume of aluminium is $V_\text{Al} = 100\,\text{cm}^3 - 39.9\,\text{cm}^3 = 60.1\,\text{cm}^3$.
Reasonableness
If the object were pure aluminium, its mass would be $270\,\text{g}$, and if it were pure copper, $896\,\text{g}$. The actual mass $520\,\text{g}$ is between these and closer to aluminium, so less than half of the volume is copper, like $39.9\,\text{cm}^3$. There is more aluminium by volume ($60.1\,\text{cm}^3$), but more copper by mass: the mass of the copper is about $357.5\,\text{g}$ and that of the aluminium about $162.3\,\text{g}$, because the density of copper is several times as great. Check:
$8.96 \cdot 39.9\,\text{g} + 2.70 \cdot 60.1\,\text{g}$$= 357.5\,\text{g} + 162.3\,\text{g}$$= 519.8\,\text{g}$$\approx 520\,\text{g}$
Photographs
- Very Blue Iceberg With Anvil Top. Eric Kilby. source, CC BY-SA 2.0
