Lesson A · 5.1, 5.3–5.4
By the end you should convert units without thinking, choose $\rho=m/V$, and write a four-mark eureka-can method.
How I start this topic
Density is how much mass is packed into a volume. A steel marble and a wooden cube can have the same mass; the steel is denser because that mass occupies less space. The equation is short. The conversions are what I mark. If you put $75\,\text{cm}^3$ straight into $\rho=m/V$ with mass in kilograms, your density will be a million times too big and you often will not notice.
5.1 · Units
Specification 5.1 wants you to use these units, not just name them. I expect the symbol and the everyday meaning.
| Quantity | Symbol | Unit | What I say in class |
|---|---|---|---|
| Temperature | $\theta$, $T$ | °C, K | Lesson C. Never leave °C in $p/T$ |
| Energy | $E$ | joule (J) | On the list; we do not do SHC here |
| Mass | $m$ | kilogram (kg) | Balance often reads grams. $\div 1000$ |
| Density | $\rho$ | kg/m³ | Water is about $1000\,\text{kg/m}^3$ |
| Length, area, volume | $l$, $A$, $V$ | m, m², m³ | cm, cm², cm³ must be converted |
| Speed, acceleration | $v$, $a$ | m/s, m/s² | On the list; $g=10\,\text{m/s}^2=10\,\text{N/kg}$ |
| Force | $F$ | newton (N) | Weight is a force. $W=mg$ |
| Pressure | $p$ | pascal (Pa) | $1\,\text{Pa}=1\,\text{N/m}^2$. Lesson B |
Conversions I want automatic
Why $\div 10^6$ for $\text{cm}^3$?
$1\,\text{m}=100\,\text{cm}$, so $1\,\text{m}^3=(100\,\text{cm})^3=1\,000\,000\,\text{cm}^3$. There are a million cubic centimetres in a cubic metre. Dividing by $100$ three times is the same as dividing by $10^6$. Students who divide by $100$ once, or by $1000$, lose the calculation mark and the sense-check.
5.3
$\rho$ is the Greek letter rho. $m$ is mass in kilograms, $V$ is volume in cubic metres, $\rho$ comes out in $\text{kg/m}^3$. Rearrange when you are asked for mass or volume: $m=\rho V$ and $V=m/\rho$.
Why it matters: two objects can have the same volume and different masses (steel vs wood), or the same mass and different volumes. Density is a property of the substance, not of the size of the lump, provided the sample is uniform.
Board example — convert first
A pebble has mass $180\,\text{g}$ and volume $75\,\text{cm}^3$. Find its density in $\text{kg/m}^3$.
$m=0.180\,\text{kg}$, $V=75\div 10^6=7.5\times 10^{-5}\,\text{m}^3$.
$$\rho = \frac{0.180}{7.5\times 10^{-5}} = 2400\,\text{kg/m}^3$$Sense-check: denser than water ($1000$), so the pebble sinks. Typical rock is $2000$–$3000\,\text{kg/m}^3$. If you forgot to convert, you would get $2.4$ and might wrongly write $\text{g/cm}^3$ when the paper asked for $\text{kg/m}^3$.
How I mark a 3-mark $\rho=m/V$
Conversion of mass and/or volume [1]. Equation stated or used [1]. Numerical answer with $\text{kg/m}^3$ [1]. A bald $2400$ with no unit is usually one mark down.
5.4 · Core practical
Every density practical has the same two measurements: mass on a balance, then volume. The volume method changes. Write the method the examiner expects, not a story about “I did an experiment.”
Callipers beat a ruler on a small cube because a millimetre error on a $2\,\text{cm}$ edge is a large percentage error.
A eureka can is a can with a spout near the top. You fill it so that water just starts to drip from the spout, then discard that overflow. When you lower the object in, a volume of water equal to the object’s volume runs out of the spout.
If the object is small enough, you can skip the can and use a measuring cylinder only: record $V_1$, lower the object, record $V_2$, $V=V_2-V_1$. The eureka can is better for a large rock that will not fit in the cylinder.
Floating objects
If $\rho$ is less than water the object will not sink, so the overflow is too small. Tie on a dense sinker. Measure the overflow for the sinker alone, then for sinker $+$ object. Subtract. The difference is the volume of the floating object.
Errors I ask you to name
4-mark “describe a method” — irregular rock
Measure mass with a balance [1]. Fill a eureka can to the spout / place a measuring cylinder under the spout [1]. Lower the rock fully underwater and collect the overflow [1]. Volume of overflow $=$ volume of rock; $\rho=m/V$ [1]. Mention repeats or a thread if you have a fifth point to spend.
Watch
Watch
1.
Convert $64\,\text{cm}^3$ to $\text{m}^3$ and $2.7\,\text{g/cm}^3$ to $\text{kg/m}^3$.
2.
A regular block is $8.0\,\text{cm}\times 4.0\,\text{cm}\times 2.0\,\text{cm}$ and has mass $173\,\text{g}$. Calculate $\rho$ in $\text{kg/m}^3$.
3.
An object floats on water. How do you find its volume with a eureka can?