Lesson 1.7 · Unit 1 · HS-PS1-3, HS-PS3-2
Density, the states of matter, and reading a heating curve
Density is the first derived unit you will use seriously, and it is also the first intensive property precise enough to identify a substance. The heating curve does a different job: it shows that adding energy to a substance does not always raise its temperature, and the flat sections are where the interesting chemistry is.
- Density is mass per unit volume, \( d = \frac{m}{V} \), usually in g/cm3 for solids and liquids and g/L for gases. The unit choice reflects how much smaller gas densities are.
- Density is intensive and is fixed for a pure substance at a given temperature, which is why it identifies a material while mass cannot.
- Volume of an irregular solid is found by displacement: the rise in water level when the object is submerged equals its volume.
- The three states differ in particle arrangement and motion. Solid: fixed positions, vibrating. Liquid: touching but free to move past one another. Gas: far apart, moving fast, negligible attraction.
- On a heating curve, the sloped sections are temperature rising within a single state, and the energy goes into faster particle motion.
- The flat sections are phase changes, and the temperature does not rise even though energy is still being supplied, because that energy is overcoming the attractions between particles rather than speeding them up.
- Water is anomalous: it is one of very few substances whose solid is less dense than its liquid, which is why ice floats and why lakes freeze from the top.
Where students lose marks: saying the temperature stays constant during melting "because no energy is being added". Energy is being added the whole time. It is going into separating particles, not into their speed, and the answer has to say so.
The problem. An irregular metal lump is placed on a balance and reads 64.8 g. A graduated cylinder holds water at 25.0 mL; with the lump submerged it reads 33.2 mL. Identify the metal from the table, and comment on the confidence the measurement supports.
| Metal | Density (g/cm3) |
|---|---|
| Aluminum | 2.70 |
| Zinc | 7.14 |
| Iron | 7.87 |
| Copper | 8.96 |
| Lead | 11.34 |
Step one: find the volume by displacement. \( 33.2 - 25.0 = 8.2 \) mL. The decimal-place rule applies to a subtraction and both readings have one decimal place, so 8.2 mL stands with two significant figures.
Step two: convert the volume unit. Density is wanted in g/cm3 and \( 1 \text{ mL} = 1 \text{ cm}^3 \) exactly, so the volume is 8.2 cm3 with no numerical change.
Step three: calculate the density.
\[ d = \frac{m}{V} = \frac{64.8 \text{ g}}{8.2 \text{ cm}^3} = 7.9024 \ldots \]Step four: apply the significant figure rule. This is a division, so the answer takes the fewest significant figures of the values used. The mass has three and the volume has two, so the answer has two: 7.9 g/cm3. Note how the imprecise displacement measurement, not the good mass measurement, sets the limit.
Step five: compare with the table. Iron at 7.87 g/cm3 rounds to 7.9 and is the closest match. Zinc at 7.14 rounds to 7.1 and copper at 8.96 rounds to 9.0, so neither is consistent with the measurement.
Step six: state the conclusion at the right strength. The sample is consistent with iron and inconsistent with the other four candidates in this table. That is not the same as proving it is iron: the table is a short list, alloys such as steel have densities in the same region, and the measurement only supports two significant figures.
Step seven: say how to strengthen it. The weak link is the volume. Using a larger sample, or a narrower graduated cylinder, or displacement into an overflow can measured on a balance, would all support a third significant figure and separate iron from a nearby alloy. A second intensive property such as melting point would settle it far more firmly than improving this one.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.
- State the formula for density and a common unit for a solid.
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\( d = \frac{m}{V} \), in g/cm3
- A sample has a mass of 22.5 g and a volume of 4.50 mL. Find its density.
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\( 22.5 \div 4.50 \). 5.00 g/mL
- Find the mass of 15.0 mL of ethanol, density 0.789 g/mL.
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\( m = dV = 0.789 \times 15.0 = 11.835 \), to three significant figures. 11.8 g
- Find the volume of 100.0 g of aluminum, density 2.70 g/cm3.
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\( V = \frac{m}{d} = \frac{100.0}{2.70} = 37.037 \), to three significant figures. 37.0 cm3
- On a heating curve, what is happening during a flat section?
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A phase change: the supplied energy is overcoming attractions between particles rather than increasing their speed
- Explain why a large steel ship floats while a small steel bolt sinks.
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Floating depends on the average density of the whole object including any space it encloses, not on the density of the material it is made from. The bolt is solid steel throughout, so its average density is steel's roughly 7.9 g/cm3, far above water's 1.00, and it sinks. The ship's hull encloses a very large volume of air, and the combined mass of steel and air divided by the total volume of the hull comes out below 1.00 g/cm3. Mass alone decides nothing, which is exactly why density rather than mass is the useful quantity. Average density including enclosed air, not the density of the material alone
- A student reports the density of a liquid as 0.92 g/mL at 20 degrees Celsius and
0.89 g/mL at 60 degrees. Explain why density changes with temperature and whether
this undermines its use as an identifying property.
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Heating gives the particles more kinetic energy, so they move further apart on average and the same mass occupies a larger volume. Mass is unchanged and volume has risen, so density falls. This does not undermine density as an identifying property, provided the temperature is stated: density is intensive, meaning it does not depend on sample size, but it is a function of temperature, so a reference value is always quoted at a specified temperature, usually 20 or 25 degrees Celsius. Comparing a measurement taken at 60 degrees against a table value for 20 degrees is the actual error to avoid. Volume expands on heating while mass is constant; the property is still identifying as long as the temperature is stated
- Explain, in terms of particles, why the temperature stays constant while ice melts
even though a burner is supplying heat throughout.
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Temperature is a measure of the average kinetic energy of the particles, so the temperature only rises when the energy supplied makes particles move faster. In a melting solid the energy is doing a different job: it is working against the attractive forces holding particles in fixed positions in the lattice, converting potential energy rather than kinetic energy. Until every particle has been freed from its lattice position the supplied energy continues to go into separation, so the average speed and therefore the temperature stay where they are. Once melting is complete the energy again goes into motion and the temperature resumes rising, which is why the curve turns upward at the end of the plateau. The energy overcomes attractions and raises potential energy, not average kinetic energy, so the temperature does not change
- Ice floats on water. Explain what this shows about the arrangement of water
molecules in the solid and why it matters for a lake in winter.
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Floating means ice is less dense than liquid water, so the same mass of water occupies more volume as a solid than as a liquid. That is the reverse of the usual pattern, and it means the molecules in ice must be held in an arrangement that is more open than the packing in the liquid. Hydrogen bonding is responsible: in the solid each molecule is locked into a fixed tetrahedral arrangement with gaps between, while in the liquid the bonds are constantly breaking and reforming and molecules can crowd closer together. For a lake this is decisive. Ice forms at the surface and stays there, insulating the water below, so the lake freezes downward slowly and liquid water persists at the bottom through winter. If ice sank, lakes would freeze solid from the bottom up. Hydrogen bonding holds the solid in a more open arrangement; the ice layer floats and insulates the liquid beneath
- Two heating curves for equal masses of different substances both show a flat section
at the melting point, but one plateau is three times longer than the other at the
same heating rate. Explain what this tells you.
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At a constant heating rate, the length of the plateau measures the energy needed to melt the sample, because time multiplied by rate is energy supplied. A plateau three times as long therefore means three times as much energy is required to melt the same mass, so that substance has roughly three times the heat of fusion per gram. Physically, its particles are held together by stronger attractions that take more energy to overcome. This is also evidence you can use elsewhere: the same substance would be expected to have a higher melting point and a higher boiling point, because all three quantities depend on the strength of the forces between particles, which unit 3 takes up as intermolecular forces. It requires about three times the energy per gram to melt, so its interparticle attractions are considerably stronger