Anomalous expansion of water | ICSE Class 9 Physics Notes

This note covers anomalous expansion of water, changes in volume and density between 0 and 10 degrees Celsius, temperature graphs, Hope’s experiment, and the consequences for freezing lakes and aquatic life.

What is anomalous expansion of water?

Temperature indicates the hotness or coldness of a body. The symbol °C means degrees Celsius, the temperature unit used here. Heat is energy transferred because of a temperature difference. Heating and cooling can change the space occupied by a substance.

Volume is the space occupied by a substance. Expansion means an increase in volume in this discussion; contraction means a decrease. Most substances expand on heating and contract on cooling. Water departs from this familiar pattern in a particular temperature interval.

Definition: Anomalous expansion of water is its unusual expansion on cooling from 4 °C to 0 °C. Conversely, water contracts on heating from 0 °C to 4 °C.

Why is this behaviour called anomalous?

Anomalous means departing from the usual pattern. The unusual feature is the direction of the volume change: cooling below 4 °C makes liquid water occupy more space, while heating towards 4 °C from below makes it occupy less space.

The interval matters as much as the words “expands” and “contracts”. Water does not contract throughout heating. Between 4 °C and 10 °C, it expands when heated. Between 10 °C and 4 °C, it contracts when cooled.

The term thermal expansion normally describes an increase in a body’s dimensions caused by a rise in temperature. Anomalous expansion identifies water’s exceptional response near its freezing point. It does not mean that heating and cooling have exchanged their meanings.

Is water freezing throughout this interval?

Freezing is the change from liquid water to solid ice. Pure water freezes at 0 °C under standard atmospheric pressure, the standard reference pressure of the surrounding air. The anomalous volume change from 4 °C towards 0 °C occurs while the water is still liquid.

Therefore, distinguish a temperature change within liquid water from a change of state. The special temperature 4 °C identifies water’s smallest volume for a given mass. It is not the temperature at which water changes into ice.

How are volume and density related?

Mass measures the quantity of matter in a sample. Density is mass per unit volume: it tells us how much mass occupies a given space. Use the same mass of water when comparing its volume or density at different temperatures.

Let ρ, the Greek letter rho, represent density; let m represent mass; and let V represent volume. Their relationship is ρ = m/V. The division sign in this expression means that density is found by dividing mass by volume.

SI means the International System of Units. The SI unit of mass is the kilogram, written kg. The SI unit of volume is the cubic metre, written m³. The SI unit of density is kilogram per cubic metre, written kg/m³.

The SI unit of temperature is the kelvin, written K. The SI unit of heat energy is the joule, written J. A temperature difference has the same numerical value in kelvins and degrees Celsius.

What changes when the mass remains constant?

For a fixed mass, an increase in volume gives a decrease in density. The same quantity of matter occupies more space. A decrease in volume gives an increase in density, because the same mass occupies less space.

Thus, between 0 °C and 4 °C, heating water decreases its volume and increases its density. Between 4 °C and 10 °C, heating increases its volume and decreases its density. The density changes follow directly from the volume changes.

Definition: Water has its maximum density at 4 °C. At this temperature, a fixed mass of liquid water has its minimum volume.

The density of water at 4 °C is given as 1.0 × 10³ kg/m³, where × means multiplication and 10³ means one thousand. This is a density value, not the mass of every possible sample of water.

Why must the comparison use a fixed mass?

If water is added to a container, its occupied volume increases because there is more water. That observation does not show thermal expansion. The temperature comparison concerns one unchanged mass, so a change in volume can be connected to its thermal behaviour.

Likewise, “denser” does not simply mean “a larger total mass”. It means more mass per unit volume. This distinction explains why a colder portion of water does not necessarily sink: its density relative to the surrounding water determines the tendency.

How does water behave during heating and cooling?

To follow a temperature change, first identify its direction and then check whether it crosses 4 °C. Treat the interval below 4 °C separately from the interval above it. This prevents a correct statement for one interval being applied to another.

What happens as liquid water is heated?

1. Start with liquid water at 0 °C and consider a fixed mass throughout the heating process.
2. As its temperature rises towards 4 °C, its volume decreases. This contraction on heating is the anomalous behaviour.
3. At 4 °C, the volume reaches its minimum and the density reaches its maximum.
4. As heating continues from 4 °C towards 10 °C, the volume increases and the density decreases.

Heating through the whole range therefore produces contraction followed by expansion. It is incorrect to describe the entire change from 0 °C to 10 °C as continuous contraction or continuous expansion. The direction reverses at 4 °C.

What happens when the direction is reversed?

Cooling from 10 °C towards 4 °C reduces volume and increases density. Continuing to cool below 4 °C reverses both trends: volume increases and density decreases. This second part is the anomalous expansion that gives the topic its name.

Note: A statement about water expanding or contracting must include the temperature interval and whether water is being heated or cooled.

A useful reasoning sequence is temperature direction, volume change, density change. For cooling below 4 °C, this becomes falling temperature, increasing volume and decreasing density. No extra mass is created during expansion, and contraction does not remove matter from the sample.

The four rows describe liquid water. They do not describe the volume change when ice melts or when water becomes ice. Keep the state of the substance explicit when explaining a cooling sequence that eventually reaches freezing.

How is the volume-temperature graph drawn and read?

A graph represents how one quantity changes with another. For the volume-temperature graph, the horizontal axis shows temperature and the vertical axis shows volume. An axis is a reference line used to locate values on a graph.

Use T to denote temperature on these graphs, measured in °C. The symbol V retains its meaning of volume. The graph refers to a fixed mass of water, so changes in the height of the curve represent changes in the space occupied by that mass.

The horizontal axis shows temperature in °C and the vertical axis shows the volume of one kilogram of water. The curve falls to a minimum near 4 °C and then rises. The temperature axis extends beyond 10 °C.

See Fig. 10.7(a) in your NCERT textbook

How should the 0 °C to 10 °C sketch be made?

1. Draw the horizontal temperature axis and mark 0 °C, 4 °C and 10 °C in their correct order.
2. Label the vertical axis “Volume of a fixed mass of water”. An exact numerical volume scale is unnecessary for a qualitative sketch.
3. Draw a smooth curve falling from 0 °C to its lowest point at 4 °C.
4. Continue the curve upwards from 4 °C towards 10 °C, showing expansion in this interval.

A qualitative sketch shows the pattern of change without claiming a full set of numerical measurements. The minimum must be above zero volume: water occupies space at 4 °C. “Minimum” means the smallest volume in the comparison, not no volume.

How does the same graph describe cooling?

Reading from left to right describes rising temperature. Reading from right to left describes falling temperature. For cooling from 10 °C to 4 °C, follow the descending volume towards the minimum. Then, from 4 °C towards 0 °C, follow the increasing volume.

The curve is not a graph of time. A point farther to the right represents a higher temperature, not necessarily a later moment in an experiment. Always identify the labelled axes before interpreting the shape or describing the direction of change.

How does the density-temperature graph differ?

The density-temperature graph uses temperature on the horizontal axis and density on the vertical axis. It describes the same behaviour as the volume graph through a different quantity. The lowest volume corresponds to the highest density for a fixed mass.

The horizontal axis shows temperature from 0 °C to 10 °C. The density curve rises to a maximum near 4 °C and then falls towards 10 °C. A vertical dashed line marks the temperature of the maximum.

See Fig. 10.7(b) in your NCERT textbook

What are the essential features of the sketch?

Label temperature T in °C along the horizontal axis. Label the vertical axis “Density, ρ”, with kg/m³ if a numerical density scale is used. Mark 4 °C clearly beneath the highest point of the curve.

From 0 °C to 4 °C, the curve rises because heating contracts the water. From 4 °C to 10 °C, it falls because heating expands the water. These are opposite trends to those on the volume graph over the same temperature intervals.

What can the graph establish without extra data?

The graph establishes that water at 4 °C is denser than water at either 0 °C or 10 °C. A qualitative sketch does not supply exact density values at every intermediate temperature. Use a supplied numerical scale when numerical readings are requested.

Opposite trends do not mean that the graphs are identical shapes turned upside down. Density depends on the reciprocal of volume, meaning one divided by volume, when mass is fixed. The important comparison here is the rising or falling trend and the common turning temperature.

For cooling, read the density curve in reverse. Density increases as temperature falls towards 4 °C from above, then decreases as temperature falls below 4 °C. This change explains the movement of water in Hope’s experiment and in a cooling lake.

How is Hope’s experiment arranged?

Hope’s experiment demonstrates water’s maximum density at 4 °C by cooling water around the middle of a vessel and observing temperatures above and below the cooled region. It links temperature changes to the upward or downward movement of water.

What apparatus is used?

The usual arrangement has a tall cylindrical vessel containing water. A cylindrical vessel has a circular horizontal shape and straight sides. Around its middle is a ring-shaped trough, an open container holding a cooling mixture of ice and salt.

A thermometer is an instrument for measuring temperature. One thermometer has its sensing bulb in the upper part of the water; another has its bulb in the lower part. The bulb is the part that responds to the surrounding water’s temperature.

The freezing mixture is the ice-and-salt mixture used to cool the middle region. It remains in the surrounding trough, separate from the water being investigated. Its purpose is to remove heat from that region of the vessel.

Draw a tall vessel containing water, with a ring-shaped trough around its middle labelled “ice and salt”. Show one thermometer bulb in the water above the trough and another below it. Label the vessel, water, upper thermometer and lower thermometer.

Why are two thermometers needed?

The water need not have the same temperature at every height during cooling. The upper and lower thermometers allow the two regions to be compared. A single reading would not show whether colder water collects above or below the middle.

Begin with water above 4 °C. Cool the middle using the surrounding mixture and observe how the two thermometer readings change. The relevant density comparison is between a cooled portion of water and the water around it.

Keep the observations separate from the explanation. Thermometers show temperatures, rather than directly measuring density. The explanation connects those readings to water’s known density changes: water becoming denser sinks, while water becoming less dense than its surroundings rises.

The arrangement separates the place where cooling occurs from the places where temperature is observed. This is why the positions of the trough and thermometer bulbs are essential labels in the diagram, rather than decorative details.

What observations and conclusions follow from Hope’s experiment?

Hope’s experiment has two stages because cooling water towards 4 °C from above increases its density, whereas cooling it below 4 °C decreases its density. The direction of movement changes when the cooled water passes through the temperature of maximum density.

What happens during the first stage?

1. Water near the middle loses heat to the cooling mixture while its temperature is above 4 °C.
2. As this water cools towards 4 °C, it contracts and becomes denser than the warmer water around it.
3. The denser cooled water sinks, bringing cooler water into the lower part of the vessel.
4. The lower thermometer falls towards 4 °C before the upper region has cooled to the same extent.

The lower region thus receives water approaching its maximum density. Cooling at the middle can first lower the temperature near the bottom because water itself moves. Heat transfer involving movement of the substance is called convection. A fluid is a substance that can flow, such as a liquid or gas.

What changes during the second stage?

When water at the cooled middle falls below 4 °C, it expands and becomes less dense than water at 4 °C. It rises into the upper region. The upper thermometer then falls towards 0 °C while the lower thermometer remains near 4 °C during this stage.

The key evidence is the difference between upper and lower readings, explained by the density change. The coldest water need not be at the bottom. Water near 0 °C can remain above denser water near 4 °C.

Do not describe the lower thermometer as permanently fixed at 4 °C under every possible condition. The observation concerns the characteristic stage of the demonstration. It shows how water redistributes while the middle region is cooled.

The conclusion follows through a complete chain: cooling changes volume, volume changes density, and density differences cause movement. Water at 4 °C collects below because it is denser than water at nearby temperatures both above and below 4 °C.

Why do lakes freeze at the surface, and what are the consequences?

The same density changes explain why lakes and ponds freeze at the top first. Water near the surface loses heat to colder air. Whether this surface water sinks depends on its temperature and density compared with the water below.

How does the water cool before surface freezing?

1. While the surface water is above 4 °C, cooling makes it contract and become denser.
2. The denser surface water sinks, while warmer, less dense water from below rises towards the surface.
3. This movement helps cool the water towards 4 °C, the temperature of maximum density.
4. When the surface water cools below 4 °C, its density decreases, so it remains above the denser water.
5. The surface water can then reach its freezing point and form ice at the top.

The change below 4 °C interrupts the earlier pattern in which cooling produced sinking. The surface can become colder than the deeper water. Aquatic life means plants and animals living in water; retaining liquid water below the surface is important for their survival.

Why is the surface ice layer significant?

Ice is solid water and is less dense than liquid water, so it floats. Ice also conducts heat poorly. Conduction is heat transfer through a material because of a temperature difference, without the bulk movement involved in convection.

A surface ice layer therefore slows further heat loss from the water beneath. It acts as an insulating layer, meaning a layer that reduces heat transfer. Insulation slows cooling; it does not supply heat or guarantee that freezing can never progress.

In the usual winter explanation for a sufficiently deep freshwater lake, water near the bottom remains around 4 °C while the surface freezes. Freshwater means water containing little dissolved salt. The liquid region below the ice provides a habitat for aquatic organisms.

How should the lake diagram be explained?

Draw a vertical section through a lake. Label ice at the surface, liquid water below it, and deeper water near 4 °C. Indicate that the upper liquid water is colder than the deeper water.

The diagram represents the winter situation being explained, rather than a claim that every lake has identical temperatures at every depth. Its main message is the order of the layers: surface ice, cold liquid water beneath, and denser water near 4 °C deeper down.

If water continued becoming denser as it cooled towards freezing, colder water would keep sinking. Lakes and ponds would freeze from the bottom upwards, destroying much of their animal and plant life. The actual anomalous behaviour has an important environmental consequence.

Keep the two physical explanations distinct: anomalous expansion explains why liquid water below 4 °C stays near the top; the lower density of ice explains why the solid ice floats. Both contribute to the surface-freezing picture.

How is ordinary thermal expansion calculated?

Linear expansion concerns length, area expansion concerns area, and volume expansion concerns volume. These calculations describe ordinary expansion and provide a comparison with water’s anomalous behaviour.

Let $l$, $A$ and $V$ denote initial length, area and volume. The symbol $\Delta$ indicates a change, and $\Delta T$ is the temperature change. The coefficients $\alpha_l$ and $\alpha_v$ describe fractional expansion per unit temperature rise.

For a small temperature change, $\Delta l=\alpha_l l\Delta T$. For volume expansion, $\Delta V=\alpha_v V\Delta T$. The coefficients are expressed in $\mathrm{K^{-1}}$. The volume coefficient generally depends on temperature, so a constant value must be appropriate to the interval used.

Derivation: How are volume and linear expansion related?

Consider a cube of initial side $l$, expanding equally in all directions. Assume $\Delta l$ is small compared with $l$.

1. The initial volume is $V=l^3$, and the increase in side length is $\Delta l=\alpha_l l\Delta T$.
2. The increase in volume is $\Delta V=(l+\Delta l)^3-l^3$. Expanding gives $\Delta V=3l^2\Delta l+3l(\Delta l)^2+(\Delta l)^3$.
3. Neglecting terms containing the square and cube of the small length change gives $\Delta V\approx3l^2\Delta l$. Dividing by the original volume gives $\frac{\Delta V}{V}\approx3\frac{\Delta l}{l}$.
4. Substituting the linear expansion expression gives $\frac{\Delta V}{V}\approx3\alpha_l\Delta T$. Compare this with $\frac{\Delta V}{V}=\alpha_v\Delta T$.

Result: $\alpha_v\approx3\alpha_l$, within the small-expansion approximation for a solid expanding equally in all directions.

Derivation: Why is area expansivity twice linear expansivity?

Consider a rectangular sheet with initial sides $a$ and $b$, both having the same linear expansion coefficient. Let $\alpha_A$ denote its coefficient of area expansion.

1. The initial area is $A=ab$. Heating produces side increases $\Delta a=\alpha_l a\Delta T$ and $\Delta b=\alpha_l b\Delta T$.
2. The area increase is $\Delta A=(a+\Delta a)(b+\Delta b)-ab$, giving $\Delta A=a\Delta b+b\Delta a+\Delta a\Delta b$.
3. Substitution gives $\Delta A=2\alpha_l ab\Delta T+\alpha_l^2ab(\Delta T)^2$. Divide by $A=ab$ and by the temperature change to obtain $\alpha_A=2\alpha_l+\alpha_l^2\Delta T$.
4. When $\alpha_l\Delta T$ is small compared with 2, neglect the final correction term. The remaining term is twice the linear coefficient.

Result: $\alpha_A\approx2\alpha_l$, within the small-expansion approximation.

How can expansion calculations be applied?

The following calculations apply the expansion relationships to changes in dimensions and density. Lengths compared in a ratio must use the same unit. In these problems, a temperature change measured in degrees Celsius has the same numerical value in kelvins.

How much heating allows an iron ring to fit?

Worked example 1. At 27 °C, an iron ring has diameter 5.231 m and a wooden wheel rim has diameter 5.243 m. Find the temperature needed for the ring to fit, using $\alpha_l=1.20\times10^{-5}\,\mathrm{K^{-1}}$.

Answer: Let $D$ be the original ring diameter and $T$ the required temperature.

Formula: $\Delta D=\alpha_l D\Delta T$; $\Delta T=\frac{\Delta D}{\alpha_l D}$; $T=27\,{}^\circ\mathrm{C}+\Delta T$.

Substitute: $\Delta D=(5.243-5.231)\,\mathrm{m}=0.012\,\mathrm{m}$. Therefore $\Delta T=\frac{0.012}{1.20\times10^{-5}\times5.231}\,\mathrm{K}\approx191.17\,\mathrm{K}$.

The required temperature is $T\approx218\,{}^\circ\mathrm{C}$. The ring must be heated to about 218 °C.

How does an expanding steel tape affect a measurement?

Worked example 2. A 1 m steel tape is correctly calibrated at 27.0 °C. At 45.0 °C, it reads 63.0 cm for a steel rod. Find the rod’s actual length at 45.0 °C and its length at 27.0 °C.

Answer: Use $\alpha_l=1.20\times10^{-5}\,\mathrm{K^{-1}}$ for both tape and rod. Let $R$ be the tape reading, $L$ the actual hot length and $L_0$ the rod’s length at 27.0 °C.

Formula: $L=R(1+\alpha_l\Delta T)$; $L_0=\frac{L}{1+\alpha_l\Delta T}$.

Substitute: $\Delta T=(45.0-27.0)\,\mathrm{K}=18.0\,\mathrm{K}$. Each tape interval expands by the factor $1+1.20\times10^{-5}\times18.0=1.000216$.

Thus $L=63.0\times1.000216\,\mathrm{cm}=63.013608\,\mathrm{cm}$, approximately 63.014 cm before rounding to the precision of the tape reading.

Cooling the rod gives $L_0=63.013608/1.000216\,\mathrm{cm}=63.0\,\mathrm{cm}$. The same expansion factor cancels because the tape and rod are both steel.

How far must a steel shaft be cooled?

Worked example 3. At 27 °C, a steel shaft has diameter 8.70 cm and a steel wheel’s central hole has diameter 8.69 cm. Only the shaft is cooled. Find its fitting temperature, taking $\alpha_l=1.20\times10^{-5}\,\mathrm{K^{-1}}$ as constant.

Answer: Let $D$ be the original shaft diameter and $T$ its final temperature.

Formula: $\Delta T=\frac{\Delta D}{\alpha_l D}$; $T=27\,{}^\circ\mathrm{C}+\Delta T$.

Substitute: $\Delta D=(8.69-8.70)\,\mathrm{cm}=-0.01\,\mathrm{cm}$. Then $\Delta T=\frac{-0.01}{1.20\times10^{-5}\times8.70}\,\mathrm{K}\approx-95.79\,\mathrm{K}$.

Hence $T\approx-68.8\,{}^\circ\mathrm{C}$, or about −69 °C. The negative temperature change represents contraction on cooling.

What happens to a hole in a heated copper sheet?

Worked example 4. A hole in a copper sheet has diameter 4.24 cm at 27.0 °C. Find the change in diameter at 227 °C, using $\alpha_l=1.70\times10^{-5}\,\mathrm{K^{-1}}$.

Answer: The hole’s diameter increases with the sheet’s dimensions. Let $D$ denote the original diameter.

Formula: $\Delta D=\alpha_l D\Delta T$.

Substitute: $\Delta T=(227-27.0)\,\mathrm{K}=200\,\mathrm{K}$. Therefore $\Delta D=1.70\times10^{-5}\times4.24\times200\,\mathrm{cm}=0.014416\,\mathrm{cm}$.

The diameter increases by approximately 0.0144 cm.

How much does a combined brass and steel rod expand?

Worked example 5. A brass rod and a steel rod, each 50 cm long and 3.0 mm in diameter, are joined together at 40.0 °C. Find the combined length increase at 250 °C when their ends are free.

Answer: Use $\alpha_b=2.0\times10^{-5}\,\mathrm{K^{-1}}$ for brass and $\alpha_s=1.2\times10^{-5}\,\mathrm{K^{-1}}$ for steel. Let $\Delta L_b$ and $\Delta L_s$ be their length increases.

Formula: $\Delta L_b=\alpha_b L_b\Delta T$; $\Delta L_s=\alpha_s L_s\Delta T$; $\Delta L=\Delta L_b+\Delta L_s$.

Substitute: $\Delta T=(250-40.0)\,\mathrm{K}=210\,\mathrm{K}$. Brass expands by $\Delta L_b=2.0\times10^{-5}\times50\times210\,\mathrm{cm}=0.210\,\mathrm{cm}$.

Steel expands by $\Delta L_s=1.2\times10^{-5}\times50\times210\,\mathrm{cm}=0.126\,\mathrm{cm}$. Adding gives $\Delta L=(0.210+0.126)\,\mathrm{cm}=0.336\,\mathrm{cm}$, or 3.36 mm.

No thermal stress develops at the junction in this arrangement because the ends are free to move and the rods can expand.

How does liquid expansion change density?

Worked example 6. Glycerine has volume expansion coefficient $\alpha_v=49\times10^{-5}\,\mathrm{K^{-1}}$. Find the fractional change in its density for a temperature rise of 30 °C.

Answer: Keep the mass fixed. Let $V_0$ and $\rho_0$ be the initial volume and density, and $V_1$ and $\rho_1$ their final values. The temperature rise is 30 K.

Formula: $V_1=V_0(1+\alpha_v\Delta T)$; $\frac{\rho_1}{\rho_0}=\frac{V_0}{V_1}$. Therefore $\frac{\rho_1-\rho_0}{\rho_0}=\frac{1}{1+\alpha_v\Delta T}-1$.

Substitute: $\alpha_v\Delta T=49\times10^{-5}\times30=0.0147$. The fractional density change is $\frac{1}{1.0147}-1\approx-0.01449$.

Density decreases by about 1.45 per cent. Neglecting higher-order small terms gives the first-order estimate $\frac{\Delta\rho}{\rho_0}\approx-\alpha_v\Delta T=-0.0147$, a decrease of about 1.47 per cent.

Glossary

• Anomalous expansion — The unusual increase in water’s volume when it cools from 4 °C towards 0 °C.
• Temperature — A measure of how hot or cold a body is compared with another body.
• Heat — Energy transferred between bodies or regions because their temperatures are different.
• Volume — The amount of space occupied by a substance or a particular sample.
• Contraction — A decrease in the volume occupied by the same mass of a substance.
• Density — Mass per unit volume, found by dividing a sample’s mass by its volume.
• Maximum density — The greatest density in a comparison, reached by liquid water at 4 °C.
• Minimum volume — The smallest volume occupied by a fixed mass of water, reached at 4 °C.
• Thermometer — An instrument used to measure temperature, including temperatures in different regions of water.
• Freezing mixture — The ice-and-salt mixture used around the middle of the vessel in Hope’s experiment.
• Convection — Transfer of heat through the actual movement of matter within a fluid.
• Freezing — The change of state in which a liquid becomes a solid.
• Insulating layer — A layer that reduces heat transfer, slowing the cooling of material beneath it.
• Aquatic life — Plants and animals living in water, including those beneath a lake’s surface ice.

Common errors and misconceptions

• Misconception: Water expands whenever it is heated. Correct: Water contracts on heating from 0 °C to 4 °C, then expands on further heating towards 10 °C.
• Misconception: Water at 0 °C has the greatest density because it is coldest. Correct: Water has maximum density at 4 °C; cooling below that temperature decreases its density.
• Misconception: Expansion increases the mass of the water. Correct: The same mass occupies a larger volume, so its density decreases.
• Misconception: Both temperature graphs have a minimum at 4 °C. Correct: The volume graph has a minimum, while the density graph has a maximum at that temperature.
• Misconception: The lower thermometer must reach 0 °C first in Hope’s experiment. Correct: Water cooled below 4 °C becomes less dense and rises, cooling the upper region towards 0 °C.
• Misconception: Anomalous expansion means that liquid water is already becoming ice between 4 °C and 0 °C. Correct: The anomaly describes a volume change within the liquid; freezing is a separate change of state.
• Misconception: A surface ice layer keeps deeper water liquid by producing heat. Correct: Ice slows heat loss because it conducts heat poorly; it does not act as a heat source.

Exam-style questions with model answers

Q1. Define anomalous expansion of water and state the temperature at which liquid water has its maximum density. [2 marks]

1. Anomalous expansion is the increase in water’s volume when it is cooled from 4 °C to 0 °C, contrary to the usual contraction on cooling.
2. Liquid water has its maximum density at 4 °C, where a fixed mass occupies its minimum volume.

Q2. A fixed mass of liquid water is heated from 0 °C to 10 °C. Describe its volume and density changes in the two intervals separated by 4 °C. [4 marks]

1. From 0 °C to 4 °C, the volume decreases: water contracts even though its temperature is rising.
2. Over this first interval, density increases because the same mass occupies less volume, reaching maximum density at 4 °C.
3. From 4 °C to 10 °C, the volume increases: water now expands on heating in the usual way.
4. Over this second interval, density decreases because the unchanged mass occupies an increasing volume as the temperature rises.

Q3. Describe labelled qualitative volume-temperature and density-temperature graphs for a fixed mass of liquid water from 0 °C to 10 °C. Include the axes, both trends and the turning temperature. Exact vertical values are not required. [6 marks]

1. For the volume graph, label the horizontal axis temperature in °C and the vertical axis volume of a fixed mass of water.
2. Draw the volume curve falling from 0 °C to 4 °C, showing that heating through this interval contracts the liquid water.
3. Mark the minimum volume at 4 °C, then draw the volume curve rising as temperature increases from 4 °C to 10 °C.
4. For the density graph, label the horizontal axis temperature in °C and the vertical axis density, meaning mass per unit volume.
5. Draw the density curve rising from 0 °C to 4 °C, showing that contraction increases density while the mass remains constant.
6. Mark maximum density at 4 °C, then draw the curve falling towards 10 °C because expansion decreases density in this interval.

Q4. In Hope’s experiment, water initially above 4 °C is cooled by an ice-and-salt trough around the middle of a tall vessel. Thermometer bulbs are above and below the trough. Explain the lower reading approaching 4 °C first, followed by the upper reading approaching 0 °C while the lower remains near 4 °C. [5 marks]

1. Water beside the trough first cools towards 4 °C from above. Its volume decreases and its density increases during this initial cooling stage.
2. The cooled, denser water sinks into the lower region, bringing its temperature down. The lower thermometer therefore approaches 4 °C first.
3. Further cooling at the middle takes some water below 4 °C. This water expands instead of continuing to contract, so its density decreases.
4. The less dense water rises into the upper region, causing the upper thermometer reading to fall towards 0 °C during continued cooling.
5. Denser water near 4 °C remains below during this stage. The contrasting readings demonstrate maximum density at 4 °C and decreasing density below it.

Q5. A sufficiently deep freshwater lake initially above 4 °C loses heat at its surface during winter under standard atmospheric pressure. Explain why freezing begins at the top and how this helps aquatic life. Include water’s density changes and the effect of surface ice. [5 marks]

1. Initially, cooling surface water towards 4 °C makes it denser. It sinks while warmer, less dense water rises, helping cool the lake.
2. After the surface water falls below 4 °C, further cooling decreases its density, so it remains above the denser water underneath.
3. The upper water reaches the freezing point, 0 °C under standard atmospheric pressure, and ice forms at the surface rather than first at the bottom.
4. Ice floats because it is less dense than liquid water. It also conducts heat poorly, so the surface layer slows further heat loss.
5. Liquid water can remain beneath the ice, with deeper water near 4 °C in the usual winter situation, providing a habitat for aquatic plants and animals.

Q6. Explain why a fixed mass of liquid water has maximum density when its volume is minimum. Define every symbol in the relationship you use and identify the relevant temperature. [3 marks]

1. The relationship is ρ = m/V, where ρ is density, m is mass and V is volume. Density means mass per unit volume.
2. With mass fixed, dividing by a smaller volume gives a greater density. Therefore, the smallest volume corresponds to the greatest density.
3. Liquid water reaches this minimum volume at 4 °C. Its density is consequently maximum at the same temperature, rather than at its freezing point.

Q7. A fixed mass of liquid water cools from 10 °C to 0 °C under standard atmospheric pressure. Describe its volume change before and after 4 °C, and distinguish the liquid anomaly from freezing. [3 marks]

1. From 10 °C to 4 °C, the water contracts as it cools, and its volume falls towards the minimum reached at 4 °C.
2. From 4 °C to 0 °C, it expands as it cools. This increase in the volume of liquid water is anomalous expansion.
3. Freezing is the change from liquid water to solid ice at 0 °C under the stated pressure, rather than the liquid volume change above that temperature.

Q8. In Hope’s experiment, the freezing mixture surrounds the middle of the water vessel. State where the two thermometer bulbs are placed and explain why two readings are needed. [2 marks]

1. Place one thermometer bulb in the upper water above the cooling trough and the other in the lower water below it.
2. Two readings reveal the different temperatures in these regions, allowing the effects of sinking and rising cooled water to be compared.

Key takeaways

• Water expands on cooling from 4 °C to 0 °C and contracts on heating through the same interval.
• A fixed mass of water has minimum volume and maximum density at the common temperature of 4 °C.
• Density is mass divided by volume, so expansion decreases density when the mass remains unchanged.
• Between 4 °C and 10 °C, liquid water follows the usual pattern of expansion on heating and contraction on cooling.
• The volume-temperature graph has a minimum at 4 °C, whereas the density-temperature graph has a maximum there.
• Hope’s experiment connects cooling at the middle of a vessel with contrasting temperatures above and below the cooled region.
• In a cooling lake, water below 4 °C remains near the surface because it is less dense than water at 4 °C.
• Surface freezing and reduced heat loss through ice allow liquid water beneath the surface to support aquatic life.

Test yourself

What does “anomalous” mean in this topic?

It means departing from the usual pattern: water expands on cooling from 4 °C towards 0 °C.

What happens to density when liquid water is heated from 0 °C to 4 °C?

Its density increases because its volume decreases while its mass remains unchanged.

Does minimum volume at 4 °C mean zero volume?

No. It means the smallest volume occupied by that fixed mass in the temperature comparison.

Which axis carries temperature on both graphs?

The horizontal axis carries temperature; the vertical axis carries volume or density.

Why does water cooled below 4 °C rise in Hope’s experiment?

It expands and becomes less dense than the water near 4 °C beneath it.

Which thermometer approaches 4 °C first in the initial stage of Hope’s experiment?

The lower thermometer approaches 4 °C first as denser cooled water sinks.

Why is “colder water always sinks” incorrect?

Below 4 °C, further cooling decreases water’s density, so it can remain above denser water.

How does the ice layer affect heat loss from water below it?

Ice conducts heat poorly, so the surface layer slows further heat loss from the underlying water.