Why a Frozen Lake Stays Liquid Underneath

Heat a steel rod and it grows slightly longer; warm the air in a tyre and its pressure rises. Raising temperature makes particles move more vigorously, and that shows up as expansion or higher gas pressure.

To describe it we need clear ideas of heat and temperature, a reliable temperature scale, and formulas for how much things expand.

This part covers heat versus temperature, the ideal gas law and absolute temperature, coefficients of expansion, and expansion problems including water's unusual behaviour.

**Heat is energy transferred because of a temperature difference, while temperature measures how hot a body is; the scales are linked by $T_F=\frac{9}{5}{T}_C+32$ and $T_K=T_C+273.15$.**

On the Celsius scale water freezes at $0°$ and boils at $100°$; on the Fahrenheit scale these are $32°$ and $212°$ — **$100$ Celsius degrees span $180$ Fahrenheit degrees.

Worked example 1 — body temperature.**

$$T_F=\frac{9}{5}(37)+32=98.6\text{ °F}{T}_K=37+273.15=310.15\text{ K}$$

Worked example 2 — a fever. $104$ °F in Celsius:

$$T_C=\frac{5}{9}(104-32)=40\text{ °C}$$

Worked example 3 — a difference. A rise of $20$ °C is a rise of $20$ K but $36$ °F.

An everyday example. A bucket of warm bathing water holds far more heat than a cup of boiling tea, even though the tea is at the higher temperature.

The substance. Heat is not something a body contains — it is energy in transit; what a body stores is internal energy.

**For a dilute gas, $PV=\mu RT$ with $R=8.314$ J/mol K; since the pressure of a gas at constant volume falls in proportion to temperature and would reach zero at $-273.15$ °C, that point is taken as absolute zero, $0$ K, giving a scale that does not depend on the gas used.**

At constant volume, $P\propto T$; at constant pressure, $V\propto T$. Different gases give straight lines that all meet the temperature axis at the same point.

Worked example 1 — a sealed container. A gas at $27$ °C ($300$ K) and $1.0\times 10^5$ Pa is heated to $127$ °C ($400$ K) at constant volume:

$$P_2=P_1\frac{T_2}{T_1}=1.0\times 10^5\times \frac{400}{300}\approx 1.33\times 10^5\text{ Pa}$$

Worked example 2 — counting moles. $1$ m${}^3$ of gas at $1.013\times 10^5$ Pa and $273$ K:

$$\mu =\frac{PV}{RT}=\frac{1.013\times 10^5\times 1}{8.314\times 273}\approx 44.6\text{ mol}$$

An everyday example. Tyre pressure rises after a long drive on a hot afternoon, as the trapped air warms at nearly constant volume.

The substance. Gas-law calculations must use kelvin — in Celsius, a gas at $0$ °C would wrongly have zero pressure.

**For a small temperature change $\Delta T$: $\Delta L=\alpha L\Delta T$, $\Delta A=\beta A\Delta T$ and $\Delta V=\gamma V\Delta T$, and for a solid that expands equally in all directions $\beta =2\alpha$ and $\gamma =3\alpha$.

Why $\gamma =3\alpha$.** A cube of side $L$ becomes $L(1+\alpha \Delta T)$, so its volume becomes

$$L^3(1+\alpha \Delta T{)}^3\approx L^3(1+3\alpha \Delta T)$$

because $\alpha \Delta T$ is so small that its square and cube can be ignored. The same reasoning with a square gives $\beta =2\alpha$.

Worked example. Steel has $\alpha =1.2\times 10^{-5}$ K${}^{-1}$:

$$\beta =2.4\times 10^{-5}{\text{ K}}^{-1}\gamma =3.6\times 10^{-5}{\text{ K}}^{-1}$$

A steel plate of area $2$ m${}^2$ warmed by $50$ K grows by

$$\Delta A=2.4\times 10^{-5}\times 2\times 50=2.4\times 10^{-3}{\text{ m}}^2$$

An everyday example. A blacksmith heats an iron tyre before fitting it on a bullock-cart wheel; as it cools it shrinks and grips the wood tightly.

The substance. A hole in a heated plate gets bigger, not smaller — it expands exactly as if it were made of the plate's material.

**Use $\Delta L=\alpha L\Delta T$ for solids, $\Delta V=\gamma V\Delta T$ for liquids and gases (with $\gamma =\frac{1}{T}$ for an ideal gas at constant pressure), and remember that water contracts when warmed from $0$ °C to $4$ °C, reaching its greatest density at $4$ °C.

Worked example 1 — railway lines.** A $30$ m steel rail is laid on a day $40$ K cooler than the hottest days it will face:

$$\Delta L=1.2\times 10^{-5}\times 30\times 40\approx 1.44\times 10^{-2}\text{ m}=14.4\text{ mm}$$

So a gap of about $1.5$ cm must be left.

Worked example 2 — a liquid. $100$ cm${}^3$ of mercury ($\gamma =1.8\times 10^{-4}$ K${}^{-1}$) warmed by $50$ K:

$$\Delta V=1.8\times 10^{-4}\times 100\times 50=0.9{\text{ cm}}^3$$

Worked example 3 — a gas. At constant pressure and $300$ K, $\gamma =\frac{1}{300}\approx 3.3\times 10^{-3}$ K${}^{-1}$ — about a hundred times larger than for steel.

Anomalous expansion of water. As a lake cools, water at $4$ °C, being densest, sinks. Below $4$ °C the colder water stays on top and freezes, forming an insulating ice layer while water below remains near $4$ °C.

An everyday example. Lakes in the high mountains freeze over in winter, yet fish survive in the liquid water beneath the ice.

The substance. Liquids have no linear coefficient — they take the container's shape, so only $\gamma$ is meaningful.