Temperature

Heat 热量 (thermal energy) flows from a higher temperature to a lower temperature. When two bodies touch, energy moves until their temperatures are equal — they reach thermal equilibrium 热平衡. At equilibrium there is no net flow of energy.

Two regions at the same temperature 温度 are in thermal equilibrium with each other — no net energy flows, even though particles still exchange energy.

Temperature decides the direction of heat flow. It is not a measure of how much thermal energy 热能 a body holds. A small cup of boiling water (100 °C) holds far less energy than a swimming pool at 25 °C, but a piece of metal put in the cup gains energy while one put in the pool loses it.

Heat flows from hot to cold until both reach the same temperature — thermal equilibrium

A thermometer measures temperature on a defined scale.

Any physical property that changes in a repeatable way with temperature can make a thermometer 温度计. Examples:

• density of a liquid — a liquid-in-glass thermometer (mercury or alcohol). As the temperature rises, the liquid expands and rises up a narrow capillary 毛细管.
• volume of a gas at constant pressure — a gas thermometer. The gas volume rises in step with the absolute temperature.
• resistance of a metal — a resistance thermometer. A metal's resistance 电阻 rises nearly in step with temperature over a wide range.
• e.m.f. of a thermocouple — a thermocouple 热电偶 is two different metals joined at two points; the electromotive force 电动势 it makes depends on the temperature difference between the joins.

Different thermometers can read slightly differently if the property does not change in a straight line; they agree only at the calibration 校准 points.

A constant-volume gas thermometer — the gas pressure is found from the height difference h

A thermal (infrared) camera maps temperature to colour: the hot fries glow bright orange, the cold drink stays dark

The thermodynamic temperature 热力学温度 (or absolute temperature 绝对温度) scale does not depend on any one substance — only on the laws of thermodynamics. Its unit is the kelvin 开尔文 (K).

Absolute zero

The lowest possible temperature is zero kelvin ($0\ \text{K}$), called absolute zero 绝对零度. There a system has its least possible internal energy 内能 — particles have no random motion to speak of. Nothing can be cooled below this.

Extrapolating the pressure–temperature line back to zero pressure gives absolute zero, about −273 °C

Celsius scale

The Celsius 摄氏度 scale $\theta$ is shifted from the thermodynamic scale by a fixed amount:

So $0\ ^{\circ}\text{C} = 273.15\ \text{K}$ and $100\ ^{\circ}\text{C} = 373.15\ \text{K}$. A kelvin and a degree Celsius are the same size, so a temperature difference of $1\ \text{K}$ equals $1\ ^{\circ}\text{C}$ — but the absolute values differ by $273.15$.

In gas-law calculations you must always use absolute temperatures in kelvin. Using °C gives wrong answers.

The specific heat capacity 比热容 $c$ of a substance is the energy 能量 needed to raise the temperature of unit mass by one kelvin:

Unit: $\text{J kg}^{-1}\ \text{K}^{-1}$.

Examples:

• water: $c \approx 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$ (high — why water is a good coolant and why oceans steady the climate).
• aluminium: $c \approx 900\ \text{J kg}^{-1}\ \text{K}^{-1}$.
• copper: $c \approx 385\ \text{J kg}^{-1}\ \text{K}^{-1}$.

Water's specific heat capacity is far higher than common solids — why it is such a good coolant

To find an unknown $c$ by experiment: supply known energy $Q$ electrically ($Q = VIt$, from the power 功率), then measure the temperature rise $\Delta T$ of a known mass 质量 $m$. Then $c = Q/(m\Delta T)$. Reduce heat loss with insulation 隔热 and use a rise of about 10 K (big enough to measure well, small enough to limit losses).

Worked example. How much energy is needed to heat $0.50\ \text{kg}$ of water from $20\ ^{\circ}\text{C}$ to $100\ ^{\circ}\text{C}$? (Specific heat capacity of water $c = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$.)

A temperature difference is the same in K and °C, so $\Delta T = 80$:

When two bodies reach thermal equilibrium with no heat lost to the surroundings, the energy gained by the colder one equals the energy lost by the hotter one:

Worked example. $0.20\ \text{kg}$ of water at $80\ ^{\circ}\text{C}$ is mixed with $0.30\ \text{kg}$ of water at $20\ ^{\circ}\text{C}$, with no heat lost. Find the final temperature.

The heat lost by the hot water equals the heat gained by the cold water (the $c$ of water cancels):

When a substance changes state (solid ↔ liquid, or liquid ↔ gas) at constant temperature, energy must be supplied (or removed) with no temperature change. This energy is the latent heat 潜热.

The specific latent heat 比潜热 $L$ is the energy to change the state of unit mass at constant temperature:

Unit: $\text{J kg}^{-1}$.

Two kinds:

• specific latent heat of fusion 熔化 $L_{\text{f}}$ — for melting or freezing (solid ↔ liquid).
• specific latent heat of vaporisation 汽化 $L_{\text{v}}$ — for boiling or condensing (liquid ↔ gas).

For water at atmospheric pressure: $L_{\text{f}} \approx 3.34 \times 10^{5}\ \text{J kg}^{-1}$ (at $0\ ^{\circ}\text{C}$); $L_{\text{v}} \approx 2.26 \times 10^{6}\ \text{J kg}^{-1}$ (at $100\ ^{\circ}\text{C}$). So $L_{\text{v}}$ is about 7 times $L_{\text{f}}$.

Heating curve: the sloped parts warm the substance ($mc\Delta T$); the flat plateaus are the phase changes ($mL$)

Worked example. A $2.0\ \text{kW}$ heater boils water already at $100\ ^{\circ}\text{C}$. How long does it take to turn $0.10\ \text{kg}$ of this water into steam? ($L_{\text{v}} = 2.26 \times 10^{6}\ \text{J kg}^{-1}$, no heat lost.)

The energy needed is $Q = mL_{\text{v}} = 0.10 \times 2.26 \times 10^{6} = 2.26 \times 10^{5}\ \text{J}$. From $Q = Pt$,

Why $L_{\text{v}} > L_{\text{f}}$

Two reasons, both from the particle picture of matter:

1. Bonds: in melting, only some of the intermolecular 分子间 bonds break; the particles stay close as a liquid. In boiling, all the bonds must break so the particles can separate. Breaking all of them needs more energy.
2. Work against the atmosphere: when a liquid turns to gas it expands hugely (vapour has about $10^{3}$ times the liquid's volume 体积), so it does work pushing back the surrounding atmospheric pressure 大气压强. That work comes from the energy supplied.

Multi-step problems

If a problem mixes temperature change and a phase change 相变 (e.g. ice at $-5\ ^{\circ}\text{C}$ warming to water at $30\ ^{\circ}\text{C}$):

1. heat the solid from $-5\ ^{\circ}\text{C}$ to $0\ ^{\circ}\text{C}$: $Q_{1} = m c_{\text{ice}} \times 5$.
2. melt at $0\ ^{\circ}\text{C}$: $Q_{2} = m L_{\text{f}}$.
3. heat the water from $0\ ^{\circ}\text{C}$ to $30\ ^{\circ}\text{C}$: $Q_{3} = m c_{\text{water}} \times 30$.

Total: $Q_{1} + Q_{2} + Q_{3}$. A phase change is at constant temperature, so use $mL$ there, not $mc\Delta T$.

Split a mixed problem into stages — warm, change state, warm — then add the energies

When a question gives heater power $P$ and asks for the time, use $Q = Pt$ (assuming no heat loss). Insulating (lagging) the container and using a small mass are common ways to improve the experiment.

• Thermal equilibrium means no net heat flow (equal temperature); the thermodynamic scale is fixed by absolute zero and the triple point.
• Convert temperatures with $T/\text{K} = \theta/^\circ\text{C} + 273.15$ and use kelvin in energy and gas equations.
• Use $Q = mc\Delta T$ for heating and $Q = mL$ for a change of state (no temperature change) — never mix the two.