题型概述：热学基础Question type: basic thermal physics
Top Gold热学是 Section 1 里内容最简单的一类：每年约 8 分，通常 1–2 道小题（每题 3–5 分）。物理只有几条公式——Q = mcΔT、Q = mL、P = kAΔθ/Δx、ΔL = αL₀ΔT、P ∝ T⁴、pV = nRT。失分几乎都来自四件事：① 辐射和气体题没把温度换成开尔文；② 能量守恒漏项（冷凝后的水还要继续降温、量热器也要吸热）；③ 线膨胀系数和体膨胀系数混用；④ 题目不给水的比热容时想不起来（约 4200 J kg⁻¹ K⁻¹）。每题做完都看一眼结果是否符合常识。Thermal physics is the simplest material in Section 1: about 8 marks a year, usually 1–2 short questions of 3–5 marks each. The physics is a handful of formulas — Q = mcΔT, Q = mL, P = kAΔθ/Δx, ΔL = αL₀ΔT, P ∝ T⁴, pV = nRT. Marks are lost to four things: ① not converting to kelvin in radiation and gas questions; ② missing a term in the energy balance (condensed water keeps cooling, the calorimeter also absorbs heat); ③ mixing up linear and volume expansion coefficients; ④ not knowing the specific heat of water when it is not given (about 4200 J kg⁻¹ K⁻¹). Always check that the answer is physically sensible.
本模块统一使用：Values used throughout: cwater = 4180 J kg⁻¹ K⁻¹, Lf = 3.34×10⁵ J kg⁻¹, Lv ≈ 2.26×10⁶ J kg⁻¹, T(K) = θ(°C) + 273.15。答案可用科学计数法输入，例如 2.25e6 或 2.25×10^6。. Answers can be typed in scientific notation, e.g. 2.25e6 or 2.25×10^6.
1比热容、潜热与"放热 = 吸热"Specific heat, latent heat and "heat lost = heat gained"
物体温度改变 ΔT 需要的热量与质量和温度变化成正比，比例系数就是比热容 c。物态变化（熔化、沸腾）时温度不变，吸收或放出的热量由潜热 L 决定。在绝热容器里混合时，热的部分放出的热量全部被冷的部分吸收——把每一项逐一列出来，一项都不能漏。The heat needed to change an object's temperature by ΔT is proportional to its mass and to ΔT; the constant is the specific heat capacity c. During a change of state (melting, boiling) the temperature stays fixed and the heat is set by the latent heat L. When things are mixed in an insulated container, all the heat given out by the hot parts is taken in by the cold parts — list every term, and miss none.
Q = mcΔT Q = mL ΣQlost = ΣQgained记住量级：水 c ≈ 4200 J kg⁻¹ K⁻¹，冰熔化 Lf ≈ 3.3×10⁵ J kg⁻¹，水汽化 Lv ≈ 2.3×10⁶ J kg⁻¹（约为 Lf 的 7 倍）。"热容量相当于 0.010 kg 水"的量热器，直接当作多了 0.010 kg 水处理。Know the sizes: water c ≈ 4200 J kg⁻¹ K⁻¹, melting ice Lf ≈ 3.3×10⁵ J kg⁻¹, boiling water Lv ≈ 2.3×10⁶ J kg⁻¹ (about 7 times Lf). A calorimeter "equivalent to 0.010 kg of water" is simply treated as an extra 0.010 kg of water.
用 2.0 kW 的加热器把 1 kg、−20 °C 的冰一直加热成 110 °C 的水蒸气。加热功率不变，所以横轴时间正比于吸收的热量。A 2.0 kW heater takes 1 kg of ice at −20 °C all the way to steam at 110 °C. The power is constant, so time on the horizontal axis is proportional to heat absorbed.
两段水平线就是潜热：熔化平台 167 s，沸腾平台 1130 s。温度不变并不代表没有吸热——热量用来拆开分子间的束缚。The two flat sections are latent heat: 167 s for melting, 1130 s for boiling. A constant temperature does not mean no heat is going in — the energy is used to break the bonds between molecules.
2热传导与热膨胀Conduction and thermal expansion
稳态热传导：单位时间通过的热量（热功率）与温差和截面积成正比，与厚度成反比，比例系数是热导率 k。几段材料首尾相接且侧面绝热时，热量无处可去，每一段的热功率都相同，所以导热差（k 小）的那段温度降得最多。Steady conduction: the heat per second (thermal power) is proportional to the temperature difference and the area and inversely proportional to the thickness; the constant is the thermal conductivity k. For sections joined end to end with insulated sides the heat has nowhere else to go, so the power is the same through every section, and the poorest conductor (smallest k) takes the biggest temperature drop.
P = kA Δθ / Δx热膨胀：长度的相对变化与温度变化成正比（线膨胀系数 α）；体积的相对变化用体膨胀系数 β。固体各方向都膨胀，β ≈ 3α。液体只给 β。Expansion: the fractional change in length is proportional to the temperature change (linear coefficient α); for volume use the volume coefficient β. A solid expands in all three directions, so β ≈ 3α. Liquids are only given a β.
ΔL = αL₀ΔT ΔV = βV₀ΔT, βsolid ≈ 3α容器装满液体再加热：容器的容积也在变大，溢出量取决于 β液 − β容器，不是 β液 本身。Heating a container filled with liquid: the container's capacity grows too, so the overflow depends on βliquid − βcontainer, not on βliquid alone.
3黑体辐射与理想气体（温度一律用 K）Black-body radiation and the ideal gas (always kelvin)
斯特藩–玻尔兹曼定律：黑体单位时间辐射的能量与表面积和绝对温度的四次方成正比；维恩位移定律：峰值波长与绝对温度成反比。题目几乎都用比值法，常数会约掉。Stefan–Boltzmann law: a black body radiates power proportional to its surface area and to the fourth power of its absolute temperature; Wien's displacement law: the peak wavelength is inversely proportional to the absolute temperature. Questions are almost always done by ratios, so the constants cancel.
P = σAT⁴ λmaxT = b ≈ 2.90×10⁻³ m K理想气体：pV = nRT。压强相同时，同一种气体的密度 ρ = pM/(RT)，所以 ρT = 常数——热空气更稀，这就是热气球的原理。Ideal gas: pV = nRT. At the same pressure the density of a given gas is ρ = pM/(RT), so ρT = constant — hot air is thinner, which is how a hot-air balloon works.
pV = nRT ρ1T1 = ρ2T2 （同压）(same p)500 °C 升到 1000 °C，辐射功率不是 2⁴ = 16 倍，而是 (1273/773)⁴ ≈ 7.4 倍。摄氏度只能用于温度"差"（ΔT），不能用于比值。Going from 500 °C to 1000 °C does not multiply the power by 2⁴ = 16 but by (1273/773)⁴ ≈ 7.4. Celsius is fine for temperature differences (ΔT), never for ratios.
| 子题型Sub-type | 识别特征How to spot it | 核心步骤Core step | 真题Past paper |
|---|---|---|---|
| 混合 / 潜热Mixing / latent heat | "steam is passed into", "ice is placed in … water" | 放热 = 吸热，逐项列出：相变 mL + 升降温 mcΔT + 量热器heat lost = heat gained, term by term: change of state mL + mcΔT + calorimeter | 2021 Q1(g); 2019 Q1(g), 2024 Q1(j) |
| 流动加热Heating a flow | "flows at … litre min⁻¹", "rate of flow of cooling water" | P = (m/t) c ΔT | 2021 Q1(f), 2020 Q1(n) |
| 串联热传导Conduction in series | "composite rod", "well insulated", k values | 各段 P 相同 ⇒ kΔθ/Δx 相同；各段温降之和 = 总温差same P in each section ⇒ same kΔθ/Δx; drops add up to the total | 2019 Q1(r) |
| 热膨胀Thermal expansion | α, β, "spills out", "difference … not to vary" | ΔV = βV₀ΔT，容器 β = 3α；溢出 = 两者之差ΔV = βV₀ΔT, container β = 3α; overflow = the difference | 2022 Q1(i); 2021 Q1(h) |
| 黑体辐射Black-body radiation | "Stefan", "Wien", T⁴, "peak wavelength", "red giant" | 比值法：P ∝ R²T⁴，λmax ∝ 1/T，T 用 Kratios: P ∝ R²T⁴, λmax ∝ 1/T, T in K | 2020 Q1(o); 2022 Q1(o) |
| 理想气体Ideal gas | "balloon", "bulbs connected", "heated to" | pV = nRT（n 守恒）或同压下 ρT = 常数，T 用 KpV = nRT (n conserved) or ρT = constant at fixed p, T in K | 2020 Q1(s); 2019 Q1(n) |