CAIE物理 热学 比热容 潜热 传导对流辐射
CAIE IGCSE Physics 0625 的考纲把热学放在第 2 部分，编号 2.1 到 2.3：粒子模型、热学性质与温度、热能传递。2026 到 2028 这一版考纲沿用同一套编号，所以整理真题时按 2.1、2.2、2.3 归档最省事。热学内容在 Paper 2 的选择题和 Paper 4 的结构题里都会出现，需要动笔计算的只有比热容和潜热两个公式，其余大量题目考的是文字解释和实例判断，把过程写清楚比算得快更值分。
Thermal physics is Section 2 of the Cambridge IGCSE Physics 0625 syllabus, and the units always appear in the same order: 2.1 kinetic particle model of matter, 2.2 thermal properties and temperature, 2.3 transfer of thermal energy. The 2026 to 2028 syllabus keeps that numbering, so past paper questions file neatly under three headings. Only two equations carry most of the numerical marks, specific heat capacity and specific latent heat; the rest of the topic rewards explanation and example recognition rather than arithmetic speed.
2.1 粒子模型与布朗运动
物质三态的差别落在粒子排列和运动方式上。固体粒子紧密排列，只在固定位置振动；液体粒子间距略大，可以互相滑过；气体粒子间距最大，运动接近自由飞行。布朗运动是考试里最常出现的间接证据：悬浮在空气中的烟粒被看不见的空气分子不停撞击，所以轨迹是无规则的折线。很多学生以为布朗运动说明分子自己在乱动，标准答案写的是分子撞击悬浮粒子，而不是有人看见了分子。气体压强来自粒子对容器壁的碰撞，体积不变时升温，粒子平均速度变大，碰撞更频繁也更有力，压强随之升高。温度升高也会让扩散加快，原因同样是粒子平均动能变大。固体里粒子没有离开原位，所以把糖块放在冷水中，糖只能靠液体一侧的扩散慢慢化开，速度比热水慢得多。答题时把布朗运动与热运动分开写，前者说的是悬浮粒子的无规则运动，后者说的是分子本身的无规则运动，这两个词互换一般不给分。
The three states of matter differ in how their particles are arranged and how they move. In a solid the particles sit close together and vibrate about fixed positions, in a liquid they can slide past one another, and in a gas the spacing is wide enough for something close to free flight. Brownian motion is the evidence examiners reach for: smoke particles in air get knocked around by air molecules nobody can see, which is why the paths look like random zigzags. Gas pressure comes from particles striking the container walls, and at constant volume a temperature rise makes the average particle faster, so collisions arrive more often and with more force. Diffusion follows the same logic, which is why a sugar cube dissolves faster in hot water than in cold. Keep Brownian motion and the thermal motion of molecules apart in an answer; swapping the two terms usually costs the mark.
温度与热能的区别
温度是分子平均动能的量度，热能与物体内全部分子的动能和势能之和有关，两者不能混用。一杯 80 °C 的水温度比一浴缸 40 °C 的水高，但浴缸里水的分子数量多得多，总热能反而更大。选择题常拿这个对比出题，问哪一份含有的热能更多，答案落在量大的那一份。写解释题时说清比较的是温度还是能量，只写水很热就下结论，通常拿不到分。计算题里把 ΔT 与 Q 分开处理也是同一个道理：温度变化量描述的是过程，Q 描述的是能量转移的多少。传热的方向由温度差决定，能量总是从高温处流向低温处，与两份物质各自含有多少热能无关，这一点常被用来解释为什么热水可以在冷水中放热而仍然很热。
Temperature measures the average kinetic energy of the molecules, while thermal energy depends on the total kinetic and potential energy of every molecule in the object, so the two are not interchangeable. A cup of water at 80 °C is hotter than a bath at 40 °C, yet the bath stores far more thermal energy because it holds so many more molecules; questions that ask which sample contains more thermal energy are really asking about quantity. Energy transfer always runs from the hotter object to the cooler one, whatever the totals involved, and answers that only mention how hot the water feels miss the point. The same separation appears in calculations, where the temperature change describes a process and Q describes the energy transferred.
2.2 温度计读数与热膨胀
温度计的题考三件事：量程、每一小格代表的值、以及读数本身。用液体温度计读数时视线要与液柱顶端齐平，温度计要留在液体里，拿出来再读会低几度。分度值决定读到哪一位，最小格是 0.5 °C 时读数写 21.5 °C，写成 21 °C 会被判读数不准。热膨胀的顺序是气体最多、液体其次、固体最少，同一次温度变化下这个排序不能写反。固体受热时体积变大，原因是粒子振动幅度变大、平均间距增大，粒子本身的体积没有变化，常见的错误是写成粒子变大了。这部分的应用题不少：双金属片在温控器里向膨胀较小的那一片弯曲，钢轨和桥梁留伸缩缝，架空电线在夏天明显下垂。液体膨胀不均匀，所以精密场合改用气体温度计或电阻温度计，读数的重复性更好。
Thermometer questions test three things: the range, the value of one small division, and the reading itself. Keep your eye level with the top of the column and leave the bulb in the liquid, because a thermometer taken out reads a few degrees low. Write the reading to the precision of the smallest division, so a 0.5 °C scale gives 21.5 °C rather than 21 °C. Gases expand most for a given temperature rise, liquids less, solids least. The particles themselves do not grow, they vibrate through a wider range, so the average spacing increases. Applications are worth knowing in detail: a bimetallic strip bends towards the metal with the smaller expansion, rails and bridges are built with gaps, and overhead cables sag on a hot afternoon. Liquid expansion is uneven enough that accurate work uses a gas or resistance thermometer instead.
2.2.2 比热容与 Q = mcΔT
比热容的定义要背准：单位质量的物质温度升高 1 摄氏度所需吸收的热量。计算公式 Q = mcΔT，其中 c 的单位是 J/(kg °C)，ΔT 只算温度变化量，和起始温度无关。CAIE 数据表给出水的比热容为 4200 J/(kg °C)，考试时直接用这个值，不要自己换成别的单位制。例题：把 2.0 kg 的水从 20 °C 加热到 100 °C，Q = 2.0 × 4200 × 80 = 672000 J，即 6.72 × 10^5 J。错得最多的是把 ΔT 直接写成终温 100，算出 840000 J，后面每一步都跟着错。功率已知时还能反推时间：同样这杯水需要 672000 J，用 1500 W 的加热器约 448 s。水的比热容偏大，所以沿海地区昼夜温差比内陆小，汽车发动机也用水做冷却液，而不是用升温快的油。反过来给热量和温差求 c 的题目也考过，步骤与正算完全一样，只是最后一步移项，单位别漏掉 J/(kg °C) 里的括号。
The definition has to be word perfect: the energy needed to raise the temperature of one kilogram of a substance by one degree Celsius. Work with Q = mcΔT, where c is in J/(kg °C) and ΔT is the temperature change only, so the starting temperature never enters the substitution. The CAIE data sheet gives water as 4200 J/(kg °C), which is the value to use. For 2.0 kg of water heated from 20 °C to 100 °C, Q = 2.0 × 4200 × 80 = 672000 J, or 6.72 × 10^5 J. Power turns the calculation into a timing question: a 1500 W heater delivers that energy in about 448 s. Water has an unusually high specific heat capacity, which is why coastal towns have smaller day-night temperature swings and why car engines are water cooled; the same equation rearranges to give c when the energy and the temperature change are supplied.
2.2.3 熔化沸腾与潜热
熔化与沸腾期间温度不变，但能量一直在进入物质，这些能量用来破坏粒子间的排列，而不是提高温度，所以叫潜热。计算用 Q = mL，冰的比熔化潜热约 3.34 × 10^5 J/kg。例题：把 0.50 kg 的 0 °C 冰全部化成 0 °C 的水，需要 Q = 0.50 × 3.34 × 10^5 = 1.67 × 10^5 J，注意这一过程温度始终是 0 °C。水的汽化潜热约 2.26 × 10^6 J/kg，比熔化潜热大一个数量级，因为沸腾要把粒子完全分开，这也是蒸汽烫伤比开水更严重的原因。蒸发只发生在液体表面，任何温度下都在进行，温度越高、表面积越大、空气流动越快，蒸发越快。出汗降温靠的就是蒸发从皮肤吸收热量。蒸发与沸腾的对比也是高频题：沸腾在沸点发生、液体内部和表面同时产生气泡，蒸发可以在任何温度下发生、只从表面带走分子。
Melting and boiling happen at a fixed temperature, yet energy keeps flowing in; it breaks the arrangement of the particles instead of raising the temperature, which is why the quantity is called latent heat. The equation is Q = mL. The specific latent heat of fusion of ice is about 3.34 × 10^5 J/kg, so melting 0.50 kg of ice at 0 °C takes 1.67 × 10^5 J, and the temperature stays at 0 °C throughout. Vaporisation is an order of magnitude larger at roughly 2.26 × 10^6 J/kg for water, since the particles must be pulled fully apart; that gap explains why steam burns are worse than boiling water. Evaporation happens at the surface at any temperature, and it speeds up with a higher temperature, a larger surface area and moving air. Boiling differs by bubbles forming throughout the liquid at the boiling point, while evaporation takes molecules from the surface alone.
升温曲线上的两个平台
加热冰的实验会画出一条温度随时间变化的曲线：升温、0 °C 平台、再升温、100 °C 平台。两个平台的斜率都是零，因为状态变化期间吸收的能量全部用于改变粒子排列。升温段的陡峭程度由比热容决定，水的升温段比冰的平缓，原因是水的比热容更大，同样的加热功率下每秒升温更少。阅卷时最常见的丢分是只写温度不变，没有交代能量去了哪里，解释分就没了。看图题还要比较两段平台的长度，0 °C 平台明显短于 100 °C 平台，对应冰的熔化潜热小于水的汽化潜热。实验里如果加热器功率不稳定，平台的形状会变形，但平台所在的温度不会变，这一条常被当成判断题来考。后一问往往要求算总能量，把升温段的 mcΔT 与平台段的 mL 分开算再相加，混成一个式子就再也算不对了。
Heating ice steadily gives a temperature-time graph with a rise, a flat section at 0 °C, a second rise and a second flat section at 100 °C. Neither plateau has a gradient, because during a change of state the energy supplied goes into rearranging particles rather than raising the temperature. The steepness of each rising section is set by specific heat capacity, so the water section is shallower than the ice section. The short plateau sits at 0 °C and the long one at 100 °C, matching the gap between the two latent heats. A drifting heater distorts the shape of the graph but not the temperatures at which the plateaus occur, and totals are found by adding mcΔT for each rising section to mL for each plateau.
2.3 传导对流与辐射
三种传递方式要按发生条件区分。传导靠粒子碰撞，金属里还有自由电子帮忙，所以金属是良导体，木头、塑料和封闭空气是绝热体，固体内部的热量基本只走传导这条路。对流只能发生在流体中，受热部分密度变小上升，周围较冷的部分下沉补充，形成循环，固体内部不会出现对流。辐射以红外线的形式传递，不需要任何介质，所以能在真空中进行，黑色粗糙表面吸收和发射辐射的能力最强，银色光滑表面最弱。一个方便的记法是：真空中唯一能传递热能的方式是辐射，太阳的能量正是这样到达地球的。保温瓶把三条路一起堵住：镀银的双层玻璃削弱辐射，夹层抽真空切断对流和传导，软木塞减少瓶口的传导。双层玻璃窗的保温思路和它一致，只是中间的空气层没有抽真空，效果差一些。
The three mechanisms are separated by the conditions they need. Conduction moves energy through particle collisions and, in metals, through free electrons, so metals conduct well while wood, plastic and trapped air do not; inside a solid it is the only route available. Convection needs a fluid that can flow, with warmer fluid rising because its density has fallen and cooler fluid sinking to take its place, which is impossible in a solid. Radiation travels as infrared, needs no medium at all and therefore crosses a vacuum, which is how energy from the Sun reaches the Earth; dull black surfaces absorb and emit best, shiny silver ones worst. A vacuum flask blocks all three routes at once. Double glazing works on the same idea with a plain air gap, so it insulates less well.
热学定义题的措辞与单位给分点
定义题按关键词给分。写比热容时必须出现单位质量和升高 1 摄氏度这两个要素，只写升高一度吸收的热量，就会丢掉质量那个关键词。写潜热要交代两件事：状态在变化，温度没有变。解释类小题先给定义，再补一句物理原因，例如金属导热好是因为内部有可以自由移动的电子，两句写全才拿得到两分。计算题的分落在公式、代入和最终值三处，最终值要带单位 J，通常保留两到三位有效数字，中间步骤不要提前四舍五入。Paper 2 的选择题常考数量级而不是精确算术，估算一杯水冷却释放的能量在 10^5 J 量级、熔化同质量冰只需要十分之一的能量，两个明显偏离的选项就能直接排除。这类题目的正确率取决于平时有没有把数值背到有感觉，而不是临场硬算。
Definition questions are marked on keywords, so precision pays. For specific heat capacity the answer must include one kilogram and one degree Celsius, or one kelvin; leave the mass out and a two-mark answer drops to one. For latent heat it has to state that the temperature stays constant while the state changes. Explanation items want the definition first and the physical reason second, since metals conduct well because their electrons are free to move. Calculations award marks for the equation, the substitution and the final value with its unit, usually to two or three significant figures. Multiple-choice items often test order of magnitude rather than arithmetic; knowing that cooling a cup of water releases around 10^5 J removes two options straight away.
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