IB物理 内能 比热容 气体定律 第一定律
热学在 IB 物理里最容易被当成背公式的一章, 也恰恰是 Paper 1B 数据处理最容易拿分的一章。这一篇按 2025 年首次考试的新大纲, 把 B.1 到 B.4 的公式、单位、典型错误和分值位置串在一起讲, 每个知识点后面跟一个能算到底的例子。
Thermal physics is often reduced to a list of formulas, yet it is also where data-handling marks in Paper 1B are easiest to secure. This article follows the syllabus first assessed in 2025 and links the equations, units, typical errors and mark locations from B.1 to B.4, with a worked example after each idea.
新版考纲里热学的位置与两卷分值
IB 物理从 2023 年 8 月开始教新大纲, 2025 年 5 月第一次考。热学不再是独立的 Thermal Physics 单元, 而是放进主题 B「物质的粒子性」。B.1 Thermal energy transfers 和 B.3 Gas laws 是 SL 与 HL 都考的内容, B.4 Thermodynamics 只出现在 HL 卷。同一主题下的 B.2 讲温室效应, B.5 讲电流与电路, 公式体系和热学不互通, 复习时不要混着记。分值上, Paper 1 占 36%, Paper 2 占 44%, 科学探究 IA 占 20%。SL 的 Paper 1 是 45 分 1 小时 30 分钟, HL 是 60 分 2 小时; SL 的 Paper 2 是 55 分 1 小时 30 分钟, HL 是 90 分 2 小时 30 分钟。这些数字来自 IBO 2025 版物理学科简章, 换考季前值得再核对自己那一年的版本。
Thermal physics sits inside Theme B of the current IB Physics course, first taught in August 2023 and first assessed in May 2025. Sections B.1 Thermal energy transfers and B.3 Gas laws are examined at both SL and HL, while B.4 Thermodynamics appears only on HL papers. The assessment split is 36 per cent for Paper 1, 44 per cent for Paper 2 and 20 per cent for the scientific investigation, with the SL papers carrying 45 and 55 marks against 60 and 90 at HL.
内能是总量, 温度是平均值
内能是系统里所有粒子随机动能与分子间势能的总和, 温度只反映平均随机动能。这个区别决定了 Paper 1A 里反复出现的一道选择题: 一池 20 摄氏度的水, 内能远大于一杯 80 摄氏度的茶, 因为粒子数量差了几个数量级。理想气体的分子间势能按模型取零, 所以它的内能只由温度决定, 等温过程中内能不变。很多学生以为温度高就等于内能大, 把两者当成一回事, 解释「冰在 0 摄氏度为什么还能吸热」时就写不出势能这一层。热平衡的判据是温度相同, 不是内能相同, 热流方向也只看温度差。
Internal energy is the total random kinetic energy plus the intermolecular potential energy of every particle, whereas temperature measures only the average random kinetic energy per particle; that difference explains a multiple-choice favourite, since a swimming pool at 20 degrees Celsius holds far more internal energy than a cup of tea at 80, simply because it contains vastly more particles. In the ideal-gas model the potential term is taken as zero, so internal energy depends on temperature alone and stays constant during an isothermal change; thermal equilibrium means equal temperature rather than equal internal energy, and heat always flows down the temperature gradient.
Q = mcΔT 与量热实验的三个误差源
升温所需能量写成 Q = mcΔT, c 是比热容, 单位 J kg^-1 K^-1, 水的 4.2 x 10^3 在 data booklet 里查得到。0.50 kg 水升高 20 K 需要 0.50 x 4200 x 20 = 42000 J, 同样的能量只够把 2.0 kg 铝加热约 23 K, 因为铝的 c 大约是 9.0 x 10^2。量热实验的误差源几乎年年考: 量热器本身吸热、装置向环境散热、温度计读到的是水温而不是样品温度。写 Method 时要把「加盖」「隔热层」「快速读数」一一对应到这三条, 而不是笼统写 reduce heat loss, 后者通常不给分。
The energy needed to raise a temperature is Q = mc delta T, with c in joules per kilogram per kelvin, and the data booklet gives 4.2 x 10^3 for water; heating 0.50 kg of water by 20 K therefore takes 42000 J, while the same energy lifts 2.0 kg of aluminium by only about 23 K because its specific heat capacity is roughly 9.0 x 10^2. Calorimetry questions usually hide three systematic problems: energy taken by the calorimeter itself, loss to the surroundings, and a thermometer reading the water rather than the sample; a good method section names the lid, the insulation and the fast reading against those three rather than offering a vague claim about heat loss.
相变平台与 Q = mL
温度不变时能量仍在进出, 这部分能量用来改变分子间势能, 对应的量是比潜热 Q = mL。水的熔化潜热约 3.3 x 10^5 J kg^-1, 汽化潜热约 2.3 x 10^6 J kg^-1, 两者相差近七倍。把 0.20 kg 的 0 摄氏度冰全部化成水需要 66000 J, 把这 0.20 kg 水从 0 加热到 100 摄氏度需要 84000 J, 再全部汽化却要 460000 J。加热曲线上的水平台就是相变区间, 平台上温度计不动, 不代表没有能量输入。常见的错误是把平台段的斜率也代进 mcΔT, 算出的 c 会大得离谱甚至为负。
During a change of state the temperature stays flat while energy keeps flowing, and that energy alters intermolecular potential energy rather than kinetic energy, so the quantity is the specific latent heat, Q = mL; water's latent heat of fusion is about 3.3 x 10^5 J kg^-1 and its latent heat of vaporisation about 2.3 x 10^6 J kg^-1, close to seven times larger. Melting 0.20 kg of ice at 0 degrees takes 66000 J, heating the same mass of water to 100 degrees takes 84000 J, and boiling it away takes 460000 J, which is why steam causes far worse burns than hot water; the flat step on a heating curve is the phase change itself, so a gradient measured across it gives a meaningless specific heat capacity.
传导速率 P = kAΔT/L 与双层玻璃
传导靠粒子碰撞, 金属里还多一条自由电子通道, 通过一块板的速率写成 P = kAΔT/L, k 是热导率, L 是厚度。若题目给出 k = 0.80 W m^-1 K^-1 的玻璃, 面积 1.5 m^2, 两侧温差 20 K, 厚度 4.0 mm, 代入得 0.80 x 1.5 x 20 / 0.004 = 6000 W。这个式子说明双层玻璃的关键不在玻璃本身, 而在中间那层密封空气: 空气的 k 比玻璃小一个数量级以上, 同样温差与厚度下传热速率大幅下降。回答「双层玻璃为什么保温」只写玻璃更厚是拿不到分的, 要点是空气是差导体, 而且被密封住无法形成对流。
Conduction passes energy through collisions between particles and, in metals, through free electrons; the rate through a slab is P = kA delta T / L, where k is the thermal conductivity and L the thickness. Taking glass with k = 0.80 W m^-1 K^-1, an area of 1.5 m^2, a 20 K difference and a 4.0 mm pane gives 0.80 x 1.5 x 20 / 0.004, or 6000 W. The point of double glazing is not thicker glass but the sealed air gap, since air conducts more than an order of magnitude less. An answer that only mentions thickness earns nothing; the mark is for a poor conductor that is trapped and therefore cannot convect.
对流只在流体里发生
对流只发生在流体里, 机制是受热部分密度变小上升, 冷的流体下沉补位, 形成循环。海风、暖气片上方的空气流动、水壶里的翻腾都属于同一套逻辑。固体里没有对流, 因为粒子被固定在晶格上, 这也是保温材料做成多孔泡沫的原因: 把空气关进小孔, 既压住传导也压住对流。很多学生把「热空气上升」当成一条独立规律背下来, 遇到冰箱冷冻室为何在顶部这类题就卡住, 其实回到密度差和浮力就能推。描述时要写出受热膨胀、密度减小、浮力大于重力、上升、冷流体补充这条链, 缺一环就少一个分点。
Convection happens only in fluids: the heated region expands, its density falls, buoyancy lifts it and cooler fluid sinks to replace it, setting up a loop; sea breezes, the air above a radiator and the rolling water in a kettle all follow that chain. Solids cannot convect because their particles are locked in a lattice, which is why insulation foam traps air in tiny pockets and blocks conduction and convection at once; students who memorise warm air rises as a standalone rule often freeze on questions about freezers or roof spaces, when a density argument would carry them through. The written answer needs the full sequence of expansion, lower density, net upward force and replacement by cooler fluid.
黑体辐射 斯特凡玻尔兹曼与维恩定律
所有物体都在辐射电磁波, 黑体是吸收率与发射率都取 1 的理想表面, 总辐射功率写成 P = eσAT^4, σ = 5.67 x 10^-8 W m^-2 K^-4。人体表面积 1.0 m^2, e 取 0.90, 皮肤 300 K 时 P 约为 0.90 x 5.67 x 10^-8 x 1.0 x 8.1 x 10^9, 即 413 W, 与代谢产热同一量级, 这也是房间偏冷时人会持续失热的原因。峰值波长由维恩定律给出, λmax T = 2.9 x 10^-3 m K, 300 K 对应 9.7 x 10^-6 m, 落在红外。T 的四次方意味着绝对温度翻倍, 辐射功率变成十六倍, 这是这类题最常被拿来做比较的一点。
Every object radiates, and a black body is the ideal surface that absorbs and emits with emissivity 1, giving a total power P = e sigma A T^4 with sigma equal to 5.67 x 10^-8 W m^-2 K^-4. For 1.0 m^2 of skin at 300 K with emissivity 0.90 the result is about 413 W, the same order as metabolic output, which is why a cold room drains heat continuously. Wien's law, lambda max times T equals 2.9 x 10^-3 m K, puts the peak at 9.7 x 10^-6 m at that temperature, firmly in the infrared. The fourth power is the comparison examiners like: double the absolute temperature and the radiated power rises sixteenfold.
pV = nRT 与必须换成开尔文
理想气体的状态方程是 pV = nRT, R = 8.31 J mol^-1 K^-1, n 由质量除以摩尔质量得到。2.0 mol 气体在 300 K 占 0.050 m^3 时, 压强为 2.0 x 8.31 x 300 / 0.050, 约 99700 Pa, 接近一个大气压。常见的错误是温度直接代摄氏度, 27 度必须写成 300 K, 否则算出的压强会小到不合物理。三条经验定律都是这个式子的特例: 等温时 pV 不变, 等压时 V/T 不变, 等容时 p/T 不变。写答案时先列出已知量并统一到 SI 单位, 再变形代入, 比直接套数字少丢分。
The equation of state for an ideal gas is pV = nRT with R equal to 8.31 J mol^-1 K^-1, and n found from mass divided by molar mass. Two moles at 300 K occupying 0.050 m^3 give a pressure of 2.0 x 8.31 x 300 / 0.050, about 99700 Pa, close to one atmosphere. Substituting degrees Celsius is the classic slip: 27 degrees must enter as 300 K, otherwise the pressure comes out nonsensically small. The three empirical gas laws are special cases, with pV fixed at constant temperature, V over T fixed at constant pressure and p over T fixed at constant volume. Listing the known quantities in SI units before rearranging saves more marks than any shortcut.
分子动理论 压强与 3/2 kT
分子动理论把压强解释成大量分子撞击容器壁的宏观结果, 模型假设分子体积可忽略、除碰撞外无相互作用、碰撞完全弹性。由此得到 pV = (1/3) N m c^2 的平均值形式, 以及每个分子的平均动能 (3/2) kT, k = 1.38 x 10^-23 J K^-1。300 K 时平均动能约 6.2 x 10^-21 J; 氮气分子质量 4.65 x 10^-26 kg, 对应的方均根速率约 520 m/s。温度翻倍时平均动能也翻倍, 速率却只乘根号二, 这一点在比较不同气体时经常写错。回答「压强为什么随温度升高」要同时说清撞击频率变高与每次撞击的动量变化变大, 只写分子运动更快通常只给一半分。
Kinetic theory turns pressure into the macroscopic result of countless collisions with the walls, under assumptions of negligible molecular volume, no forces except during contact and perfectly elastic collisions. The model yields pV as one third of N times m times the mean square speed, and an average kinetic energy per molecule of (3/2) kT with k equal to 1.38 x 10^-23 J K^-1. At 300 K that energy is about 6.2 x 10^-21 J, and for nitrogen molecules of 4.65 x 10^-26 kg the root mean square speed is roughly 520 m/s. Doubling the temperature doubles the average kinetic energy but multiplies the speed by only the square root of two. Explaining a pressure rise needs both a higher collision rate and a larger momentum change per collision, since faster molecules alone is a half answer.
热力学第一定律的符号约定
HL 的 B.4 用 Q = ΔU + W 表述能量守恒, 约定 W 是气体对外做的功, 因此膨胀时 W 为正。体积不变时 W 为零, 无论压强怎么变; 等压过程 W = pΔV; 等温过程中理想气体 ΔU = 0, 吸收的热量全部转成对外做功。算一笔完整的账: 气体吸热 800 J, 同时对外做功 300 J, 则 ΔU = 500 J, 对 2.0 mol 单原子气体意味着升温约 20 K, 因为 ΔU = (3/2) nRΔT。p-V 图上过程曲线下的面积就是功, 闭合循环围出的面积是净功, 方向决定正负。符号约定每年都有人弄反, 动笔前在草稿角上写一行「W 是气体对外做功」, 比事后检查省时间。
Higher Level section B.4 writes conservation of energy as Q = delta U + W, with W defined as work done by the gas, so expansion counts positive. At constant volume W is zero however the pressure changes, at constant pressure W equals p delta V, and for an isothermal change of an ideal gas delta U is zero so all the heat becomes work. A full example: 800 J of heat in and 300 J of work out leave 500 J of internal energy, which for 2.0 mol of monatomic gas means a rise of about 20 K through delta U = (3/2) nR delta T. On a p-V diagram the area under the curve is the work, the enclosed area of a cycle is the net work, and the direction of travel fixes the sign. Write the convention down before starting, because reversing it is the most expensive slip in this section.
熵与第二定律 为什么热不会倒流
第二定律给过程规定方向, 孤立系统的熵不会减少, 写成 ΔS 大于等于零。可逆等温过程中熵变用 ΔS = Q/T 计算, 800 J 的热量在 300 K 下传入, 熵增约 2.7 J K^-1。这条限制直接压住热机效率: 任何循环热机都必须向冷库排出部分热量, 效率不可能达到百分之百, 理论上限由卡诺循环给出。解释热量为何不自发从冷物体流向热物体, 用熵增比写「不可能」三个字更有说服力。很多学生把熵当成混乱度的形容词, 解释题里拿不出可计算的量, 建议始终落到 ΔS 与一条可逆路径上。
The second law gives thermal processes a direction: the entropy of an isolated system never decreases, so delta S is greater than or equal to zero; for a reversible isothermal transfer the change is Q over T, and 800 J entering at 300 K raises entropy by about 2.7 J K^-1. That restriction caps heat-engine efficiency, since every cycle must reject some heat to a cold reservoir and no engine reaches one hundred per cent, with the Carnot cycle setting the ceiling. Explaining why heat does not flow spontaneously from cold to hot is far stronger with an entropy argument than with a flat statement that it is impossible; treating entropy as a vague synonym for disorder leaves explanation answers without a quantity, so keep the working on delta S and a reversible path.
数据处理与三周复习顺序
数据处理的分值集中在 Paper 1B 和 IA, 常考有效数字、绝对与百分比不确定度、以及从图像斜率反推物理量。作图时两轴都要有量和单位, 斜率的单位由纵轴除以横轴得到, 这一步写错会连带后面全错。三周可以这样排: 第一周把 B.1 的四个公式和三种传热机制连成一条线, 第二周攻 B.3 的气体定律与分子动理论, 第三周按 HL 补 B.4 并做两套整卷。每次做完题, 把用错的单位记在同一张纸上, 开尔文、立方米、帕斯卡这三样失分最多。考前一周只看这张纸和自己的错题, 比再刷十套新题有用。
Marks for data handling cluster in Paper 1B and the internal assessment, mostly significant figures, absolute and percentage uncertainty, and a gradient converted into a physical quantity; both axes need quantity and unit, and the gradient carries the unit of the vertical axis divided by that of the horizontal one, so an error there poisons everything that follows. A three-week plan can run from the four equations and three transfer mechanisms of B.1, through the gas laws and kinetic model of B.3, to the HL thermodynamics of B.4 plus two full papers. Keep a single sheet of unit mistakes, since kelvin, cubic metres and pascals account for most lost marks; in the final week, reread that sheet and your own errors instead of starting ten new papers.
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