High School Physics

Thermodynamics and Heat热力学与热

Thermodynamics is the study of thermal energy, heat transfer, and the limits of converting heat into useful work. This guide builds from the microscopic picture of temperature through the three modes of heat transfer (conduction, convection, radiation), the calorimetry equation $Q = mc\Delta T$, phase changes and latent heat $Q = mL$, the ideal gas law $PV = nRT$, the four laws of thermodynamics, and finally heat-engine efficiency $\eta = W/Q_H$. Worked examples use real numbers throughout. Note: Alberta Physics 20/30 does not include a dedicated thermodynamics unit.热力学(thermodynamics,热力学)研究热能、热传递以及将热转化为有用功的极限。本指南从温度(temperature,温度)的微观图像出发,历经三种热传递模式(传导、对流、辐射)、量热方程 $Q = mc\Delta T$、相变与潜热 $Q = mL$、理想气体定律 $PV = nRT$、热力学四大定律,直至热机效率(efficiency,效率)$\eta = W/Q_H$。全部例题均用真实数字演算。注:阿尔伯塔 Physics 20/30 中不设专门的热力学单元。

7 sections7 节内容 US NGSS · ON · BC (AB: gap noted)US NGSS · ON · BC(AB:缺口已注) KaTeX math: $Q=mc\Delta T$, $PV=nRT$KaTeX 数学:$Q=mc\Delta T$,$PV=nRT$

How to use this guide如何使用本指南

Thermodynamics is strong in three of the four curricula mapped here: NGSS anchors it with HS-PS3-4 (heat transfer and the second law) and HS-PS3-1 (thermal energy bookkeeping); Ontario SPH3U folds thermal energy and heat into Strand D alongside work and energy; BC Physics 11 explicitly names "thermal equilibrium and specific heat capacity" as Content. Alberta Physics 20/30 is the exception: heat is treated at the Science 10 level and there is no dedicated thermodynamics unit in the provincial physics program. AB students should still work through this guide for AP/IB readiness, but should not expect Alberta-specific exam questions at this level. The table below maps each curriculum to the sections most relevant to it.热力学在本指南对照的四套大纲中,三套有扎实内容:NGSS 以 HS-PS3-4(热传递与第二定律)和 HS-PS3-1(热能核算)为锚;安大略 SPH3U 将热能与热嵌入 D 单元,与功、能量并列;BC Physics 11 明确把"热平衡与比热容"列为内容。阿尔伯塔 Physics 20/30 是例外:热在 Science 10 层面涉及,省级物理项目中无专门的热力学单元。AB 学生仍应通读本指南以备 AP/IB,但不必期待省级考试在此层面出题。下表将各课纲对应到最相关的节。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 §1, §2, §3, §4, §6 — the HS-PS3-4 investigation core (heat transfer, thermal equilibrium, calorimetry) and HS-PS3-1 (energy bookkeeping in a system)§1、§2、§3、§4、§6 —— HS-PS3-4 探究核心(热传递、热平衡、量热法)与 HS-PS3-1(系统能量核算) §7 (heat engines / Carnot): NGSS keeps efficiency qualitative (HS-PS3-3 mentions "efficiency" as a constraint for energy-conversion devices, but does not require quantitative $\eta = W/Q_H$)§7(热机/卡诺循环):NGSS 仅定性处理效率(HS-PS3-3 提到"效率"作为约束,但不要求定量 $\eta = W/Q_H$) ngss_hs_ps_extract.md — HS-PS3-4 + HS-PS3-1 PE verbatim— HS-PS3-4 + HS-PS3-1 表现期望原文
🇨🇦 ON Grade 11 — SPH3U安大略 11 年级 — SPH3U §1 through §6 in full (Strand D covers work, energy, power, and thermal energy/heat). §7 at a conceptual level for D3.12 (nuclear plant thermodynamics)§1 至 §6 完整学习(D 单元涵盖功、能量、功率及热能/热)。§7 概念层面适用于 D3.12(核电站热力学) Quantitative heat-engine cycle calculations (§7 going-deeper): Ontario D3 does not require Carnot-cycle algebra定量热机循环计算(§7 深入):安大略 D3 不要求卡诺循环代数 science_11-12_physics_extract.md — SPH3U Strand D, D3.1, D3.12— SPH3U D 单元,D3.1,D3.12
🇨🇦 BC Grade 11 — Physics 11BC 11 年级 — Physics 11 §1, §2, §3, §4 are core (thermal equilibrium and specific heat capacity, "application of law of conservation of energy"). §5 gas laws and §6 first law are reinforcing§1、§2、§3、§4 为核心(热平衡与比热容,"能量守恒定律的应用")。§5 气体定律与 §6 第一定律起强化作用 §7 (heat engines / efficiency cycles): BC Physics 11 mentions "power and efficiency" generally but does not require the full Carnot treatment§7(热机/效率循环):BC Physics 11 泛泛提及"功率与效率",不要求完整卡诺处理 physics_11-12_extract.md — Physics 11 Content: thermal equilibrium, specific heat, power/efficiency— Physics 11 内容:热平衡、比热容、功率/效率
🇨🇦 AB — Physics 20/30阿尔伯塔 — Physics 20/30 No provincial expectation at Physics 20/30 level. Work through all 7 sections for AP/IB preparation and conceptual fluency省级 Physics 20/30 层面无要求。为 AP/IB 准备和概念流利度,通读全部 7 节 No sections are provinicial exam content for AB Physics 20/30; treat this as enrichment toward AP/IB本指南各节均非 AB Physics 20/30 省考内容;作为 AP/IB 拓展内容对待 physics_20-30_extract.md — confirmed: no thermodynamics unit in Physics 20 or 30— 已确认:Physics 20 或 30 中无热力学单元

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后,用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise three formulas: $Q = mc\Delta T$ (sensible heat), $Q = mL$ (latent heat), and $PV = nRT$ (ideal gas). Know the three heat-transfer modes (conduction, convection, radiation) by name and mechanism. State the first law ($\Delta U = Q - W$) and the key second-law consequence: heat flows spontaneously from hot to cold, never the reverse. Read every cram-cheat box; skip the going-deeper derivations.背熟三个公式:$Q = mc\Delta T$(显热)、$Q = mL$(潜热)、$PV = nRT$(理想气体)。按名称与机制记住三种热传递模式(传导、对流、辐射)。陈述第一定律($\Delta U = Q - W$)及第二定律的关键推论:热自发从热处流向冷处,反之不然。读每个速记框,跳过深入推导。

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If you are going for the top mark如果你目标顶分

Precision matters: $Q$ is energy transferred as heat; it is not the same as internal energy $U$ or temperature $T$. For calorimetry problems, always set heat lost by the hot object equal to heat gained by the cold one, with careful signs. For gas-law problems, convert temperature to kelvin before substituting. For thermodynamics laws, distinguish the zeroth (thermal equilibrium), first (energy conservation), and second (entropy / direction of heat flow) laws clearly. Work through the efficiency derivation in §7 until you understand why $\eta < 1$ is unavoidable.精确性至关重要:$Q$ 是以热形式传递的能量;它不等于内能 $U$ 或温度 $T$。对量热问题,始终令热物体失热等于冷物体得热,符号须仔细。对气体定律问题,代入前先将温度换算为开尔文。对热力学定律,清晰区分第零定律(热平衡)、第一定律(能量守恒)与第二定律(熵/热流方向)。在 §7 中推导效率,直到理解为何 $\eta < 1$ 是不可避免的。


Temperature and Thermal Energy温度与热能

Three concepts, three distinctions.三个概念,三组区分。
  • Temperature $T$温度 $T$ — a measure of the average random kinetic energy of the particles in a substance. SI unit: kelvin (K); $T_K = T_C + 273.15$.— 物质中粒子平均随机动能的量度。国际单位:开尔文(K);$T_K = T_C + 273.15$。
  • Thermal energy $U$热能 $U$ — the total internal kinetic and potential energy of all particles in a substance. A larger object at the same temperature has more thermal energy than a smaller one.— 物质中所有粒子内部动能与势能的总和。同温度下,较大的物体比较小的物体含有更多热能。
  • Heat $Q$热量 $Q$ — energy transferred between objects because of a temperature difference. Heat is not a property an object possesses; it is energy in transit. SI unit: joule (J).— 因温差而在物体之间转移的能量。热不是物体的属性;它是传递中的能量。国际单位:焦耳(J)。
$$ T_K = T_C + 273.15 \qquad (0\,^\circ\text{C} = 273.15\,\text{K}) $$ NGSS HS-PS3-1 emphasises thermal energy as one of the energy forms in a system; HS-PS3-4 centres on the thermal-equilibrium outcome of heat transfer. Ontario D3 and BC Physics 11 both require distinguishing heat from temperature.NGSS HS-PS3-1 将热能列为系统中的能量形式之一;HS-PS3-4 以热传递的热平衡结果为核心。安大略 D3 与 BC Physics 11 都要求区分热与温度。
Worked Example 1 · Kelvin conversion例题 1 · 开尔文换算

Normal body temperature is $37.0\ ^\circ\text{C}$ and the temperature of liquid nitrogen is $-196\ ^\circ\text{C}$. Convert both to kelvin and find the difference.正常体温为 $37.0\ ^\circ\text{C}$,液氮温度为 $-196\ ^\circ\text{C}$。将两者换算为开尔文并求差值。

Add 273.15 to each Celsius value.各摄氏值加 273.15。

$$ T_{\text{body}} = 37.0 + 273.15 = 310.15\,\text{K} \approx 310\,\text{K}. $$ $$ T_{\text{N}_2} = -196 + 273.15 = 77.15\,\text{K} \approx 77\,\text{K}. $$

Temperature difference.温差。 $\Delta T = 310 - 77 = 233\,\text{K}$. Note: a temperature difference of $1\,\text{K}$ equals a difference of $1\,^\circ\text{C}$, so the $233\,\text{K}$ gap also equals $233\,^\circ\text{C}$.$\Delta T = 310 - 77 = 233\,\text{K}$。注意:$1\,\text{K}$ 的温度等于 $1\,^\circ\text{C}$ 的差,故 $233\,\text{K}$ 的温差也等于 $233\,^\circ\text{C}$。

Which statement correctly distinguishes temperature from thermal energy?哪句话正确区分了温度与热能?
§1 · Q1
Temperature and thermal energy are different names for the same quantity温度与热能是同一量的不同名称
Temperature measures average particle KE; thermal energy is the total internal energy of all particles温度衡量粒子平均动能;热能是所有粒子内能的总和
Thermal energy is the same as heat; both describe energy in transit热能与热量相同;两者都描述传递中的能量
A hotter object always has more thermal energy than a cooler one较热的物体总比较冷的物体含有更多热能
Temperature is the average random KE per particle. Thermal energy is the total of all particles' internal energies — a bathtub of warm water has more thermal energy than a cup of boiling water, even though the cup is at a higher temperature.温度是每个粒子平均随机动能。热能是所有粒子内能的总和 — 一浴缸温水含有的热能比一杯沸水更多,尽管那杯水温度更高。
Temperature (average KE per particle) and thermal energy (total internal energy) are related but distinct. A larger mass at the same temperature has more thermal energy. Heat is energy in transit, not a stored quantity.温度(每粒子平均动能)与热能(总内能)相关但不同。同温度下,质量较大的物体含有更多热能。热是传递中的能量,而非储存量。
What is $-40\ ^\circ\text{C}$ in kelvin?$-40\ ^\circ\text{C}$ 对应多少开尔文?
§1 · Q2
$-313\,\text{K}$
$40\,\text{K}$
$233\,\text{K}$
$313\,\text{K}$
$T_K = T_C + 273.15 = -40 + 273 = 233\,\text{K}$. (Incidentally, $-40$ is the one temperature where Celsius and Fahrenheit agree.)$T_K = T_C + 273.15 = -40 + 273 = 233\,\text{K}$。(顺带一提,$-40$ 是摄氏与华氏唯一相等的温度。)
Add $273$ (or $273.15$) to convert $^\circ$C to K. Kelvin can never be negative.将摄氏加 $273$(或 $273.15$)换算为开尔文。开尔文永远不为负。
Going deeper — absolute zero and the kinetic theory basis of temperature深入 — 绝对零度与温度的动理论基础

The kinetic theory of gases identifies temperature with the average translational kinetic energy of gas molecules: $\bar{KE} = \tfrac{3}{2} k_B T$, where $k_B = 1.38 \times 10^{-23}\ \text{J/K}$ is Boltzmann's constant. At absolute zero ($T = 0\,\text{K}$) this average kinetic energy reaches its minimum; quantum mechanics forbids it from reaching exactly zero (zero-point energy), but classically $T = 0$ is the theoretical lower bound. The kelvin scale is defined so that $T \ge 0$ always. This is why gas-law calculations must use kelvin: the ideal gas law $PV = nRT$ is derived assuming $T$ is the absolute temperature; plugging in Celsius gives nonsense when $T_C < 0$.气体动理论将温度认同为气体分子平均平动动能:$\bar{KE} = \tfrac{3}{2} k_B T$,其中 $k_B = 1.38 \times 10^{-23}\ \text{J/K}$ 是玻尔兹曼常数。在绝对零度($T = 0\,\text{K}$),此平均动能达到最小值;量子力学禁止其精确为零(零点能),但经典上 $T = 0$ 是理论下限。开尔文量程正是为使 $T \ge 0$ 恒成立而定义。这就是为什么气体定律计算必须使用开尔文:理想气体定律 $PV = nRT$ 的推导假设 $T$ 是绝对温度;当 $T_C < 0$ 时代入摄氏温度会得到无意义结果。


Heat Transfer: Conduction, Convection, and Radiation热传递:传导、对流与辐射

Three mechanisms — know the medium each requires.三种机制 — 记住各自所需的介质。
  • Conduction (传导)传导 — direct particle-to-particle energy transfer through a solid (or still fluid). Requires contact. Metals conduct well (free electrons); wood and air conduct poorly.— 通过固体(或静止流体)直接进行的粒子间能量传递。需要接触。金属导热良好(自由电子);木材和空气导热差。
  • Convection (对流)对流 — heat transfer by bulk movement of a fluid (liquid or gas). Hot fluid rises, cool fluid sinks, forming a convection current. Requires a fluid; cannot occur in a solid or vacuum.— 通过流体(液体或气体)整体运动传递热量。热流体上升,冷流体下沉,形成对流循环。需要流体;不能在固体或真空中发生。
  • Radiation (辐射)辐射 — energy transfer by electromagnetic waves (infrared, visible, etc.). Does not require a medium; operates in vacuum. All objects above $0\,\text{K}$ emit thermal radiation; hotter objects emit more and at shorter wavelengths (Wien's law).— 通过电磁波(红外线、可见光等)传递能量。需要介质;可在真空中进行。高于 $0\,\text{K}$ 的所有物体都发射热辐射;温度越高的物体发射量越多,波长越短(维恩定律)。
NGSS HS-PS3-4 asks you to investigate heat transfer as producing a "more uniform energy distribution" — thermal equilibrium. Ontario D3.1 names "a variety of energy transfers and transformations." BC Physics 11 elaborates "thermal equilibrium" as an application of conservation of energy.NGSS HS-PS3-4 要求你探究热传递如何产生"更均匀的能量分布" — 即热平衡。安大略 D3.1 点名"多种能量转移与转化"。BC Physics 11 将"热平衡"细化为能量守恒的应用。
Worked Example 2 · Identifying heat-transfer modes例题 2 · 识别热传递模式

For each situation, identify the dominant mode of heat transfer and justify why: (a) a metal spoon left in hot soup gets hot; (b) a room heater warms air that rises to the ceiling; (c) the Sun warms Earth's surface.对每种情况,识别主要热传递模式并说明理由:(a) 置于热汤中的金属勺变热;(b) 室内暖气加热空气,空气上升至天花板;(c) 太阳温暖地球表面。

(a) Conduction.(a) 传导。 The spoon is in direct contact with the soup; energetic liquid molecules collide with spoon atoms, transferring energy atom-to-atom up the handle. Metal's free electrons accelerate the process.勺子与热汤直接接触;高能液体分子与勺子原子碰撞,能量沿勺柄逐原子传递。金属的自由电子加速了这一过程。

(b) Convection.(b) 对流。 Air heated by the radiator expands, becomes less dense, and rises. Cooler denser air from above sinks to replace it, creating a circulation loop that distributes heat throughout the room.被暖气加热的空气膨胀,密度变小,上升。上方较冷较密的空气下沉填补,形成在整个房间中分配热量的循环流。

(c) Radiation.(c) 辐射。 The Sun is separated from Earth by $\approx 150 \times 10^6$ km of near-vacuum. Neither conduction (no direct contact) nor convection (no fluid) can operate across that gap — only electromagnetic radiation (primarily infrared and visible light) can traverse it.太阳与地球之间相隔约 $150 \times 10^6$ km 的近真空。传导(无直接接触)与对流(无流体)都无法跨越这段距离 — 只有电磁辐射(主要是红外线和可见光)才能穿越。

Which mode of heat transfer can operate through a vacuum?哪种热传递模式可以在真空中进行?
§2 · Q1
Conduction only仅传导
Radiation only仅辐射
Convection only仅对流
Both conduction and convection传导与对流皆可
Radiation travels as electromagnetic waves and requires no medium. Conduction needs direct particle contact; convection needs a fluid. Only radiation works in a vacuum (e.g., solar energy reaching Earth).辐射以电磁波传播,不需要介质。传导需要粒子直接接触;对流需要流体。只有辐射在真空中有效(例如太阳能到达地球)。
Conduction requires contact between particles; convection requires a fluid. Neither works in a vacuum. Only electromagnetic radiation can travel through empty space.传导需要粒子间接触;对流需要流体。两者在真空中均无效。只有电磁辐射能穿越真空。
Hot soup cools in a bowl. The warmer liquid near the sides rises and cooler soup sinks in the middle. This is an example of:热汤在碗中冷却。靠近碗壁的较热液体上升,中间较冷的汤下沉。这是以下哪种热传递的例子?
§2 · Q2
Conduction传导
Radiation辐射
Convection对流
Conduction and radiation equally传导与辐射各半
Bulk fluid movement driven by density differences from temperature gradients is convection. The hot-rises-cool-sinks circulation is the defining feature.由温度梯度引起的密度差驱动的流体整体运动是对流。热升冷降的循环是其定义特征。
When fluid moves in bulk due to temperature-driven density differences, the mechanism is convection. Conduction is particle-to-particle; radiation is via electromagnetic waves.当流体因温度驱动的密度差整体运动时,机制是对流。传导是粒子间传递;辐射通过电磁波。
Going deeper — Fourier's law of conduction and Stefan-Boltzmann radiation深入 — 傅里叶导热定律与斯特藩-玻尔兹曼辐射定律

Fourier's law (conduction rate): the rate of heat flow through a slab of material is $\dot{Q} = kA\Delta T / d$, where $k$ is thermal conductivity (W m$^{-1}$ K$^{-1}$), $A$ is cross-sectional area, $\Delta T$ is the temperature difference across thickness $d$. Insulators have small $k$ (e.g. air: $0.025$; wood: $\approx 0.1$); metals have large $k$ (e.g. copper: $400$). This is the physics behind double-pane windows and wall insulation R-values.傅里叶定律(导热速率):通过材料平板的热流速率为 $\dot{Q} = kA\Delta T / d$,其中 $k$ 是导热系数(W m$^{-1}$ K$^{-1}$),$A$ 是截面积,$\Delta T$ 是厚度 $d$ 两侧的温差。绝热材料 $k$ 小(如空气:$0.025$;木材:$\approx 0.1$);金属 $k$ 大(如铜:$400$)。这是双层窗户和墙体隔热 R 值背后的物理。

Stefan-Boltzmann law (radiation power): a perfect blackbody emits power $P = \sigma A T^4$, where $\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ is the Stefan-Boltzmann constant and $T$ is in kelvin. The $T^4$ dependence means doubling absolute temperature increases radiated power 16-fold. Net power exchanged between an object at $T$ and surroundings at $T_0$ is $P_{\text{net}} = \sigma A (T^4 - T_0^4)$. This drives everything from incandescent bulbs to infrared cameras to Earth's energy balance.斯特藩-玻尔兹曼定律(辐射功率):理想黑体辐射功率为 $P = \sigma A T^4$,其中 $\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{ K}^{-4}$ 是斯特藩-玻尔兹曼常数,$T$ 为开尔文。$T^4$ 的依赖关系意味着绝对温度加倍,辐射功率增大 16 倍。物体(温度 $T$)与环境(温度 $T_0$)之间的净交换功率为 $P_{\text{net}} = \sigma A (T^4 - T_0^4)$。这驱动了从白炽灯到红外摄像机再到地球能量平衡的一切。


Specific Heat and Calorimetry比热容与量热法

The central equation of this unit.本单元的核心方程。 $$ Q = mc\Delta T $$
  • $Q$ = heat transferred (J); positive if absorbed, negative if released.= 传递的热量(J);吸热为正,放热为负。
  • $m$ = mass (kg).= 质量(kg)。
  • $c$ = specific heat capacity (J kg$^{-1}$ K$^{-1}$ or J kg$^{-1}$ $^\circ$C$^{-1}$); how much energy is needed to raise $1\,\text{kg}$ of the substance by $1\,\text{K}$. Water: $4186\ \text{J kg}^{-1}\text{K}^{-1}$; aluminium: $900$; copper: $385$.= 比热容(J kg$^{-1}$ K$^{-1}$ 或 J kg$^{-1}$ $^\circ$C$^{-1}$);将 $1\,\text{kg}$ 物质升温 $1\,\text{K}$ 所需的能量。水:$4186\ \text{J kg}^{-1}\text{K}^{-1}$;铝:$900$;铜:$385$。
  • $\Delta T$ = $T_f - T_i$ (K or $^\circ$C; the same magnitude).= $T_f - T_i$(K 或 $^\circ$C;数值相同)。
Calorimetry: in a closed system, heat lost by the hot object equals heat gained by the cold one: $|Q_{\text{lost}}| = |Q_{\text{gained}}|$, i.e. $m_1 c_1 |\Delta T_1| = m_2 c_2 |\Delta T_2|$. This is conservation of energy. BC Physics 11 explicitly calls thermal-equilibrium problems "an application of law of conservation of energy (e.g., calorimeter)." NGSS HS-PS3-1 similarly asks you to "calculate the change in the energy of one component" using system energy bookkeeping.量热法:在封闭系统中,热物体失去的热量等于冷物体获得的热量:$|Q_{\text{失}}| = |Q_{\text{得}}|$,即 $m_1 c_1 |\Delta T_1| = m_2 c_2 |\Delta T_2|$。这是能量守恒。BC Physics 11 明确把热平衡问题称为"能量守恒定律的应用(如量热计)"。NGSS HS-PS3-1 同样要求你用系统能量核算"计算某一成分的能量变化"。
Worked Example 3 · Mixing hot and cold water例题 3 · 冷热水混合

$0.30\,\text{kg}$ of water at $85\,^\circ\text{C}$ is poured into $0.50\,\text{kg}$ of water at $15\,^\circ\text{C}$. Assuming no heat loss to the container, find the final equilibrium temperature. (Use $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{ K}^{-1}$.)$0.30\,\text{kg}$ 的 $85\,^\circ\text{C}$ 热水倒入 $0.50\,\text{kg}$ 的 $15\,^\circ\text{C}$ 冷水。假设无热量散失至容器,求最终平衡温度。(取 $c_{\text{水}} = 4186\ \text{J kg}^{-1}\text{ K}^{-1}$。)

Set heat lost $=$ heat gained.令失热 $=$ 得热。 Let $T_f$ be the final temperature.设最终温度为 $T_f$。

$$ m_{\text{hot}} c (T_{\text{hot}} - T_f) = m_{\text{cold}} c (T_f - T_{\text{cold}}). $$

Since both are water, $c$ cancels:两者均为水,$c$ 约去:

$$ 0.30(85 - T_f) = 0.50(T_f - 15). $$

Expand and solve.展开求解。

$$ 25.5 - 0.30 T_f = 0.50 T_f - 7.5 \;\Longrightarrow\; 33 = 0.80 T_f \;\Longrightarrow\; T_f = 41.25\,^\circ\text{C} \approx 41\,^\circ\text{C}. $$

Sanity-check.合理性核验。 $41\,^\circ\text{C}$ lies between $15$ and $85\,^\circ\text{C}$, as it must. It is closer to $15\,^\circ\text{C}$ because there is more cold water ($0.50\,\text{kg}$ vs $0.30\,\text{kg}$). ✓$41\,^\circ\text{C}$ 介于 $15$ 与 $85\,^\circ\text{C}$ 之间,这是必然的。它更靠近 $15\,^\circ\text{C}$,因为冷水更多($0.50\,\text{kg}$ 对 $0.30\,\text{kg}$)。✓

How much heat is required to raise $2.0\,\text{kg}$ of water from $20\,^\circ\text{C}$ to $70\,^\circ\text{C}$? ($c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$)将 $2.0\,\text{kg}$ 的水从 $20\,^\circ\text{C}$ 升至 $70\,^\circ\text{C}$ 需要多少热量?($c_{\text{水}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$)
§3 · Q1
$84\,\text{kJ}$
$167\,\text{kJ}$
$293\,\text{kJ}$
$419\,\text{kJ}$
$Q = mc\Delta T = 2.0 \times 4186 \times (70 - 20) = 2.0 \times 4186 \times 50 = 418\,600\,\text{J} \approx 419\,\text{kJ}$.$Q = mc\Delta T = 2.0 \times 4186 \times (70 - 20) = 2.0 \times 4186 \times 50 = 418\,600\,\text{J} \approx 419\,\text{kJ}$。
$Q = mc\Delta T$. Here $\Delta T = 70 - 20 = 50\,\text{K}$, $m = 2.0\,\text{kg}$, $c = 4186\ \text{J kg}^{-1}\text{K}^{-1}$. Multiply all three.$Q = mc\Delta T$。此处 $\Delta T = 70 - 20 = 50\,\text{K}$,$m = 2.0\,\text{kg}$,$c = 4186\ \text{J kg}^{-1}\text{K}^{-1}$。三者相乘。
A $0.50\,\text{kg}$ copper block ($c = 385\ \text{J kg}^{-1}\text{K}^{-1}$) at $200\,^\circ\text{C}$ is dropped into $1.0\,\text{kg}$ of water ($c = 4186\ \text{J kg}^{-1}\text{K}^{-1}$) at $20\,^\circ\text{C}$. What happens to the final temperature compared to the water's initial temperature?一块 $0.50\,\text{kg}$ 的铜块($c = 385\ \text{J kg}^{-1}\text{K}^{-1}$)在 $200\,^\circ\text{C}$ 时投入 $1.0\,\text{kg}$、$20\,^\circ\text{C}$ 的水中($c = 4186\ \text{J kg}^{-1}\text{K}^{-1}$)。与水的初温相比,最终温度如何变化?
§3 · Q2
The final temperature is slightly above $20\,^\circ\text{C}$ (water gains a little heat)最终温度略高于 $20\,^\circ\text{C}$(水获得少量热量)
The final temperature is exactly $110\,^\circ\text{C}$ (midpoint)最终温度恰好为 $110\,^\circ\text{C}$(中点)
The final temperature equals $200\,^\circ\text{C}$ (copper dominates)最终温度等于 $200\,^\circ\text{C}$(铜主导)
The final temperature is below $20\,^\circ\text{C}$最终温度低于 $20\,^\circ\text{C}$
Water's large $c$ and large mass mean it can absorb the copper's heat with only a small temperature rise. Heat capacity of water: $1.0 \times 4186 = 4186\ \text{J/K}$; of copper: $0.50 \times 385 = 193\ \text{J/K}$. The water's heat capacity is $\approx 22\times$ larger, so most of the temperature change falls on the copper. The final $T_f \approx 24\,^\circ\text{C}$.水的高比热容和大质量使其能以仅小幅升温吸收铜的热量。水的热容:$1.0 \times 4186 = 4186\ \text{J/K}$;铜:$0.50 \times 385 = 193\ \text{J/K}$。水的热容约大 $22$ 倍,故大部分温度变化发生在铜上。最终 $T_f \approx 24\,^\circ\text{C}$。
The final temperature depends on the heat capacities ($mc$) of both objects, not just their initial temperatures. Water has a very high $c$, so it changes temperature little even when absorbing substantial heat.最终温度取决于两者的热容($mc$),而非仅取决于初温。水的比热容很高,即使吸收大量热量,温度变化也很小。

Phase Changes and Latent Heat相变与潜热

Heat without temperature change.无温度变化的热量。 $$ Q = mL $$
  • $L$ = specific latent heat (J kg$^{-1}$); the energy per unit mass to change phase at constant temperature.= 比潜热(J kg$^{-1}$);在恒温下每千克物质发生相变所需的能量。
  • Fusion (melting / freezing)熔化(熔化/凝固): $L_f$ for water $= 334\,000\ \text{J kg}^{-1} = 334\ \text{kJ kg}^{-1}$.:水的 $L_f = 334\,000\ \text{J kg}^{-1} = 334\ \text{kJ kg}^{-1}$。
  • Vaporisation (boiling / condensing)汽化(沸腾/冷凝): $L_v$ for water $= 2\,260\,000\ \text{J kg}^{-1} = 2260\ \text{kJ kg}^{-1}$.:水的 $L_v = 2\,260\,000\ \text{J kg}^{-1} = 2260\ \text{kJ kg}^{-1}$。
  • Temperature stays constant during a phase change相变期间温度保持不变: all the heat goes into breaking (or forming) intermolecular bonds, not into raising kinetic energy.:所有热量用于断裂(或形成)分子间键,而非提高动能。
The heating curve shows $T$ vs $Q$: sloped segments (sensible heat, $Q = mc\Delta T$) alternate with flat plateaux (latent heat, $Q = mL$). NGSS HS-PS3-2 connects energy at the macroscopic scale to energy "associated with the motion of particles (objects) and energy associated with the relative positions of particles" — exactly the KE vs PE distinction during a phase change.加热曲线显示 $T$ 对 $Q$:斜线段(显热,$Q = mc\Delta T$)与平台段(潜热,$Q = mL$)交替出现。NGSS HS-PS3-2 将宏观尺度上的能量与"与粒子运动相关的能量"和"与粒子相对位置相关的能量"相联——正是相变期间的动能与势能之分。
Worked Example 4 · Melting ice and warming water例题 4 · 冰融化再升温

How much heat is needed to convert $0.200\,\text{kg}$ of ice at $0\,^\circ\text{C}$ into water at $50\,^\circ\text{C}$? ($L_f = 334\ \text{kJ kg}^{-1}$; $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$.)将 $0.200\,\text{kg}$ 的 $0\,^\circ\text{C}$ 冰变为 $50\,^\circ\text{C}$ 的水需要多少热量?($L_f = 334\ \text{kJ kg}^{-1}$;$c_{\text{水}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$。)

Step 1 — melt the ice (phase change at $0\,^\circ\text{C}$).第 1 步 — 冰融化($0\,^\circ\text{C}$ 相变)。

$$ Q_1 = mL_f = 0.200 \times 334\,000 = 66\,800\,\text{J}. $$

Step 2 — warm the liquid water from $0$ to $50\,^\circ\text{C}$.第 2 步 — 液态水从 $0$ 升至 $50\,^\circ\text{C}$。

$$ Q_2 = mc\Delta T = 0.200 \times 4186 \times 50 = 41\,860\,\text{J}. $$

Total heat required.总需热量。

$$ Q_{\text{total}} = Q_1 + Q_2 = 66\,800 + 41\,860 = 108\,660\,\text{J} \approx 109\,\text{kJ}. $$

Note:注意: melting alone requires $66.8\,\text{kJ}$, while warming the resulting water requires only $41.9\,\text{kJ}$. The latent heat is substantial.仅融化就需 $66.8\,\text{kJ}$,而将融化后的水升温仅需 $41.9\,\text{kJ}$。潜热相当可观。

During a phase change (e.g. boiling), a substance absorbs heat but its temperature does not rise. Where does that energy go?在相变期间(如沸腾),物质吸收热量但温度不升高。这些能量去了哪里?
§4 · Q1
It is lost to the surroundings as waste heat以废热形式散失到环境中
It increases the average kinetic energy of the particles它增加了粒子的平均动能
It breaks intermolecular bonds, increasing potential energy without changing average KE它断裂分子间键,增加势能而不改变平均动能
It is stored as nuclear energy inside the atoms它以核能形式储存在原子内部
Temperature is proportional to average random KE. If $T$ does not change, neither does the average KE. The absorbed energy goes into overcoming the attractive forces between molecules (potential energy), allowing them to move apart — that is what a phase transition is.温度与平均随机动能成正比。若 $T$ 不变,平均动能也不变。吸收的能量用于克服分子间的引力(势能),使分子得以分离 — 这正是相变的本质。
During a phase change, temperature is constant, so KE does not change. The energy goes into breaking intermolecular bonds (potential energy increases), not into kinetic energy.相变期间温度恒定,故动能不变。能量用于断裂分子间键(势能增加),而非动能。
How much heat is released when $0.50\,\text{kg}$ of steam at $100\,^\circ\text{C}$ condenses to liquid water at $100\,^\circ\text{C}$? ($L_v = 2260\ \text{kJ kg}^{-1}$)$0.50\,\text{kg}$ 的 $100\,^\circ\text{C}$ 水蒸气冷凝为 $100\,^\circ\text{C}$ 液态水时释放多少热量?($L_v = 2260\ \text{kJ kg}^{-1}$)
§4 · Q2
$167\,\text{kJ}$
$1130\,\text{kJ}$
$2260\,\text{kJ}$
$4520\,\text{kJ}$
$Q = mL_v = 0.50 \times 2260 = 1130\,\text{kJ}$. (Released, so negative in the sign convention, but the magnitude is $1130\,\text{kJ}$.)$Q = mL_v = 0.50 \times 2260 = 1130\,\text{kJ}$。(释放,按符号约定为负,但大小为 $1130\,\text{kJ}$。)
$Q = mL_v = 0.50\,\text{kg} \times 2260\,\text{kJ kg}^{-1} = 1130\,\text{kJ}$. The full $L_v$ applies to the whole mass.$Q = mL_v = 0.50\,\text{kg} \times 2260\,\text{kJ kg}^{-1} = 1130\,\text{kJ}$。$L_v$ 适用于全部质量。
Going deeper — the full heating curve for water and why steam burns are worse than boiling-water burns深入 — 水的完整加热曲线以及为何蒸汽烫伤比沸水烫伤更严重

The heating curve for $1\,\text{kg}$ of water, starting from $-20\,^\circ\text{C}$ ice, shows five segments: (1) ice warming to $0\,^\circ\text{C}$ ($Q = m c_{\text{ice}} \Delta T$, $c_{\text{ice}} = 2090\ \text{J kg}^{-1}\text{K}^{-1}$); (2) flat plateau at $0\,^\circ\text{C}$ ($Q = mL_f = 334\,\text{kJ}$); (3) liquid water warming from $0$ to $100\,^\circ\text{C}$ ($Q = 418.6\,\text{kJ}$); (4) flat plateau at $100\,^\circ\text{C}$ ($Q = mL_v = 2260\,\text{kJ}$); (5) steam warming above $100\,^\circ\text{C}$.从 $-20\,^\circ\text{C}$ 冰开始,$1\,\text{kg}$ 水的加热曲线呈现五个段落:(1) 冰升温至 $0\,^\circ\text{C}$($Q = m c_{\text{冰}} \Delta T$,$c_{\text{冰}} = 2090\ \text{J kg}^{-1}\text{K}^{-1}$);(2) $0\,^\circ\text{C}$ 平台($Q = mL_f = 334\,\text{kJ}$);(3) 液态水从 $0$ 升至 $100\,^\circ\text{C}$($Q = 418.6\,\text{kJ}$);(4) $100\,^\circ\text{C}$ 平台($Q = mL_v = 2260\,\text{kJ}$);(5) 水蒸气在 $100\,^\circ\text{C}$ 以上升温。

The vaporisation plateau is the largest segment by far ($2260\,\text{kJ}$ vs $334\,\text{kJ}$ for melting). When steam at $100\,^\circ\text{C}$ touches skin, it first releases $2260\,\text{kJ kg}^{-1}$ as it condenses, then releases another $418.6\,\text{kJ}$ as the resulting liquid cools to body temperature — far more energy deposited than an equal mass of boiling water at $100\,^\circ\text{C}$, which only delivers the liquid-cooling portion. This is why steam scalds are so severe.汽化平台是迄今最长的段落($2260\,\text{kJ}$,而熔化仅 $334\,\text{kJ}$)。当 $100\,^\circ\text{C}$ 的水蒸气接触皮肤时,首先在冷凝时释放 $2260\,\text{kJ kg}^{-1}$,在生成的液体冷却至体温时释放另外 $418.6\,\text{kJ}$ — 比等质量的 $100\,^\circ\text{C}$ 沸水释放的能量多得多,因为沸水只提供液体冷却部分。这就是为什么蒸汽烫伤如此严重。


Thermal Expansion and the Gas Laws热膨胀与气体定律

The ideal gas law unifies Boyle, Charles, and Gay-Lussac.理想气体定律统一了玻意耳、查理与盖-吕萨克定律。 $$ PV = nRT $$
  • $P$ = absolute pressure (Pa); $V$ = volume (m$^3$); $n$ = amount of substance (mol);= 绝对压力(Pa);$V$ = 体积(m$^3$);$n$ = 物质的量(mol);
  • $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ (universal gas constant); $T$ = absolute temperature in kelvin (never Celsius here).(普适气体常数);$T$ = 开尔文绝对温度(此处绝不用摄氏)。
  • Special cases:特例: Boyle's law ($T$ constant: $P_1 V_1 = P_2 V_2$); Charles's law ($P$ constant: $V \propto T$); Gay-Lussac's law ($V$ constant: $P \propto T$). All follow from $PV = nRT$ with one variable fixed.玻意耳定律($T$ 不变:$P_1 V_1 = P_2 V_2$);查理定律($P$ 不变:$V \propto T$);盖-吕萨克定律($V$ 不变:$P \propto T$)。均由 $PV = nRT$ 固定一个变量得出。
  • Linear thermal expansion:线性热膨胀: $\Delta L = \alpha L_0 \Delta T$, where $\alpha$ is the linear expansion coefficient (K$^{-1}$). Bridges have expansion joints because metals expand significantly with temperature.其中 $\alpha$ 是线膨胀系数(K$^{-1}$)。桥梁设有伸缩缝,因为金属随温度显著膨胀。
Always convert temperature to kelvin before substituting into any gas law. The combined gas law $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ is the most useful form for "before-and-after" problems.代入任何气体定律前,务必将温度换算为开尔文。联合气体定律 $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ 是"前后对比"问题最实用的形式。
Worked Example 5 · Compressed gas heated at constant volume例题 5 · 恒容加热的压缩气体

A sealed rigid container holds gas at $1.00 \times 10^5\,\text{Pa}$ and $20\,^\circ\text{C}$. It is heated to $120\,^\circ\text{C}$. Find the new pressure. (Volume is constant.)一个密封刚性容器中的气体处于 $1.00 \times 10^5\,\text{Pa}$ 和 $20\,^\circ\text{C}$。将其加热至 $120\,^\circ\text{C}$。求新压力。(体积不变。)

Convert to kelvin first.首先换算为开尔文。 $T_1 = 20 + 273 = 293\,\text{K}$; $T_2 = 120 + 273 = 393\,\text{K}$.$T_1 = 20 + 273 = 293\,\text{K}$;$T_2 = 120 + 273 = 393\,\text{K}$。

Gay-Lussac's law (constant $V$).盖-吕萨克定律($V$ 不变)。

$$ \frac{P_1}{T_1} = \frac{P_2}{T_2} \;\Longrightarrow\; P_2 = P_1 \frac{T_2}{T_1} = (1.00 \times 10^5) \times \frac{393}{293} \approx 1.34 \times 10^5\,\text{Pa}. $$

Interpretation.解读。 The pressure rises by $34\%$ because the absolute temperature increased by $34\%$ (from $293$ to $393\,\text{K}$). This is why you should never throw sealed containers into a fire.压力上升 $34\%$,因为绝对温度升高了 $34\%$(从 $293$ 到 $393\,\text{K}$)。这就是为什么绝不应将密封容器投入火中。

A gas at $300\,\text{K}$ and $2.0 \times 10^5\,\text{Pa}$ has its volume halved while the temperature is held constant. What is the new pressure? (Boyle's law)$300\,\text{K}$、$2.0 \times 10^5\,\text{Pa}$ 的气体在温度不变的情况下体积减半。新压力是多少?(玻意耳定律)
§5 · Q1
$4.0 \times 10^5\,\text{Pa}$
$1.0 \times 10^5\,\text{Pa}$
$2.0 \times 10^5\,\text{Pa}$
$3.0 \times 10^5\,\text{Pa}$
Boyle's law: $P_1 V_1 = P_2 V_2$. Halving $V$ doubles $P$: $P_2 = 2 \times 2.0 \times 10^5 = 4.0 \times 10^5\,\text{Pa}$.玻意耳定律:$P_1 V_1 = P_2 V_2$。体积减半,压力加倍:$P_2 = 2 \times 2.0 \times 10^5 = 4.0 \times 10^5\,\text{Pa}$。
At constant temperature, $PV =$ const. If $V$ halves, $P$ doubles.温度不变时,$PV =$ 常数。若 $V$ 减半,$P$ 加倍。
Why must temperature be in kelvin when using the ideal gas law $PV = nRT$?使用理想气体定律 $PV = nRT$ 时,为何温度必须用开尔文?
§5 · Q2
Kelvin is easier to remember than Celsius开尔文比摄氏更好记
The gas constant $R$ was defined for kelvin only气体常数 $R$ 只为开尔文定义
Celsius temperatures are always negative摄氏温度总是负的
The law is derived from kinetic theory where $T$ is the absolute temperature; using Celsius gives meaningless results when $T_C < 0$该定律由动理论推导,其中 $T$ 是绝对温度;当 $T_C < 0$ 时代入摄氏温度会给出无意义结果
Kinetic theory derives $PV = nRT$ with $T$ as the absolute (kelvin) temperature, proportional to average KE. Celsius is an arbitrary scale with a non-physical zero point; at $T_C = -273\,^\circ\text{C}$ the kelvin value is zero (absolute zero), but plugging in $-273$ as a Celsius number into $PV = nRT$ would give negative pressure, which is physically impossible.动理论以绝对(开尔文)温度 $T$ 推导 $PV = nRT$,$T$ 与平均动能成正比。摄氏是一个非物理零点的任意量程;在 $T_C = -273\,^\circ\text{C}$ 时开尔文值为零(绝对零度),但将 $-273$ 作为摄氏数代入 $PV = nRT$ 会得到负压力,这在物理上是不可能的。
The ideal gas law is derived with absolute temperature. Celsius has a zero point at $-273\,\text{K}$, not at true zero thermal energy. Using Celsius yields unphysical results for sub-zero temperatures.理想气体定律用绝对温度推导。摄氏零点在 $-273\,\text{K}$,而非真正的热能零点。对零摄氏以下的温度使用摄氏会得到非物理结果。
Going deeper — deriving the combined gas law and connecting to kinetic theory深入 — 推导联合气体定律并与动理论相联

For a fixed amount of gas ($n$ constant), $PV = nRT$ gives $PV / T = nR = \text{const}$, which is the combined gas law: $P_1 V_1 / T_1 = P_2 V_2 / T_2$. Setting one variable constant recovers Boyle's ($T$ fixed), Charles's ($P$ fixed), or Gay-Lussac's ($V$ fixed) law. The kinetic derivation: pressure arises from molecules bouncing off walls. The average translational kinetic energy per molecule is $\tfrac{3}{2} k_B T$, so $P = \tfrac{2}{3} \frac{N}{V} \bar{KE} = \frac{Nk_B T}{V}$. Since $Nk_B = nR$, this gives $PV = nRT$. The ideal gas law is therefore not empirical — it follows from Newton's laws applied to molecular collisions.对固定量的气体($n$ 不变),$PV = nRT$ 给出 $PV / T = nR = \text{常数}$,即联合气体定律:$P_1 V_1 / T_1 = P_2 V_2 / T_2$。固定一个变量即还原为玻意耳定律($T$ 固定)、查理定律($P$ 固定)或盖-吕萨克定律($V$ 固定)。动理论推导:压力源于分子对器壁的碰撞。每个分子平均平动动能为 $\tfrac{3}{2} k_B T$,故 $P = \tfrac{2}{3} \frac{N}{V} \bar{KE} = \frac{Nk_B T}{V}$。由于 $Nk_B = nR$,得 $PV = nRT$。理想气体定律因此不是经验性的 — 它由牛顿定律应用于分子碰撞而来。


The Laws of Thermodynamics热力学定律

Four laws — each a separate physical truth.四大定律 — 各是独立的物理真理。
  • Zeroth law (热平衡定律):第零定律(热平衡定律): if A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A is in equilibrium with C. This defines temperature as a state function and justifies thermometers.若 A 与 B 热平衡,B 与 C 热平衡,则 A 与 C 亦热平衡。这将温度定义为状态函数,并为温度计的使用提供依据。
  • First law (能量守恒):第一定律(能量守恒): $$ \Delta U = Q - W $$ The change in internal energy equals heat added to the system minus work done by the system. Energy is conserved; it cannot be created or destroyed.内能的变化等于系统吸收的热量减去系统对外做的功。能量守恒;不能被创造或消灭。
  • Second law (熵增定律):第二定律(熵增定律): heat flows spontaneously only from a hot object to a cold one; no process can convert heat completely into work without another change in the system. Entropy (disorder) of an isolated system never decreases. This is the law NGSS HS-PS3-4 directly investigates.热只会自发从热物体流向冷物体;没有任何过程能在不引起其他变化的情况下将热完全转化为功。孤立系统的熵(无序度)不会减小。这正是 NGSS HS-PS3-4 直接探究的定律。
  • Third law (绝对零度不可达):第三定律(绝对零度不可达): it is impossible to reach absolute zero in a finite number of steps. (Qualitative HS awareness; quantitative treatment is university-level.)不可能通过有限步骤达到绝对零度。(高中阶段定性了解;定量处理属大学层面。)
NGSS HS-PS3-4 centres on the second law: mixing substances at different temperatures produces a "more uniform energy distribution" — entropy increases. Ontario D3.12 ties the laws of thermodynamics to nuclear-plant energy conversion.NGSS HS-PS3-4 以第二定律为核心:将不同温度的物质混合会产生"更均匀的能量分布" — 熵增加。安大略 D3.12 将热力学定律与核电站能量转化相联。
Worked Example 6 · Applying the first law例题 6 · 应用第一定律

A gas absorbs $500\,\text{J}$ of heat and simultaneously does $200\,\text{J}$ of work by expanding against a piston. What is the change in the gas's internal energy?气体吸收 $500\,\text{J}$ 热量,同时通过膨胀推动活塞对外做 $200\,\text{J}$ 的功。气体内能的变化量是多少?

Apply the first law.应用第一定律。 $Q = +500\,\text{J}$ (heat added to system); $W = +200\,\text{J}$ (work done by system).$Q = +500\,\text{J}$(系统吸热);$W = +200\,\text{J}$(系统对外做功)。

$$ \Delta U = Q - W = 500 - 200 = +300\,\text{J}. $$

Interpretation.解读。 The gas's internal energy increases by $300\,\text{J}$. Of the $500\,\text{J}$ that flowed in, $200\,\text{J}$ was used to do mechanical work (push the piston) and $300\,\text{J}$ raised the thermal energy of the gas itself. This is energy bookkeeping — exactly what NGSS HS-PS3-1 requires ("calculate the change in the energy of one component in a system").气体内能增加 $300\,\text{J}$。流入的 $500\,\text{J}$ 中,$200\,\text{J}$ 用于做机械功(推动活塞),$300\,\text{J}$ 提升了气体本身的热能。这是能量核算 — 正是 NGSS HS-PS3-1 所要求的("计算系统中某一成分的能量变化")。

Which law of thermodynamics explains why heat flows spontaneously from a hot cup of coffee to the cooler room, but never the reverse?热力学哪条定律解释了为何热量自发从热咖啡流向较冷的房间,而不会反过来?
§6 · Q1
Zeroth law第零定律
First law第一定律
Second law第二定律
Third law第三定律
The second law states that heat flows spontaneously from hot to cold (entropy of the isolated system increases). The first law only says energy is conserved — it would not forbid the reverse flow from cold to hot, which the second law does forbid.第二定律指出热自发从热处流向冷处(孤立系统的熵增加)。第一定律仅说能量守恒 — 它不禁止从冷到热的反向流动,而第二定律明确禁止。
The first law (conservation of energy) would not rule out heat flowing from cold to hot, since energy could still be conserved in that case. It is the second law (entropy never decreases in an isolated system) that forbids the spontaneous cold-to-hot direction.第一定律(能量守恒)不会排除热从冷流向热,因为那种情况下能量仍可守恒。是第二定律(孤立系统熵不减小)禁止了从冷到热的自发方向。
A gas releases $300\,\text{J}$ of heat to its surroundings and has $100\,\text{J}$ of work done on it (compressed). What is $\Delta U$?气体向外界释放 $300\,\text{J}$ 热量,并受外界压缩对气体做 $100\,\text{J}$ 的功。$\Delta U$ 是多少?
§6 · Q2
$-200\,\text{J}$
$+200\,\text{J}$
$-400\,\text{J}$
$+400\,\text{J}$
$Q = -300\,\text{J}$ (heat leaves system); work done on the system means $W = -100\,\text{J}$ (work done by the system is negative when the surroundings compress it). $\Delta U = Q - W = -300 - (-100) = -200\,\text{J}$.$Q = -300\,\text{J}$(系统放热);对系统做功意味着 $W = -100\,\text{J}$(外界压缩时系统对外做功为负)。$\Delta U = Q - W = -300 - (-100) = -200\,\text{J}$。
Be careful with signs. Heat released: $Q < 0$. Work done on the gas (compression): $W < 0$ in the convention $\Delta U = Q - W$ (since $W$ is work done by the gas, and compression means the gas does negative work). $\Delta U = (-300) - (-100) = -200\,\text{J}$.注意符号。放热:$Q < 0$。对气体做功(压缩):在 $\Delta U = Q - W$ 约定下 $W < 0$(因 $W$ 是气体对外做的功,压缩时气体做负功)。$\Delta U = (-300) - (-100) = -200\,\text{J}$。
Going deeper — entropy and why the second law has a preferred direction in time深入 — 熵以及为何第二定律有时间的优先方向

Newton's laws of motion are time-reversible — the equations look the same if you replace $t$ with $-t$. Yet we never see broken eggs reassemble or heat flow spontaneously from cold to hot. The second law introduces a preferred direction. Statistically, entropy $S = k_B \ln \Omega$ (Boltzmann's formula), where $\Omega$ is the number of microstates (microscopic arrangements) consistent with the observed macrostate. The second law says $\Omega$ increases for irreversible processes: the chance of all gas molecules spontaneously moving to one side of a box, creating a low-entropy state, is vanishingly small. This probabilistic interpretation shows the second law is not absolute in the way Newton's laws are — it reflects overwhelmingly probable behaviour of $\sim 10^{23}$ molecules, not a microscopic force. The NGSS HS-PS3-4 investigation of mixing two temperatures is a macroscopic demonstration of this entropy-increase principle.牛顿运动定律具有时间反演对称性 — 将 $t$ 替换为 $-t$ 方程形式不变。然而我们从未见过破碎的鸡蛋自动复原,或热量自发从冷处流向热处。第二定律引入了时间的优先方向。从统计角度,熵 $S = k_B \ln \Omega$(玻尔兹曼公式),其中 $\Omega$ 是与观测到的宏观状态一致的微观状态数。第二定律指出,对不可逆过程 $\Omega$ 增大:所有气体分子自发移动到箱子一侧、造成低熵状态的概率极小。这种概率性解释表明第二定律不像牛顿定律那样绝对 — 它反映约 $10^{23}$ 个分子的绝倒性概率行为,而非微观力。NGSS HS-PS3-4 中混合两种温度的探究活动,正是这一熵增原理的宏观演示。


Heat Engines and Efficiency热机与效率

Curriculum note.课纲提示。 NGSS HS-PS3-3 mentions "efficiency" as a constraint for energy-conversion devices; HS-PS3-4 investigates the second-law underpinning qualitatively. Ontario SPH3U Strand D includes efficiency as a required concept (D3); D3.12 explicitly connects thermodynamic laws to nuclear-plant heat-engine operation. BC Physics 11 names "power and efficiency" as Content. Alberta Physics 20/30 has no thermodynamics unit — use this section for AP/IB preparation.NGSS HS-PS3-3 将"效率"列为能量转化装置的设计约束;HS-PS3-4 定性探究第二定律的基础。安大略 SPH3U D 单元将效率列为必学概念(D3);D3.12 明确将热力学定律与核电站热机运行相联。BC Physics 11 将"功率与效率"列为内容。阿尔伯塔 Physics 20/30 没有热力学单元 — 本节仅用于 AP/IB 准备。
A heat engine converts heat into work — but never completely.热机将热转化为功 — 但绝不能完全转化。 $$ \eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H} $$
  • $Q_H$ = heat absorbed from the hot reservoir; $Q_C$ = heat rejected to the cold reservoir; $W = Q_H - Q_C$ = net work output.= 从热源吸收的热量;$Q_C$ = 排向冷源的热量;$W = Q_H - Q_C$ = 净功输出。
  • Efficiency $\eta$ (效率)效率 $\eta$ is always between 0 and 1 (never 100%) because the second law requires that some heat $Q_C$ must be rejected.始终介于 0 和 1 之间(永不为 100%),因为第二定律要求必须排出部分热量 $Q_C$。
  • Carnot efficiency (theoretical maximum):卡诺效率(理论最大值): $$ \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \qquad (T \text{ in kelvin}) $$ No real engine can exceed this. A higher $T_H$ or lower $T_C$ both increase the theoretical limit.任何实际热机均不能超越此值。更高的 $T_H$ 或更低的 $T_C$ 都能提高理论上限。
NGSS HS-PS3-3 and BC Physics 11 both mention "efficiency" as a design constraint for energy-conversion devices. Ontario D3.12 connects heat-engine efficiency to nuclear-plant thermodynamics.NGSS HS-PS3-3 与 BC Physics 11 都将"效率"列为能量转化装置的设计约束。安大略 D3.12 将热机效率与核电站热力学相联。
Worked Example 7 · Steam turbine efficiency例题 7 · 蒸汽轮机效率

A steam turbine takes in $Q_H = 8000\,\text{J}$ of heat from steam and produces $2400\,\text{J}$ of mechanical work per cycle. (a) Find the thermal efficiency. (b) Find the heat rejected to the cold reservoir. (c) What is the Carnot efficiency if $T_H = 500\,\text{K}$ and $T_C = 300\,\text{K}$?一台蒸汽轮机每循环从蒸汽吸入 $Q_H = 8000\,\text{J}$ 热量并产生 $2400\,\text{J}$ 机械功。(a) 求热效率。(b) 求排向冷源的热量。(c) 若 $T_H = 500\,\text{K}$,$T_C = 300\,\text{K}$,卡诺效率是多少?

(a) Thermal efficiency.(a) 热效率。

$$ \eta = \frac{W}{Q_H} = \frac{2400}{8000} = 0.30 = 30\%. $$

(b) Heat rejected.(b) 排出热量。 By energy conservation: $Q_C = Q_H - W = 8000 - 2400 = 5600\,\text{J}$.由能量守恒:$Q_C = Q_H - W = 8000 - 2400 = 5600\,\text{J}$。

(c) Carnot efficiency.(c) 卡诺效率。

$$ \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{500} = 1 - 0.60 = 40\%. $$

Interpretation.解读。 The real turbine achieves $30\%$, less than the $40\%$ Carnot maximum — as the second law demands. The $5600\,\text{J}$ per cycle ejected as waste heat is unavoidable; this is what makes all heat engines thermodynamically imperfect.实际轮机达到 $30\%$,低于 $40\%$ 的卡诺最大值 — 正如第二定律所要求的。每循环排出的 $5600\,\text{J}$ 废热是不可避免的;这使所有热机在热力学上都是不完美的。

A heat engine absorbs $5000\,\text{J}$ and rejects $3500\,\text{J}$. What is its efficiency?一台热机吸入 $5000\,\text{J}$ 并排出 $3500\,\text{J}$。效率是多少?
§7 · Q1
$70\%$$70\%$
$30\%$$30\%$
$43\%$$43\%$
$100\%$$100\%$
$W = Q_H - Q_C = 5000 - 3500 = 1500\,\text{J}$. $\eta = W/Q_H = 1500/5000 = 0.30 = 30\%$. Equivalently, $\eta = 1 - Q_C/Q_H = 1 - 3500/5000 = 0.30$.$W = Q_H - Q_C = 5000 - 3500 = 1500\,\text{J}$。$\eta = W/Q_H = 1500/5000 = 0.30 = 30\%$。等价地,$\eta = 1 - Q_C/Q_H = 1 - 3500/5000 = 0.30$。
Work output $= Q_H - Q_C = 1500\,\text{J}$. Efficiency $= W / Q_H$, not $W / Q_C$.功输出 $= Q_H - Q_C = 1500\,\text{J}$。效率 $= W / Q_H$,不是 $W / Q_C$。
Why can no heat engine ever reach 100% efficiency?为何任何热机都无法达到 100% 效率?
§7 · Q2
Because friction always wastes some energy因为摩擦总会浪费一些能量
Because engineers have not yet built a perfect engine因为工程师尚未建造完美的发动机
Because the first law prevents it因为第一定律阻止了这一点
Because the second law requires that some heat must be rejected; complete conversion of heat to work would decrease entropy因为第二定律要求必须有部分热量排向冷源;将热完全转化为功会使熵减少
100% efficiency would mean converting all absorbed heat into work with zero heat rejection, which would be a spontaneous decrease in entropy — forbidden by the second law. Even a perfectly frictionless Carnot engine is limited to $\eta = 1 - T_C / T_H < 1$ (as long as $T_C > 0\,\text{K}$).100% 效率意味着将所有吸收的热量转化为功而不排出任何热量,这将是熵的自发减少 — 被第二定律所禁止。即使是完全无摩擦的卡诺发动机也受限于 $\eta = 1 - T_C / T_H < 1$(只要 $T_C > 0\,\text{K}$)。
Friction is one cause of inefficiency in real engines, but even a theoretical frictionless engine cannot reach 100% — the second law sets a thermodynamic ceiling regardless of engineering quality.摩擦是实际发动机低效的原因之一,但即使理论上的无摩擦发动机也无法达到 100% — 无论工程质量如何,第二定律都设定了热力学上限。
Going deeper — the Carnot cycle and why efficiency is bounded深入 — 卡诺循环以及为何效率有上限

Carnot proved (1824) that no engine operating between two heat reservoirs at $T_H$ and $T_C$ can be more efficient than a reversible engine cycling between those same temperatures. His ideal cycle has four steps: (1) isothermal expansion at $T_H$ (absorbs $Q_H$); (2) adiabatic expansion (no heat, temperature drops to $T_C$); (3) isothermal compression at $T_C$ (rejects $Q_C$); (4) adiabatic compression back to $T_H$. The efficiency is $\eta_{\text{Carnot}} = 1 - T_C / T_H$, which equals 100% only if $T_C = 0\,\text{K}$ (absolute zero) — unreachable by the third law. For a steam turbine with $T_H = 800\,\text{K}$ and $T_C = 300\,\text{K}$, the Carnot limit is $1 - 300/800 = 62.5\%$; real turbines reach $35$–$45\%$ due to friction, heat losses, and non-ideal gas behaviour. Ontario D3.12 connects this physics to nuclear-plant operation: fission releases heat that drives a steam cycle, and waste heat is always rejected, limiting efficiency exactly as the second law predicts.卡诺于 1824 年证明:在温度为 $T_H$ 和 $T_C$ 的两个热源之间运行的任何发动机,其效率都不能超过在同样温度之间循环的可逆发动机。他的理想循环有四个步骤:(1) 在 $T_H$ 下等温膨胀(吸收 $Q_H$);(2) 绝热膨胀(无热交换,温度降至 $T_C$);(3) 在 $T_C$ 下等温压缩(排出 $Q_C$);(4) 绝热压缩回 $T_H$。效率为 $\eta_{\text{Carnot}} = 1 - T_C / T_H$,仅当 $T_C = 0\,\text{K}$(绝对零度)时才等于 100% — 而这由第三定律保证不可达。对于 $T_H = 800\,\text{K}$、$T_C = 300\,\text{K}$ 的蒸汽轮机,卡诺上限为 $1 - 300/800 = 62.5\%$;实际轮机因摩擦、热损失和非理想气体行为只能达到 $35$–$45\%$。安大略 D3.12 将此物理与核电站运行相联:裂变释放热量驱动蒸汽循环,废热始终被排出,效率正如第二定律所预测的那样受限。


Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline解题前的纪律
  • Always convert temperature to kelvin.始终将温度换算为开尔文。 Before substituting into $PV = nRT$, $Q = mc\Delta T$, or $\eta_{\text{Carnot}} = 1 - T_C/T_H$, add 273.15 to Celsius. The most common error in thermodynamics problems is using Celsius directly in a gas-law or Carnot equation.在代入 $PV = nRT$、$Q = mc\Delta T$ 或 $\eta_{\text{Carnot}} = 1 - T_C/T_H$ 之前,将摄氏加 273.15。热力学题中最常见的错误是直接将摄氏温度代入气体定律或卡诺方程。
  • Identify the heat-transfer type first.先识别热传递类型。 Conduction (contact), convection (fluid bulk movement), radiation (electromagnetic waves). These are concept marks in every curriculum.传导(接触)、对流(流体整体运动)、辐射(电磁波)。这些在每门课程中都是概念分。
  • Watch the sign of $Q$ in calorimetry.在量热法中注意 $Q$ 的符号。 Heat lost by the hot body equals heat gained by the cold body: $Q_{\text{lost}} + Q_{\text{gained}} = 0$, or equivalently $m_1 c_1 (T_f - T_1) + m_2 c_2 (T_f - T_2) = 0$.热体失热等于冷体得热:$Q_{\text{失}} + Q_{\text{得}} = 0$,即 $m_1 c_1 (T_f - T_1) + m_2 c_2 (T_f - T_2) = 0$。
Phase changes and multi-step heating problems相变与多步加热问题
  • Never apply $Q = mc\Delta T$ across a phase change.绝不在跨越相变时应用 $Q = mc\Delta T$。 At a phase change, $\Delta T = 0$ but $Q \ne 0$. Split the problem: sensible heat ($Q = mc\Delta T$) before and after; latent heat ($Q = mL$) during.在相变时 $\Delta T = 0$ 但 $Q \ne 0$。将问题分段:相变前后用显热($Q = mc\Delta T$);相变期间用潜热($Q = mL$)。
  • Steam burns vs boiling-water burns.蒸汽烫伤与沸水烫伤。 Steam at $100\,^\circ\text{C}$ releases $L_v = 2260\,\text{kJ/kg}$ on condensing plus the sensible heat of cooling — far more total energy than the same mass of liquid water at $100\,^\circ\text{C}$.$100\,^\circ\text{C}$ 的蒸汽冷凝时释放 $L_v = 2260\,\text{kJ/kg}$,加上冷却时的显热 — 远比同质量的 $100\,^\circ\text{C}$ 液态水释放的总能量多。
Thermodynamics laws (§6–§7)热力学定律(§6–§7)
  • First law sign convention matters.第一定律符号约定很重要。 $\Delta U = Q - W$: $Q$ is positive when heat flows into the system; $W$ is positive when the system does work on surroundings. Some textbooks use $\Delta U = Q + W$ (with $W$ = work done on system) — know which convention your course uses.$\Delta U = Q - W$:热流系统时 $Q$ 为正;系统对外界做功时 $W$ 为正。有些教材使用 $\Delta U = Q + W$($W$ 为对系统做的功)— 了解你的课程使用哪种约定。
  • Efficiency is always $W / Q_H$, not $W / Q_C$.效率始终是 $W / Q_H$,而非 $W / Q_C$。 The heat input is in the denominator. And $\eta < 1$ always — if you get $\eta > 1$, check signs and which quantities you used.分母是热输入量。且 $\eta < 1$ 恒成立 — 若得 $\eta > 1$,检查符号和所用的量。
Answer hygiene作答规范
  • Include units throughout.全程带单位。 J, kJ, kg, K, Pa, m$^3$, mol — keep them in every line. Unit errors are one of the easiest marks to lose.J、kJ、kg、K、Pa、m$^3$、mol — 每行都带。单位错误是最容易失分的地方之一。
  • Sanity-check the final temperature in calorimetry.在量热法中核查最终温度的合理性。 The equilibrium temperature must lie between the initial temperatures of the two objects. If it does not, you have a sign error somewhere.平衡温度必须介于两物体的初温之间。若不在,则某处有符号错误。
  • For NGSS HS-PS3-4 investigations:对于 NGSS HS-PS3-4 探究活动: State the second-law conclusion explicitly — that thermal equilibrium produces a more uniform energy distribution, consistent with entropy increase.明确陈述第二定律的结论 — 热平衡产生更均匀的能量分布,与熵增加一致。

Flashcards闪卡

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Temperature vs thermal energy?温度与热能?
Temperature = average KE per particle. Thermal energy (热能) = total internal energy of all particles.温度 = 每粒子平均动能。热能 = 所有粒子的内能总和。
Celsius to kelvin?摄氏换开尔文?
$$T_K = T_C + 273.15$$
Three modes of heat transfer (热传递)?三种热传递模式?
Conduction (传导) — contact; Convection (对流) — fluid bulk; Radiation (辐射) — EM waves, no medium.传导 — 接触;对流 — 流体整体;辐射 — 电磁波,无需介质。
Specific heat capacity (比热容) equation?比热容方程?
$$Q = mc\Delta T$$
Calorimetry (量热法) principle?量热法原理?
Heat lost = heat gained: $m_1 c_1 |\Delta T_1| = m_2 c_2 |\Delta T_2|$失热 = 得热:$m_1 c_1 |\Delta T_1| = m_2 c_2 |\Delta T_2|$
Latent heat (潜热) equation?潜热方程?
$$Q = mL$$ $T$ constant during phase change (相变)相变期间 $T$ 不变
$L_v$ of water?水的 $L_v$?
$$L_v = 2260\,\text{kJ/kg}$$ (much larger than $L_f = 334\,\text{kJ/kg}$)(远大于 $L_f = 334\,\text{kJ/kg}$)
Ideal gas law (理想气体定律)?理想气体定律?
$$PV = nRT$$ $T$ in kelvin always$T$ 始终为开尔文
Combined gas law?联合气体定律?
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
First law of thermodynamics (热力学第一定律)?热力学第一定律?
$$\Delta U = Q - W$$ $Q$ absorbed = $+$; $W$ done by system = $+$$Q$ 吸入 $= +$;系统做功 $= +$
Second law (热力学第二定律) direction of heat flow?第二定律:热流方向?
Heat flows spontaneously hot → cold only. Entropy (熵) of an isolated system never decreases.热只自发从热处流向冷处。孤立系统的熵不减小。
Heat engine efficiency (效率)?热机效率?
$$\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}$$
Carnot efficiency (卡诺效率)?卡诺效率?
$$\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$$ theoretical maximum; $T$ in kelvin理论最大值;$T$ 用开尔文
Why is $\eta < 1$ always?为何 $\eta < 1$ 恒成立?
Second law requires heat rejection $Q_C > 0$; complete heat-to-work conversion would decrease entropy.第二定律要求排热 $Q_C > 0$;将热完全转化为功会使熵减少。

Practice Quiz综合测验

What is $25\,^\circ\text{C}$ in kelvin?$25\,^\circ\text{C}$ 等于多少开尔文?
Q1
$25\,\text{K}$$25\,\text{K}$
$248\,\text{K}$$248\,\text{K}$
$298\,\text{K}$$298\,\text{K}$
$-248\,\text{K}$$-248\,\text{K}$
$T_K = T_C + 273 = 25 + 273 = 298\,\text{K}$.$T_K = T_C + 273 = 25 + 273 = 298\,\text{K}$。
Add 273 (or 273.15) to convert $^\circ$C to K. Kelvin is always positive.加 273(或 273.15)将摄氏换算为开尔文。开尔文始终为正。
A hot metal rod is plunged into cool water in an insulated container. Which statement is correct at thermal equilibrium?将一根热金属棒插入绝热容器中的冷水。在热平衡时,哪句陈述正确?
Q2
The metal and water reach the same final temperature; heat lost by metal equals heat gained by water金属与水达到相同的最终温度;金属失热等于水得热
The water temperature rises until it equals the original metal temperature水温上升直到等于金属的原始温度
The final temperature is the average of the two initial temperatures最终温度是两个初温的平均值
The total thermal energy increases due to the heat transfer由于热传递,总热能增加
At thermal equilibrium, both reach the same temperature. By conservation of energy (NGSS HS-PS3-4 / BC thermal equilibrium), heat lost by the hot rod equals heat gained by the cooler water: $Q_{\text{lost}} = Q_{\text{gained}}$.在热平衡时,两者达到相同温度。由能量守恒(NGSS HS-PS3-4 / BC 热平衡),热棒失热等于冷水得热:$Q_{\text{失}} = Q_{\text{得}}$。
The final temperature lies between the two initial temperatures, not at either extreme. Energy is conserved, not created — total thermal energy stays the same (in the insulated container).最终温度介于两初温之间,而非任何极端值。能量守恒,不被创造 — 总热能保持不变(在绝热容器中)。
How much heat is needed to warm $0.500\,\text{kg}$ of aluminium ($c = 900\ \text{J kg}^{-1}\text{K}^{-1}$) from $20\,^\circ\text{C}$ to $120\,^\circ\text{C}$?将 $0.500\,\text{kg}$ 铝($c = 900\ \text{J kg}^{-1}\text{K}^{-1}$)从 $20\,^\circ\text{C}$ 加热至 $120\,^\circ\text{C}$ 需要多少热量?
Q3
$9\,\text{kJ}$$9\,\text{kJ}$
$18\,\text{kJ}$$18\,\text{kJ}$
$54\,\text{kJ}$$54\,\text{kJ}$
$45\,\text{kJ}$$45\,\text{kJ}$
$Q = mc\Delta T = 0.500 \times 900 \times (120-20) = 0.500 \times 900 \times 100 = 45\,000\,\text{J} = 45\,\text{kJ}$.$Q = mc\Delta T = 0.500 \times 900 \times (120-20) = 0.500 \times 900 \times 100 = 45\,000\,\text{J} = 45\,\text{kJ}$。
$Q = mc\Delta T$. Here $\Delta T = 100\,\text{K}$, $m = 0.500\,\text{kg}$, $c = 900\ \text{J kg}^{-1}\text{K}^{-1}$. Multiply all three: $0.500 \times 900 \times 100$.$Q = mc\Delta T$。此处 $\Delta T = 100\,\text{K}$,$m = 0.500\,\text{kg}$,$c = 900\ \text{J kg}^{-1}\text{K}^{-1}$。三者相乘:$0.500 \times 900 \times 100$。
A thermometer reads $20\,^\circ\text{C}$ inside and $-10\,^\circ\text{C}$ outside. Heat flows from inside to outside through the wall. What mechanism is dominant inside the wall (which is solid)?室内温度计读数 $20\,^\circ\text{C}$,室外 $-10\,^\circ\text{C}$。热量通过墙壁从室内流向室外。在墙壁(固体)内部,主要热传递机制是什么?
Q4
Convection对流
Conduction传导
Radiation辐射
Evaporation蒸发
Through a solid, heat is transferred by conduction (传导): energetic particles collide with and pass energy to neighbouring particles, with no bulk material movement. Convection requires fluid flow; radiation occurs but is minor compared to conduction in a solid wall.通过固体,热量由传导传递:高能粒子与相邻粒子碰撞并传递能量,无物料整体流动。对流需要流体流动;辐射存在但在固体墙中与传导相比微乎其微。
Convection requires a fluid. In a solid wall, the dominant mechanism is conduction — particle-to-particle energy transfer.对流需要流体。在固体墙中,主要机制是传导 — 粒子间能量传递。
A sample of ice at $0\,^\circ\text{C}$ is being melted. As it absorbs heat, what happens to its temperature?一块 $0\,^\circ\text{C}$ 的冰正在熔化。当它吸收热量时,其温度如何变化?
Q5
It rises continuously持续上升
It drops slightly, then rises先略降,再上升
It stays at $0\,^\circ\text{C}$ until all ice has melted保持在 $0\,^\circ\text{C}$,直到所有冰融化
It rises to $100\,^\circ\text{C}$ immediately立即升至 $100\,^\circ\text{C}$
During a phase change (相变), all the absorbed heat goes into breaking intermolecular bonds (latent heat, 潜热), not into increasing kinetic energy. So temperature remains constant at the melting point until the phase change is complete.在相变期间,所有吸收的热量用于断裂分子间键(潜热),而非增加动能。因此温度在相变完成前保持在熔点不变。
During melting, $T$ stays constant because the energy goes into breaking intermolecular bonds (latent heat), not into kinetic energy. Only after all the ice melts does $T$ begin to rise.熔化期间 $T$ 保持不变,因为能量用于断裂分子间键(潜热),而非动能。只有在所有冰融化之后 $T$ 才开始上升。
A gas at $400\,\text{K}$ and volume $2.0\,\text{L}$ is cooled at constant pressure to $200\,\text{K}$. What is the new volume? (Charles's law)$400\,\text{K}$、体积 $2.0\,\text{L}$ 的气体在恒压下冷却至 $200\,\text{K}$。新体积是多少?(查理定律)
Q6
$1.0\,\text{L}$$1.0\,\text{L}$
$4.0\,\text{L}$$4.0\,\text{L}$
$2.0\,\text{L}$$2.0\,\text{L}$
$0.50\,\text{L}$$0.50\,\text{L}$
Charles's law: $V \propto T$ at constant $P$. $V_2 = V_1 (T_2/T_1) = 2.0 \times (200/400) = 2.0 \times 0.50 = 1.0\,\text{L}$. Temperature halves, so volume halves.查理定律:恒压下 $V \propto T$。$V_2 = V_1 (T_2/T_1) = 2.0 \times (200/400) = 2.0 \times 0.50 = 1.0\,\text{L}$。温度减半,故体积减半。
Charles's law: $V_1/T_1 = V_2/T_2$ at constant pressure. Temperatures must be in kelvin. $V_2 = V_1 \times T_2/T_1 = 2.0 \times 200/400 = 1.0\,\text{L}$.查理定律:恒压下 $V_1/T_1 = V_2/T_2$。温度必须用开尔文。$V_2 = V_1 \times T_2/T_1 = 2.0 \times 200/400 = 1.0\,\text{L}$。
A gas absorbs $800\,\text{J}$ of heat and does $300\,\text{J}$ of work by expanding. What is the change in its internal energy?气体吸收 $800\,\text{J}$ 热量并通过膨胀做 $300\,\text{J}$ 的功。其内能的变化量是多少?
Q7
$1100\,\text{J}$$1100\,\text{J}$
$500\,\text{J}$$500\,\text{J}$
$-500\,\text{J}$$-500\,\text{J}$
$300\,\text{J}$$300\,\text{J}$
First law (热力学第一定律): $\Delta U = Q - W = 800 - 300 = 500\,\text{J}$. The gas's internal energy increases by $500\,\text{J}$.第一定律:$\Delta U = Q - W = 800 - 300 = 500\,\text{J}$。气体内能增加 $500\,\text{J}$。
First law: $\Delta U = Q - W$. $Q = +800\,\text{J}$ (absorbed); $W = +300\,\text{J}$ (done by gas). $\Delta U = 800 - 300 = 500\,\text{J}$.第一定律:$\Delta U = Q - W$。$Q = +800\,\text{J}$(吸收);$W = +300\,\text{J}$(气体做功)。$\Delta U = 800 - 300 = 500\,\text{J}$。
A heat engine operating between $T_H = 600\,\text{K}$ and $T_C = 300\,\text{K}$ achieves $35\%$ efficiency. Which statement is true?一台在 $T_H = 600\,\text{K}$ 和 $T_C = 300\,\text{K}$ 之间运行的热机效率为 $35\%$。哪句陈述正确?
Q8
This engine violates the second law because $\eta > 0$此热机因 $\eta > 0$ 而违反第二定律
This engine is more efficient than the Carnot limit for these temperatures此热机效率超过这些温度下的卡诺极限
This engine is less efficient than the Carnot limit ($50\%$) for these temperatures, as expected此热机效率低于这些温度下的卡诺极限($50\%$),符合预期
No heat engine can operate between these temperatures任何热机都不能在这些温度之间运行
Carnot efficiency: $\eta_{\text{Carnot}} = 1 - T_C/T_H = 1 - 300/600 = 50\%$. The real engine achieves $35\% < 50\%$ — below the Carnot limit, exactly as the second law requires. No real engine can exceed Carnot.卡诺效率:$\eta_{\text{Carnot}} = 1 - T_C/T_H = 1 - 300/600 = 50\%$。实际热机达到 $35\% < 50\%$ — 低于卡诺极限,正如第二定律所要求的。任何实际热机都不能超过卡诺效率。
Calculate Carnot first: $\eta_C = 1 - 300/600 = 50\%$. The real engine at $35\%$ is below this — consistent with the second law. A real engine exceeding Carnot would violate the second law.先计算卡诺效率:$\eta_C = 1 - 300/600 = 50\%$。实际热机为 $35\%$,低于此值 — 符合第二定律。超过卡诺效率的实际热机将违反第二定律。
How much heat (in kJ) is released when $0.100\,\text{kg}$ of steam at $100\,^\circ\text{C}$ condenses and then cools to $20\,^\circ\text{C}$? ($L_v = 2260\,\text{kJ/kg}$; $c_{\text{water}} = 4.186\,\text{kJ kg}^{-1}\text{K}^{-1}$) 🇺🇸 NGSS HS-PS3-1$0.100\,\text{kg}$ 的 $100\,^\circ\text{C}$ 水蒸气冷凝后再冷却至 $20\,^\circ\text{C}$ 释放多少热量(kJ)?($L_v = 2260\,\text{kJ/kg}$;$c_{\text{水}} = 4.186\,\text{kJ kg}^{-1}\text{K}^{-1}$)🇺🇸 NGSS HS-PS3-1
Q9
$259\,\text{kJ}$$259\,\text{kJ}$
$226\,\text{kJ}$$226\,\text{kJ}$
$33\,\text{kJ}$$33\,\text{kJ}$
$420\,\text{kJ}$$420\,\text{kJ}$
Step 1: condensation at $100\,^\circ\text{C}$: $Q_1 = mL_v = 0.100 \times 2260 = 226\,\text{kJ}$. Step 2: cooling liquid from $100$ to $20\,^\circ\text{C}$: $Q_2 = mc\Delta T = 0.100 \times 4.186 \times 80 = 33.5\,\text{kJ}$. Total $= 226 + 33.5 \approx 259\,\text{kJ}$.第 1 步:$100\,^\circ\text{C}$ 冷凝:$Q_1 = mL_v = 0.100 \times 2260 = 226\,\text{kJ}$。第 2 步:液体从 $100$ 降至 $20\,^\circ\text{C}$ 冷却:$Q_2 = mc\Delta T = 0.100 \times 4.186 \times 80 = 33.5\,\text{kJ}$。合计 $= 226 + 33.5 \approx 259\,\text{kJ}$。
Two steps: (1) condensation releases $mL_v$; (2) cooling the resulting water releases $mc\Delta T$. Add both contributions.两步:(1) 冷凝释放 $mL_v$;(2) 冷却液态水释放 $mc\Delta T$。将两部分相加。
A heat engine absorbs $10\,000\,\text{J}$ per cycle and operates between $T_H = 500\,\text{K}$ and $T_C = 250\,\text{K}$. The Carnot efficiency sets the upper bound. What is the maximum work the engine could produce per cycle?一台热机每循环吸入 $10\,000\,\text{J}$,在 $T_H = 500\,\text{K}$ 与 $T_C = 250\,\text{K}$ 之间运行。卡诺效率设定上限。该热机每循环最多能产生多少功?
Q10
$10\,000\,\text{J}$$10\,000\,\text{J}$
$2\,500\,\text{J}$$2\,500\,\text{J}$
$7\,500\,\text{J}$$7\,500\,\text{J}$
$5\,000\,\text{J}$$5\,000\,\text{J}$
Carnot efficiency: $\eta_C = 1 - 250/500 = 0.50 = 50\%$. Max work $= \eta_C \times Q_H = 0.50 \times 10\,000 = 5\,000\,\text{J}$. The other $5\,000\,\text{J}$ must be rejected as waste heat.卡诺效率:$\eta_C = 1 - 250/500 = 0.50 = 50\%$。最大功 $= \eta_C \times Q_H = 0.50 \times 10\,000 = 5\,000\,\text{J}$。另外 $5\,000\,\text{J}$ 必须以废热形式排出。
First find Carnot efficiency: $\eta_C = 1 - T_C/T_H = 1 - 250/500 = 0.50$. Then max work $= \eta_C \times Q_H = 5\,000\,\text{J}$.首先求卡诺效率:$\eta_C = 1 - T_C/T_H = 1 - 250/500 = 0.50$。则最大功 $= \eta_C \times Q_H = 5\,000\,\text{J}$。

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成,再勾选每一项。

0 / 11 mastered已掌握 0 / 11

What This Feeds Into本单元的去向

Thermodynamics and heat connect to several HS Physics units. The Work, Energy and Power guide defines the mechanical work and energy framework that the first law extends; the gas laws connect to kinetic theory and pressure-volume work. At the AP and IB level, thermodynamics is a dedicated topic. The IB Physics HL guide covers the thermodynamics topic in depth. Note that no AP Thermodynamics unit exists in this repo, as thermodynamics is integrated into AP Physics 2 and AP Chemistry rather than Physics 1/C, so no feeder link is possible.热力学与热与若干 HS Physics 单元相联。功、能与功率指南定义了第一定律所扩展的机械功与能量框架;气体定律与动理论及压容功相联。在 AP 和 IB 层面,热力学是专门主题。IB Physics HL 指南深入涵盖热力学主题。注意本仓库中不存在 AP 热力学单元,因为热力学整合在 AP Physics 2 和 AP Chemistry 中,而非 Physics 1/C,因此无法提供衔接链接。

Within High School Physics.在 HS Physics 内部。

The Work, Energy and Power guide builds $W = F\Delta x$, $KE = \tfrac{1}{2}mv^2$, and gravitational PE — the mechanical energy side of the first law. The Waves and Sound unit connects standing waves and resonance (cavity modes) to concepts that appear in thermodynamic systems. The Laws of Thermodynamics in §6 are the conceptual bridge to any AP or IB course that treats heat engines, refrigerators, and entropy quantitatively.功、能与功率指南构建了 $W = F\Delta x$、$KE = \tfrac{1}{2}mv^2$ 及重力势能 — 第一定律的机械能一侧。波与声单元将驻波和共振(腔体模式)与热力学系统中出现的概念相联。§6 的热力学定律是通往任何定量处理热机、制冷机和熵的 AP 或 IB 课程的概念桥梁。

Across the IB feeders in this repo.本仓库中的 IB 衔接单元。

No AP thermodynamics unit exists in this repo (thermodynamics is in AP Physics 2 and AP Chemistry, not AP Physics 1/C). The IB Physics HL curriculum covers thermodynamics quantitatively; if an IB Physics HL Study Guide is available in this repo, it extends the second-law and heat-engine content of §6–§7 with Carnot cycle diagrams, entropy change calculations ($\Delta S = Q/T$), and P-V diagrams for ideal-gas processes. For IB Chemistry HL students, the enthalpy and entropy coverage in IB Chemistry also builds on the thermodynamic foundation laid here.本仓库中不存在 AP 热力学单元(热力学在 AP Physics 2 和 AP Chemistry 中,而非 AP Physics 1/C)。IB Physics HL 课程定量涵盖热力学;若本仓库中有 IB Physics HL 学习指南,它将以卡诺循环图、熵变计算($\Delta S = Q/T$)和理想气体过程的 P-V 图,扩展 §6–§7 的第二定律和热机内容。对于 IB Chemistry HL 学生,IB Chemistry 中的焓与熵内容也建立在此处奠定的热力学基础上。