Mastering Physical Changes Application Questions (Matter 1.2.1) | 掌握物理变化应用题（物质1.2.1）

📚 Mastering Physical Changes Application Questions (Matter 1.2.1) | 掌握物理变化应用题（物质1.2.1）

Physical changes are transformations where no new substances are formed – think of melting, boiling, dissolving, or thermal expansion. In IGCSE Physics, the topic ‘Matter 1.2.1 Physical Changes’ includes not only the particle model but also density, changes of state, and energy transfers. Exam application questions often combine several of these ideas, asking you to calculate energy, interpret graphs, or analyse experimental data. This article guides you through the most effective techniques to handle such questions accurately and efficiently, with paired explanations in English and Chinese.

物理变化是指不生成新物质的变化，例如熔化、沸腾、溶解或热膨胀。在 IGCSE 物理中，“物质 1.2.1 物理变化”不仅涉及粒子模型，还包括密度、状态变化和能量转移。考试应用题常常结合多个概念，要求计算能量、解读图线或分析实验数据。本文将带你掌握最有效的技巧，准确高效地应对这些题目，并提供中英对照讲解。

1. Identifying Physical Changes in Context | 在语境中识别物理变化

The first step in many application questions is to decide whether a process is a physical change. Look for clues: change of state (melting, freezing, boiling, condensation, sublimation), change in shape or size, dissolving, or expansion. No new chemical substance is produced, so the particles themselves remain the same – only their arrangement or energy changes. A classic trick question is to present ‘dissolving sugar in tea’ (physical) versus ‘baking a cake’ (chemical). If mass appears to be ‘lost’ during boiling, remember it is just water vapour escaping; the total mass of the closed system is conserved.

许多应用题的第一步是判断过程是否为物理变化。寻找线索：状态变化（熔化、凝固、沸腾、冷凝、升华）、形状或大小改变、溶解或膨胀。没有新的化学物质生成，因此粒子本身没有变化——只是排列或能量不同。常见的陷阱题有“把糖溶在茶里”（物理）与“烘焙蛋糕”（化学）。如果沸腾时质量似乎“减少”，要记住那只是水蒸气逸散；封闭体系的总质量是守恒的。

2. Mass Conservation in Physical Processes | 物理过程中的质量守恒

Mass is always conserved during a physical change. When 50 g of ice melts, you obtain exactly 50 g of liquid water. When a metal block expands, its mass stays the same even though its volume increases. This principle is the foundation for many calculations. For example, if you are asked to find the volume of water produced after melting ice, first confirm the mass, then apply the density formula. Always remind yourself: ‘mass before = mass after’. In exam questions, watch out for situations where a container is open and some vapour escapes – the mass of the remaining substance changes, but the total mass of water substance (liquid + vapour) is conserved.

物理变化中质量总是守恒的。50 g 冰熔化后，你恰好得到 50 g 液态水。金属块膨胀时，尽管体积增大了，但质量保持不变。这一原理是许多计算的基础。例如，如果题目要求计算冰熔化后水的体积，先确认质量，再运用密度公式。要时刻提醒自己：“变化前质量 = 变化后质量”。考试中要留意，若容器敞开且有蒸气逸出，则剩余物质的质量会改变，但水这一物质的总质量（液态 + 气态）仍是守恒的。

3. Density Calculations and State Changes | 密度计算与状态变化

Density links mass and volume: ρ = m / V. Since mass is fixed during a state change, any change in volume causes a corresponding change in density. For example, ice (density about 0.92 g/cm³) floats on water (density 1.00 g/cm³) because it expands on freezing. A common question: ‘An ice cube of mass 180 g has density 0.90 g/cm³. Calculate its volume. If it melts and the water formed has density 1.0 g/cm³, what is the new volume?’ The answer: V_ice = 180 / 0.90 = 200 cm³; V_water = 180 / 1.0 = 180 cm³. Notice the volume decreases by 20 cm³, explaining why the water level in a glass of ice water stays constant as it melts (the submerged part already displaces that volume).

密度将质量与体积联系起来：ρ = m / V。因为状态变化时质量不变，体积的任何改变都会引起密度的相应变化。例如，冰（密度约 0.92 g/cm³）浮在水（密度 1.00 g/cm³）上，因为水结冰时膨胀。常见考题：“一块质量 180 g 的冰密度为 0.90 g/cm³，求其体积。若它熔化后水的密度为 1.0 g/cm³，新体积是多少？”答案：V_冰 = 180 / 0.90 = 200 cm³；V_水 = 180 / 1.0 = 180 cm³。注意体积减小了 20 cm³，这也解释了为什么一杯冰水在冰块熔化时液面高度保持不变（冰块浸没部分早已排开了该体积的水）。

4. Thermal Expansion and Contraction Problems | 热胀冷缩问题

When a solid or liquid is heated, its particles vibrate more vigorously and move slightly apart, causing expansion. Application questions often involve linear expansion of rails, bimetallic strips, or volume expansion in thermometers. The formula for linear expansion is ΔL = α L₀ Δθ, where α is the coefficient of linear expansion. For simple IGCSE problems, you may just need to explain why gaps are left in railway tracks or why a bimetallic strip bends. If a calculation is given, the key is to identify the original length L₀, the temperature change Δθ (in °C or K), and the correct coefficient α. Always use the same length unit throughout. Remember that cooling causes contraction, and the same formula applies with a negative Δθ.

固体或液体受热时，粒子振动加剧并稍稍分开，引起膨胀。应用题常涉及铁轨的线性膨胀、双金属片或温度计的体积膨胀。线性膨胀公式为 ΔL = α L₀ Δθ，其中 α 是线膨胀系数。对于简单的 IGCSE 问题，你可能只需解释为什么铁轨连接处要留缝隙，或者双金属片为何弯曲。如果需要进行计算，关键是确定原始长度 L₀、温度变化 Δθ（单位 °C 或 K）以及正确的系数 α。整个过程必须采用统一的长度单位。记住冷却会引起收缩，此时公式中的 Δθ 为负值即可。

5. Energy for State Changes: Specific Latent Heat | 物态变化能量：比潜热

A change of state happens at constant temperature and requires energy. The energy needed to melt a solid or boil a liquid without temperature change is given by Q = m L, where L is the specific latent heat. For melting, use L_f (fusion); for boiling, use L_v (vaporisation). A typical question: ‘How much energy is needed to melt 0.25 kg of ice at 0 °C? (L_f = 3.34 × 10⁵ J/kg)’ Solution: Q = 0.25 × 3.34 × 10⁵ = 8.35 × 10⁴ J. When a substance freezes or condenses, it releases the same amount of energy. Always check that the mass is in kg and the latent heat in J/kg. If the mass is given in grams, convert to kg by dividing by 1000.

状态变化在恒定温度下发生，并需要能量。在不改变温度的情况下熔化固体或沸腾液体所需的能量由 Q = m L 给出，其中 L 是比潜热。熔化时用 L_f（熔解），沸腾时用 L_v（汽化）。典型题目：“将 0.25 kg、0 °C 的冰熔化需要多少能量？（L_f = 3.34 × 10⁵ J/kg）”解答：Q = 0.25 × 3.34 × 10⁵ = 8.35 × 10⁴ J。当物质凝固或液化时，会放出相同的能量。务必确认质量以 kg 为单位，比潜热以 J/kg 为单位。若质量以克给出，则除以 1000 换算为 kg。

6. Interpreting Heating and Cooling Curves | 解读加热与冷却曲线

A heating curve plots temperature against time (or energy supplied) for a substance being heated. The flat sections (plateaux) represent changes of state where energy is absorbed but temperature does not change. The length of a plateau is proportional to the specific latent heat – a longer plateau for vaporisation means L_v is larger than L_f. The sloped sections correspond to temperature changes and reflect the specific heat capacity. To calculate the energy supplied during a whole process, break it into segments: for each sloped segment use Q = m c Δθ; for each flat segment use Q = m L. Add them up. Remember to read the graph carefully: the time axis may represent minutes, and the heating power might be constant, so you can deduce that a longer plateau requires more energy.

加热曲线描绘了物质受热时温度与时间（或供能）的关系。平坦的部分（平台）代表状态变化，此时能量被吸收但温度不变。平台的长度与比潜热成正比——汽化时的平台更长，意味着 L_v 大于 L_f。倾斜段对应温度变化，反映比热容。计算整个过程的能量供给时，将其拆分为多个阶段：每个倾斜段用 Q = m c Δθ；每个平台段用 Q = m L。最后相加。记住仔细阅读图线：时间轴可能以分钟为单位，加热功率可能是恒定的，因此可以推断较长平台需要的能量更多。

7. Specific Heat Capacity Calculations | 比热容计算

The energy required to change the temperature of a substance without a change of state is given by Q = m c Δθ, where c is the specific heat capacity. Typical values: water has c = 4200 J/(kg °C), ice ≈ 2100 J/(kg °C). When solving problems, identify the initial and final temperatures and calculate Δθ = θ_final − θ_initial (the magnitude matters for energy, so take the positive difference). If a 0.50 kg iron block (c = 450 J/(kg °C)) cools from 80 °C to 20 °C, the energy released is Q = 0.50 × 450 × (80−20) = 0.50 × 450 × 60 = 13 500 J. Always include units and check that the specific heat capacity value matches the substance and its state.

在不发生状态变化的情况下，改变物质温度所需要的能量由 Q = m c Δθ 给出，其中 c 是比热容。典型值：水的 c = 4200 J/(kg °C)，冰约为 2100 J/(kg °C)。解题时，确定初始和末态温度，计算 Δθ = θ_末 − θ_初（大小即为变化量，能量计算取正值）。若一个 0.50 kg 的铁块（c = 450 J/(kg °C)）从 80 °C 冷却到 20 °C，释放能量为 Q = 0.50 × 450 × (80−20) = 0.50 × 450 × 60 = 13 500 J。务必标注单位，并确认比热容数值与物质及其状态匹配。

8. Mixing Problems and Thermal Equilibrium | 混合问题与热平衡

When hot and cold substances are mixed in an insulated container, heat lost by the hot part equals heat gained by the cold part until they reach the same final temperature θ_f. The principle is: m₁ c₁ (θ₁ − θ_f) = m₂ c₂ (θ_f − θ₂). If the substances are the same (e.g., hot water and cold water), c cancels out and the final temperature becomes a weighted average. In a typical problem: 200 g of water at 80 °C is mixed with 300 g of water at 20 °C. Find the final temperature. Since c is the same, 0.20 × (80 − θ_f) = 0.30 × (θ_f − 20). Solve to get θ_f ≈ 44 °C. Always use mass in kg for consistency with specific heat capacity units, though here it cancels. Watch out for phase changes: if the mixture reaches melting or boiling point, you may need to include latent heat terms.

当热水和冷水在绝热容器中混合时，高温部分失去的热量等于低温部分获得的热量，直到它们达到相同的末温 θ_f。原理是：m₁ c₁ (θ₁ − θ_f) = m₂ c₂ (θ_f − θ₂)。如果物质相同（如热水与冷水），c 可约去，最终温度成为加权平均值。典型问题：将 200 g、80 °C 的水与 300

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）