📚 Year 9 Edexcel Science: Case Study Practice in Action | Year 9 Edexcel 科学：案例分析实战演练
Case studies are a brilliant way to sharpen your scientific thinking. In this article, we work through real-world scenarios drawn from the Edexcel Year 9 curriculum, blending biology, chemistry and physics. Each case study models how to analyse data, spot patterns and apply core principles – exactly the skills you need to ace the exam.
案例分析是锻炼科学思维的绝佳方式。本文依据 Edexcel 九年级课程设计了多个真实情境，融合生物、化学与物理。每个案例都示范了如何分析数据、发现规律并运用核心原理——这些正是你在考试中脱颖而出的关键技能。
1. Case Study 1 – Analysing a Sprinter’s Motion | 案例一：短跑运动员的运动分析
Alice runs a 100 m race. Her coach records the time taken to cover each 20 m segment. The data are shown in the table below.
爱丽丝参加 100 米赛跑，教练记录了她每 20 米区间所用时间。数据见下表。
| Distance interval (m) | Time (s) |
|---|---|
| 0 – 20 | 3.2 |
| 20 – 40 | 2.8 |
| 40 – 60 | 2.9 |
| 60 – 80 | 3.1 |
| 80 – 100 | 3.4 |
Calculate the average speed for the whole race. Use the formula: speed = distance ÷ time. Total time = 3.2 + 2.8 + 2.9 + 3.1 + 3.4 = 15.4 s. Average speed = 100 m ÷ 15.4 s ≈ 6.49 m/s. Notice that her speed was highest in the 20–40 m interval (20 m ÷ 2.8 s ≈ 7.14 m/s) and lowest in the final 20 m, showing fatigue.
计算全程平均速度。使用公式：速度 = 距离 ÷ 时间。总时间 = 3.2 + 2.8 + 2.9 + 3.1 + 3.4 = 15.4 秒。平均速度 = 100 米 ÷ 15.4 秒 ≈ 6.49 米/秒。注意她在 20–40 米区间速度最快（20 米 ÷ 2.8 秒 ≈ 7.14 米/秒），最后 20 米最慢，表明出现了疲劳。
2. Case Study 2 – Interpreting a Force–Extension Graph | 案例二：力–伸长图像解读
A student hangs masses from a spring and records the extension. The data are plotted, showing a straight line up to a load of 6 N. At 7 N the line begins to curve. Explain what this tells us about the spring’s behaviour.
一名学生将不同质量的砝码悬挂在弹簧下，记录伸长量。数据绘成图像，在 6 牛顿的负荷以内为直线，7 牛顿时开始弯曲。请解释这说明了弹簧怎样的行为。
The straight region obeys Hooke’s Law: extension is directly proportional to force (F = kx). The gradient gives the spring constant, k. Beyond 6 N, the spring exceeds its elastic limit; it undergoes plastic deformation and will not return to its original length when the load is removed.
直线区域遵循胡克定律：伸长量与力成正比（F = kx）。斜率给出劲度系数 k。超过 6 牛顿后，弹簧超出了弹性极限，发生塑性形变，卸去负荷后将无法恢复到原始长度。
3. Case Study 3 – Food Chains and DDT Bioaccumulation | 案例三：食物链与 DDT 生物累积
In a lake ecosystem, researchers measured DDT concentration (in parts per million, ppm) in different organisms: water (0.00005 ppm), phytoplankton (0.04 ppm), zooplankton (0.5 ppm), small fish (2 ppm), and large predatory fish (25 ppm). Explain the trend using ecological concepts.
在某湖泊生态系统中，研究人员测定了不同生物体内的 DDT 浓度（单位 ppm）：水体（0.00005 ppm），浮游植物（0.04 ppm），浮游动物（0.5 ppm），小鱼（2 ppm），大型掠食性鱼类（25 ppm）。请用生态学概念解释这一趋势。
DDT is a persistent pesticide that is not easily broken down. Phytoplankton absorb it from water, accumulating a slightly higher concentration. At each trophic level, the toxin becomes more concentrated because consumers eat many organisms from the level below. This is biomagnification. The top predator (large fish) shows the highest concentration, which can cause reproductive failure.
DDT 是一种持久性农药，不易分解。浮游植物从水中吸收 DDT，积累稍高的浓度。由于消费者摄食大量低营养级生物，毒素在每一营养级都会进一步浓缩，这就是生物放大作用。处于最高营养级的掠食性鱼类浓度最高，可能导致繁殖失败。
4. Case Study 4 – Energy Transfer in a Falling Object | 案例四：落体的能量转化
A ball of mass 0.5 kg is dropped from a height of 10 m. Calculate its gravitational potential energy (GPE) at the top and estimate its speed just before hitting the ground (assuming no air resistance). Use g = 10 N/kg.
一个质量为 0.5 千克的小球从 10 米高处落下。计算小球在最高点的重力势能（GPE），并估算它触地前的速度（不计空气阻力）。取 g = 10 牛/千克。
GPE = mgh = 0.5 × 10 × 10 = 50 J. As it falls, GPE converts to kinetic energy (KE). Just before impact, KE = 50 J = (1/2) m v². So v² = (2 × 50) / 0.5 = 200, giving v = √200 ≈ 14.1 m/s. This assumes all potential energy turns into kinetic energy.
重力势能 = mgh = 0.5 × 10 × 10 = 50 焦耳。下落过程中，重力势能转化为动能（KE）。触地瞬间，动能 = 50 焦耳 = (1/2) m v²，因此 v² = (2 × 50) / 0.5 = 200，v = √200 ≈ 14.1 米/秒。这假设了全部势能都转化为动能。
5. Case Study 5 – Rate of Reaction and Temperature | 案例五：反应速率与温度
A student investigates the reaction between magnesium ribbon and excess hydrochloric acid. She records the volume of hydrogen gas produced every 10 seconds at 20 °C and at 40 °C. At 20 °C, the reaction produces 24 cm³ of gas in the first minute; at 40 °C it produces 60 cm³ in the same time. Explain this observation using particle theory.
一名学生研究镁条与过量盐酸的反应。她分别在 20 °C 和 40 °C 下记录每 10 秒产生的氢气体积。20 °C 时，第一分钟产生 24 cm³ 氢气；40 °C 时，相同时间产生 60 cm³。请用粒子理论解释该观察结果。
The increase in temperature gives reactant particles greater kinetic energy. They move faster and collide more frequently. More importantly, a larger fraction of collisions now have energy equal to or greater than the activation energy. Both factors increase the frequency of successful collisions, so the rate of reaction is higher at 40 °C.
温度升高使反应物粒子获得更大的动能，运动加快，碰撞更频繁。更重要的是，此时有更多比例的碰撞能达到或超过活化能。这两个因素均增加了有效碰撞频率，因而 40 °C 下反应速率更高。
6. Case Study 6 – Neutralisation and Temperature Change | 案例六：中和反应与温度变化
A student mixes 25 cm³ of sodium hydroxide solution with 25 cm³ of hydrochloric acid in a polystyrene cup. The temperature rises from 21.0 °C to 28.5 °C. Calculate the temperature change and explain why polystyrene is used.
一名学生在聚苯乙烯杯中混合 25 cm³ 氢氧化钠溶液与 25 cm³ 盐酸。温度从 21.0 °C 升高至 28.5 °C。计算温度变化并解释为何使用聚苯乙烯。
Temperature change ΔT = 28.5 – 21.0 = 7.5 °C. The reaction between an acid and an alkali is exothermic, releasing heat. Polystyrene is a good thermal insulator; it reduces heat loss to the surroundings, making the measured temperature change more accurate.
温度变化 ΔT = 28.5 – 21.0 = 7.5 °C。酸与碱的中和反应是放热反应，释放热量。聚苯乙烯是良好的绝热材料，可减少热量散失到周围环境，使测得的温度变化更准确。
7. Case Study 7 – Electrical Circuits: Finding an Unknown Resistance | 案例七：电路分析：求未知电阻
A pupil builds a series circuit with a 12 V battery, an ammeter and an unknown resistor R. The ammeter reads 0.4 A. Calculate the resistance of R. Then, what would happen to the current if a second identical resistor is added in series?
一名学生用 12 伏电池、电流表和一个未知电阻 R 组成串联电路。电流表读数为 0.4 安。计算 R 的阻值。如果再串联一个相同的电阻，电流将如何变化？
Using Ohm’s Law: R = V / I = 12 V / 0.4 A = 30 Ω. When a second 30 Ω resistor is added in series, total resistance doubles to 60 Ω. The current becomes I = 12 V / 60 Ω = 0.2 A, half the original value. This makes sense because for a fixed voltage, current is inversely proportional to resistance.
运用欧姆定律：R = V / I = 12 V / 0.4 A = 30 Ω。再串联一个 30 Ω 电阻后，总电阻加倍至 60 Ω。电流变为 I = 12 V / 60 Ω = 0.2 A，即原来的一半。这很合理：电压固定时，电流与电阻成反比。
8. Case Study 8 – Diffusion and Surface Area | 案例八：扩散与表面积
A cube of agar jelly containing universal indicator is cut into a large 2 cm × 2 cm × 2 cm cube and eight 1 cm × 1 cm × 1 cm cubes. All cubes are placed in hydrochloric acid. Explain why the smaller cubes decolourise faster.
一块含通用指示剂的琼脂凝胶被切成一个 2 cm × 2 cm × 2 cm 的大方块和八个 1 cm × 1 cm × 1 cm 的小方块。所有方块都浸入盐酸中。解释为什么小方块褪色更快。
Diffusion rate depends on surface area to volume ratio (SA:V). The large cube has SA = 24 cm², V = 8 cm³, SA:V = 3:1. Each small cube has SA = 6 cm², V = 1 cm³, SA:V = 6:1. The much larger surface area relative to volume allows acid to diffuse into the smaller cubes more quickly, so the indicator changes colour faster.
扩散速率取决于表面积与体积比（SA:V）。大方块表面积为 24 cm²，体积为 8 cm³，SA:V = 3:1。每个小方块表面积为 6 cm²，体积为 1 cm³，SA:V = 6:1。相对体积而言更大的表面积使酸能更快地扩散进入小方块，因此指示剂变色更快。
9. Case Study 9 – Inheritance and Pedigree Analysis | 案例九：遗传与系谱分析
Study the pedigree chart of a family with a recessive genetic disorder. Two unaffected parents have three children: one affected daughter and two unaffected sons. Determine the genotypes of the parents and explain the probability of their next child being affected.
研究某隐性遗传病家族的系谱图。一对表型正常的父母有三个孩子：一个患病的女儿和两个正常的儿子。确定父母的基因型，并解释他们下一个孩子患病的概率。
Since the disorder is recessive, an affected individual must have two recessive alleles (let’s use ‘aa’). The parents are unaffected but must each carry one recessive allele to have an ‘aa’ child, so both are heterozygous (Aa). For each pregnancy, the probability of an ‘aa’ offspring is 25 % (using a Punnett square: Aa × Aa → 1 AA : 2 Aa : 1 aa). The sex of the child does not matter because the disorder is not sex-linked.
由于该疾病为隐性遗传，患者必须有两个隐性等位基因（设为 “aa”）。父母表型正常，但为了生下 “aa” 的孩子，必然各携带一个隐性等位基因，因此两人均为杂合子（Aa）。每次怀孕，后代为 “aa” 的概率是 25%（由旁氏表得出：Aa × Aa → 1 AA : 2 Aa : 1 aa）。孩子的性别无关紧要，因为该病非伴性遗传。
10. Case Study 10 – Chemical Calculations and Conservation of Mass | 案例十：化学计算与质量守恒
When 5.6 g of iron filings react with excess copper sulfate solution, 6.4 g of copper metal is produced. What mass of iron(II) sulfate is formed? Explain your reasoning using the law of conservation of mass.
5.6 克铁屑与过量硫酸铜溶液反应，生成 6.4 克铜。问生成的硫酸亚铁的质量是多少？请用质量守恒定律解释你的推理。
The total mass of reactants equals the total mass of products. The reaction is: Fe + CuSO₄ → FeSO₄ + Cu. We know the mass of Fe reacted (5.6 g) and mass of Cu formed (6.4 g). The copper sulfate was in excess, but the copper sulfate solution contains water; we cannot simply add masses of solutions. Instead, mass of FeSO₄ = mass of Fe + mass of CuSO₄ that reacted – mass of Cu. However, a simpler approach: the iron atoms are replaced by copper atoms. Gain in mass of solid = mass of Cu – mass of Fe = 6.4 – 5.6 = 0.8 g. This gain is the difference between the sulfate portion and water, but to find FeSO₄ mass directly, note that moles of Fe = 5.6 / 56 = 0.1 mol, so moles of FeSO₄ = 0.1 mol, mass = 0.1 × 152 = 15.2 g. At Year 9 level, the key point is that mass is conserved; the iron atoms are simply replaced by copper, and the sulfate ions remain in solution. A measured evaporation of the solution would yield 15.2 g of FeSO₄.
反应物总质量等于生成物总质量。反应方程式：Fe + CuSO₄ → FeSO₄ + Cu。已知铁的质量为 5.6 克，生成的铜为 6.4 克。硫酸铜过量，但其溶液含水，不能简单加和。铁原子被铜原子置换。固体质量增加 = 6.4 – 5.6 = 0.8 克。从摩尔数入手：铁的物质的量 = 5.6 / 56 = 0.1 摩尔，所以硫酸亚铁也为 0.1 摩尔，质量 = 0.1 × 152 = 15.2 克。在九年级层面，重点是质量守恒：铁原子被铜原子取代，硫酸根离子留在溶液中。将溶液蒸干可获得 15.2 克硫酸亚铁。
11. Putting It All Together – A Mixed Science Scenario | 案例十一：综合科学情境实战
Read the following scenario: A camper boils water using a gas stove (propane combustion: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O). The water absorbs 33.6 kJ of energy and warms from 20 °C to 100 °C. The camper also notices that a nearby pond has green algae blooms. Explain the science behind these observations, linking energy transfer, chemical reactions, and ecology.
阅读以下情境：露营者用燃气炉（丙烷燃烧：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O）烧水。水吸收了 33.6 千焦能量，从 20 °C 加热至 100 °C。露营者还注意到附近池塘出现藻类大量繁殖。请解释这些现象背后的科学原理，联系能量传递、化学反应和生态学。
Combustion of propane is an exothermic reaction; chemical energy stored in bonds is released as heat. The heat transferred raises the water’s temperature: energy = mass × specific heat capacity × temperature rise (Q = mcΔT). Assuming 200 g of water, c = 4.2 J/g°C, ΔT = 80 °C, Q ≈ 67.2 kJ, but the water only absorbed 33.6 kJ, implying some heat was lost to the surroundings. The algae bloom may be caused by eutrophication: nutrient runoff (perhaps from washing) enters the pond, fuelling rapid algal growth. The algae block sunlight and deplete oxygen, harming aquatic life. This scenario links chemistry (combustion, energy), physics (heat transfer) and biology (ecosystems).
丙烷燃烧是放热反应；化学键中储存的化学能以热的形式释放。传递的热量使水温升高：能量 = 质量 × 比热容 × 温度变化（Q = mcΔT）。假设有 200 克水，c = 4.2 J/g°C，ΔT = 80 °C，Q ≈ 67.2 kJ，但水只吸收了 33.6 kJ，说明部分热量散失到环境中。藻类大量繁殖可能由富营养化引起：营养物质（或许是洗涤水）流入池塘，促使藻类迅速生长。藻类遮挡阳光并消耗氧气，危害水生生物。该情境将化学（燃烧、能量）、物理（热传递）与生物（生态系统）串联起来。
12. Final Tips for Case Study Success | 案例题高分技巧总结
Always identify the relevant scientific concept first. State the formula or principle clearly before plugging in numbers. Use data from the case study – quote figures directly. Show your working step by step. Finally, link your conclusion back to the real‑world context to earn those evaluation marks.
首先确定相关的科学概念。代入数值前，先把公式或原理清楚地写出来。直接引用案例中的数据，一步一步展示计算过程。最后，将结论与真实情境联系起来，这样才能拿到评价分。
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