📚 Year 10 CAIE Science: Case Study Practice | Year 10 CAIE 科学：案例分析实战演练
Case studies are at the heart of the CAIE IGCSE Science examinations. They challenge you to move beyond recalling facts and to apply your understanding in unfamiliar contexts. In this article, we will work through eight carefully crafted case studies spanning Chemistry, Physics and Biology. Each one will develop your skills in data analysis, experimental design, evaluation and drawing evidence-based conclusions.
案例分析是CAIE IGCSE科学考试的核心。它们要求你不再仅仅回忆事实，而是将理解应用到不熟悉的情境中。本文将带你深入八个精心设计的案例，覆盖化学、物理和生物学。每个案例都会培养你的数据分析、实验设计、评价以及基于证据得出结论的能力。
1. Why Case Studies Matter | 为什么案例分析很重要？
In Paper 4 (Extended) or Paper 2 (Multiple Choice) of the IGCSE Science exams, you will often encounter scenarios where you must interpret graphs, tables and experimental procedures. Case study practice trains you to think like a scientist: to identify variables, spot anomalies, evaluate reliability and suggest improvements. Strong performance in these questions can significantly boost your final grade.
在IGCSE科学考试的试卷4（扩展）或试卷2（选择题）中，你经常会遇到需要解读图表、表格和实验步骤的场景。案例分析练习训练你像科学家一样思考：识别变量、发现异常值、评价可靠性并提出改进建议。在这些题目上表现出色可以大大提升你的最终成绩。
Effective case study technique involves reading the stem carefully, noting the command words ('describe', 'explain', 'calculate', 'suggest') and linking your answer directly to the evidence provided. Always use scientific terminology precisely. For example, instead of writing 'the reaction got faster', state 'the initial rate of reaction increased' and support it with data.
有效的案例分析技巧包括仔细阅读题干，注意指令词（“描述”、“解释”、“计算”、“建议”），并将答案直接与所给证据联系起来。始终准确使用科学术语。例如，不要写“反应变快了”，而要写“初始反应速率增加”并用数据支持。
2. Case Study 1: Analysing Water Purity (Chemistry) | 案例1：分析水的纯度（化学）
A student tested three water samples labelled X, Y and Z. She measured the boiling point, electrical conductivity and performed a test with silver nitrate solution acidified with nitric acid. The results are shown in the table below.
一名学生测试了标记为X、Y和Z的三种水样。她测量了沸点、电导率，并用酸化硝酸银溶液进行了测试。结果如下表所示。
| Test | Sample X | Sample Y | Sample Z |
|---|---|---|---|
| Boiling point / °C | 100.0 | 102.3 | 100.0 |
| Electrical conductivity | Very low | High | Moderate |
| Silver nitrate + nitric acid | No precipitate | White precipitate | Faint white precipitate |
Use the data to identify which sample is pure water. Explain your reasoning by referring to the concept that pure substances have fixed melting and boiling points and that dissolved ionic compounds affect both boiling point and conductivity.
利用数据确定哪个样品是纯水。请解释你的推理，并提及纯物质具有固定的熔点和沸点以及溶解的离子化合物会影响沸点和电导率的概念。
Sample X boils at exactly 100 °C and shows very low conductivity, indicating an absence of dissolved ions. The silver nitrate test is negative, confirming no chloride ions are present. Sample Y boils above 100 °C, conducts electricity well and gives a heavy white precipitate, showing it contains dissolved salts. Sample Z has a sharp boiling point at 100 °C but conducts moderately, suggesting a small amount of ionic impurity; the faint precipitate supports this. Therefore, Sample X is the purest water.
样品X的沸点恰好为100 °C，电导率非常低，表明没有溶解的离子。硝酸银测试结果为阴性，证实不存在氯离子。样品Y的沸点高于100 °C，导电性好，并产生大量白色沉淀，说明它含有溶解的盐。样品Z的沸点精确为100 °C，但导电性中等，表明含有少量离子杂质；轻微的沉淀也支持这一点。因此，样品X是最纯净的水。
Always state that pure water is a covalent compound that does not conduct electricity, whereas dissolved ions such as Na⁺ and Cl⁻ carry charge. A rise in boiling point is called boiling point elevation and is a colligative property linked to the number of dissolved particles.
一定要说明纯水是共价化合物，不导电，而溶解的离子如Na⁺和Cl⁻会携带电荷。沸点升高被称为沸点上升，是一种依数性质，与溶解颗粒的数量有关。
3. Case Study 2: Investigating Motion on a Slope (Physics) | 案例2：研究斜坡上的运动（物理）
A dynamics trolley was released from rest on a ramp at a height of 0.40 m. The distance travelled and the time taken were recorded using a light gate and a data logger. The results are presented in the table.
一辆动力小车从斜坡上0.40米高处静止释放。使用光门和数据记录器记录了行进的距离和所用时间。结果如下表所示。
| Distance s / m | Time t / s |
|---|---|
| 0.00 | 0.00 |
| 0.20 | 0.45 |
| 0.40 | 0.63 |
| 0.60 | 0.77 |
| 0.80 | 0.89 |
| 1.00 | 1.00 |
Calculate the average speed of the trolley between 0.20 m and 0.80 m. Use the equation:
计算小车在0.20 m到0.80 m之间的平均速度。使用公式：
average speed = Δs / Δt
The change in distance Δs = 0.80 m – 0.20 m = 0.60 m. The change in time Δt = 0.89 s – which distance was covered; it is not necessarily the instantaneous speed at the midpoint.
距离变化Δs = 0.80 m – 0.20 m = 0.60 m。时间变化Δt = 0.89 s – 0.45 s = 0.44 s。因此平均速度 = 0.60 / 0.44 ≈ 1.36 m/s。这个值代表距离覆盖的平均速率，不一定是中点处的瞬时速度。
To investigate whether the trolley was accelerating uniformly, the student could calculate the speed over each 0.20 m interval and see if it increases by equal amounts per unit time. A velocity–time graph could then be plotted; a straight line through the origin would indicate constant acceleration.
为了研究小车是否匀加速运动，学生可以计算每段0.20 m间隔的速度，看看它是否每单位时间增加相等的量。然后可以绘制速度-时间图；一条通过原点的直线将表明加速度恒定。
A significant source of error in this experiment is the reaction time when releasing the trolley, which can be reduced by using an electromagnetic release system. Frictional forces between the trolley wheels and the ramp also cause energy loss, so the measured speeds may be slightly lower than the theoretical values predicted by energy conservation.
本实验的一个重要误差来源是释放小车时的反应时间，可以通过使用电磁释放系统来减少。小车车轮与斜坡之间的摩擦力也会导致能量损失，因此测得的速度可能略低于能量守恒预测的理论值。
4. Case Study 3: Enzyme Activity and Temperature (Biology) | 案例3：酶活性与温度（生物）
An investigation was carried out into the effect of temperature on the activity of amylase, an enzyme that breaks down starch into maltose. Starch agar plates were prepared with wells cut into them. Amylase solution was placed in each well, and the plates were incubated at different temperatures for 10 minutes. After incubation, iodine solution was poured over the plates, and the diameter of the clear zone around each well (where starch was digested) was measured.
进行了一项温度对淀粉酶活性影响的探究。淀粉酶是一种将淀粉分解为麦芽糖的酶。准备了淀粉琼脂平板，并在上面打孔。将淀粉酶溶液放入每个孔中，平板在不同温度下培养10分钟。培养后，将碘液倒在平板上，测量每个孔周围透明圈（淀粉被消化处）的直径。
| Temperature / °C | Diameter of clear zone / mm |
|---|---|
| 10 | 4 |
| 20 | 11 |
| 30 | 18 |
| 40 | 22 |
| 50 | 17 |
| 60 | 5 |
Explain the relationship between temperature and enzyme activity as shown by the data. Use your knowledge of enzyme structure and denaturation.
用数据解释温度与酶活性之间的关系。运用你关于酶结构和变性的知识。
The clear zone diameter increases from 4 mm at 10 °C to a maximum of 22 mm at 40 °C, indicating that the rate of starch digestion rises as temperature increases. This is because molecules have more kinetic energy, so collisions between enzyme and substrate are more frequent and energetic. Above 40 °C, the diameter decreases, falling to 5 mm at 60 °C. At high temperatures, the enzyme’s active site is permanently altered due to breaking of bonds (denaturation), so the substrate no longer fits.
透明圈直径从10 °C时的4 mm增加到40 °C时的最大22 mm，表明淀粉消化速率随温度升高而增加。这是因为分子具有更多的动能，因此酶与底物之间的碰撞更频繁、更有力。高于40 °C后，直径减小，60 °C时降至5 mm。在高温下，酶的活性位点因键的断裂而发生永久性改变（变性），因此底物不再适配。
The optimum temperature for this amylase is around 40 °C. A control well containing boiled enzyme would show no clear zone, confirming that activity depends on an intact protein structure. When describing such investigations, it is essential to mention that the iodine test identifies the presence of starch: a blue-black colour appears where starch remains undigested.
这种淀粉酶的最适温度大约是40 °C。含煮沸酶的对照孔将不出现透明圈，证实活性依赖于完整的蛋白质结构。在描述此类探究时，必须提及碘液测试可鉴定淀粉的存在：在未消化的淀粉处会出现蓝黑色。
5. Case Study 4: Electrical Circuit Fault Diagnosis (Physics) | 案例4：电路故障诊断（物理）
A student built a simple series circuit containing a battery, a switch, two identical filament lamps L₁ and L₂, and an ammeter. When the switch was closed, neither lamp lit and the ammeter read zero. A voltmeter was then connected across various components with the switch closed. The readings are given below.
一名学生搭建了一个简单的串联电路，包含电池、开关、两个相同的白炽灯L₁和L₂以及一个电流表。闭合开关后，两个灯都不亮，电流表读数为零。然后在开关闭合的情况下，用电压表跨接各个元件，读数如下。
| Voltmeter connected across | Reading / V |
|---|---|
| Battery terminals | 6.0 |
| Lamp L₁ | 6.0 |
| Lamp L₂ | 0.0 |
| Closed switch | 0.0 |
Deduce the location and nature of the fault. Explain your reasoning fully.
推断故障的位置和性质。充分解释你的推理过程。
The battery gives a terminal voltage of 6.0 V, so the power supply is working. The ammeter reads zero (no current), indicating a break in the series circuit. When the voltmeter is placed across L₁, it reads 6.0 V, which is the full battery voltage. This means that the voltmeter is effectively completing the circuit across a break. Since a good lamp has a low resistance, the only way to get the full 6.0 V across L₁ is if L₁ has a very high resistance or is open circuit. L₂ shows 0.0 V, confirming that no current flows and the break is not there. Therefore, lamp L₁ has a broken filament (open circuit).
电池端电压为6.0 V，因此电源正常。电流表读数为零（无电流），表明串联电路存在断路。当电压表跨接L₁时，读数为6.0 V，即全部电池电压。这意味着电压表实际上跨接在断路处完成了电路。由于正常灯泡电阻较低，要在L₁两端得到全部6.0 V的唯一可能是L₁具有极高的电阻或者处于断路状态。L₂显示0.0 V，证实没有电流流通，且故障不在那里。因此，灯泡L₁的灯丝断裂（断路）。
In a series circuit, if one component fails as an open circuit, the entire current stops. The full supply voltage appears across the faulty component when measured with a high-resistance voltmeter. This diagnostic method is commonly used to find blown fuses or broken wires.
在串联电路中，如果一个元件断路，整个电流停止。用高电阻电压表测量时，全部电源电压会出现在故障元件两端。这种诊断方法常用于查找熔断的保险丝或断开的导线。
6. Case Study 5: Food Tests and Nutrient Identification (Biology/Chemistry) | 案例5：食物测试与营养物质鉴定（生物/化学）
A sample of an unknown food was ground with water and filtered. The filtrate was divided into four test tubes and the following tests were carried out. The observations are recorded.
将一份未知食物加水研磨并过滤。滤液分装在四个试管中，进行了以下测试。观察到的情况记录如下。
| Test | Observation |
|---|---|
| Benedict's solution, heated | Brick-red precipitate formed |
| Iodine solution | Remained yellow-brown |
| Biuret test (sodium hydroxide + copper sulfate) | Purple colour developed |
| Ethanol emulsion test, poured into water | Cloudy white emulsion formed |
State which nutrients are present in the food sample. Give the evidence for each.
说明食物样品中含有哪些营养物质。给出每种营养物质的证据。
The brick-red precipitate with Benedict's solution indicates the presence of reducing sugars, such as glucose. Iodine solution remaining yellow-brown shows that starch is absent. The development of a purple colour in the Biuret test confirms the presence of proteins (peptide bonds). A cloudy white emulsion with ethanol indicates the presence of lipids (fats/oils).
与本尼迪克特试剂产生的砖红色沉淀表明存在还原糖，例如葡萄糖。碘液仍为黄棕色说明不含淀粉。双缩脲测试中出现紫色证实存在蛋白质（肽键）。加入乙醇后产生浑浊白色乳状液表明存在脂质（脂肪/油）。
When writing answers, remember that Benedict's test detects reducing sugars (not all carbohydrates) and must be heated. The Biuret test requires an alkaline environment and a few drops of dilute copper sulfate. The ethanol test works because lipids are soluble in ethanol but form a fine suspension in water, scattering light. Never miss the key detail that the sample must be in solution for Benedict's and Biuret tests.
书写答案时请记住，本尼迪克特测试检测的是还原糖（不是所有碳水化合物），并且需要加热。双缩脲测试需要碱性环境和几滴稀硫酸铜。乙醇乳液测试有效是因为脂质溶于乙醇，但在水中形成微细悬浮液，散射光。绝不要遗漏关键细节：进行本尼迪克特和双缩脲测试时，样品必须呈溶液状态。
7. Case Study 6: Acid–Base Titration and Calculation (Chemistry) | 案例6：酸碱滴定与计算（化学）
A student titrated 25.0 cm³ of hydrochloric acid (HCl) with sodium hydroxide solution (NaOH) of concentration 0.100 mol/dm³. Three consistent burette readings were obtained, as shown below. The indicator used was phenolphthalein.
一名学生用浓度为0.100 mol/dm³的氢氧化钠溶液（NaOH）滴定25.0 cm³的盐酸（HCl）。获得了三次一致的滴定管读数，如下所示。使用的指示剂是酚酞。
| Titration | Final burette reading / cm³ | Initial burette reading / cm³ | Volume used / cm³ |
|---|---|---|---|
| Rough | 23.80 | 0.00 | 23.80 |
| 1 | 22.50 | 0.00 | 22.50 |
| 2 | 22.40 | 0.00 | 22.40 |
| 3 | 22.45 | 0.00 | 22.45 |
Calculate the mean titre using the concordant results (within 0.20 cm³). Hence determine the concentration of the hydrochloric acid in mol/dm³. The equation for the reaction is:
使用一致的读数（相差在0.20 cm³以内）计算平均滴定体积。由此确定盐酸的浓度，单位为mol/dm³。反应方程式为：
HCl + NaOH → NaCl + H₂O
oles of NaOH used = concentration × volume (in dm³) = 0.100 × (22.45/1000) = 0.002245 mol. The mole ratio of HCl to NaOH is 1:1, so moles of HCl = 0.002245 mol. Concentration of HCl = moles / volume in
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