📚 Case Study Practical Exercises for Year 10 Eduqas Science | 案例分析实战演练（Year 10 Eduqas 科学）
Case studies in science are not just about memorising facts—they train you to think like a scientist. In the Eduqas Year 10 Science curriculum, you will encounter practical scenarios that require you to interpret data, evaluate methods, and apply your understanding of Biology, Chemistry, and Physics to unfamiliar contexts. This article presents a series of guided case study exercises to help you build confidence, sharpen your analytical skills, and prepare effectively for your examinations. Work through each example carefully, paying attention to the language of measurement, variables, and scientific reasoning.
科学中的案例分析不仅仅是记忆事实——它们训练你像科学家一样思考。在Eduqas十年级科学课程中，你会遇到需要解读数据、评估方法并将你对生物、化学和物理的理解应用于陌生情境的实践场景。本文提供一系列有指导的案例分析练习，帮助你建立信心、提高分析能力，并有效备战考试。请仔细钻研每一个示例，注意测量的表述、变量和科学推理。
1. Introduction to Case Studies in Science | 科学案例研究导论
In Eduqas Science, case studies typically present a real-life or experimental situation, accompanied by tables of results, graphs, or descriptions of methods. You are expected to identify variables, suggest improvements, calculate means, and draw conclusions based on evidence. The key is to approach each case systematically: read the context, extract the data, link it to scientific principles, and then communicate your ideas clearly. This skill is assessed in both the practical components and the written examinations.
在Eduqas科学中，案例研究通常呈现一个现实生活或实验情境，并附有结果表格、图表或方法描述。你需要识别变量、提出改进建议、计算平均值并根据证据得出结论。关键是要系统地处理每一个案例：阅读背景，提取数据，将其与科学原理联系起来，然后清晰地表达你的想法。这项技能在实践部分和笔试中都会得到评估。
2. Case Study 1: Investigating the Effect of Temperature on Enzyme Activity | 案例1：探究温度对酶活性的影响
A student investigated how temperature affects the activity of the enzyme amylase, which breaks down starch. Amylase solution was added to starch suspension at five different temperatures: 10 °C, 20 °C, 30 °C, 40 °C, and 50 °C. Every 30 seconds, a drop of the mixture was tested with iodine solution. The time taken for the iodine to stop turning blue-black (indicating starch had been fully digested) was recorded. The results are shown in the table below.
一名学生研究了温度如何影响分解淀粉的酶（淀粉酶）的活性。将淀粉酶溶液分别加入置于五个不同温度（10 °C、20 °C、30 °C、40 °C 和 50 °C）下的淀粉悬浮液中。每隔30秒，取一滴混合液用碘液测试。记录碘液不再变蓝黑色（表明淀粉已完全被消化）所需的时间。结果如下表所示。
| Temperature / °C | Time taken for starch to disappear / s |
|---|---|
| 10 | 180 |
| 20 | 90 |
| 30 | 45 |
| 40 | 30 |
| 50 | 150 |
Analyse the data by calculating the rate of reaction at each temperature. Rate can be expressed as 1 / time (s⁻¹). Identify the optimum temperature. Explain why the time increases again at 50 °C, referring to enzyme denaturation. Suggest one way to improve the experiment, such as using a water bath to maintain constant temperatures and repeating each measurement three times for reliability.
分析数据时，计算每个温度下的反应速率。速率可表示为 1 / 时间（s⁻¹）。找出最适温度。解释为什么在50 °C时所需时间再次增加，并提及酶的变性。提出一个改进实验的方法，例如使用水浴以保持恒温，并每个测量重复三次以确保可靠性。
For example, the rate at 30 °C is 1/45 ≈ 0.0222 s⁻¹, while at 50 °C the rate drops to 1/150 ≈ 0.0067 s⁻¹. This shows that the enzyme begins to lose its shape at higher temperatures, reducing the rate of starch breakdown. The independent variable is temperature, the dependent variable is the time for starch digestion (or rate), and control variables include enzyme concentration, starch concentration, and pH.
例如，30 °C时的速率是 1/45 ≈ 0.0222 s⁻¹，而50 °C时的速率降至 1/150 ≈ 0.0067 s⁻¹。这表明酶在较高温度下开始失去其形状，降低了淀粉分解的速率。自变量是温度，因变量是淀粉消化所需的时间（或速率），控制变量包括酶浓度、淀粉浓度和pH值。
3. Case Study 2: Determining the Concentration of an Unknown Acid by Titration | 案例2：通过滴定确定未知酸的浓度
A titration was carried out to find the concentration of a sample of hydrochloric acid. 25.0 cm³ of the acid was placed in a conical flask, and a few drops of phenolphthalein indicator were added. Sodium hydroxide solution of concentration 0.100 mol/dm³ was added from a burette until the solution turned permanently pink. Three titre readings were obtained: 23.80 cm³, 23.70 cm³, and 23.90 cm³.
进行了一次滴定实验，以确定某盐酸样品的浓度。将25.0 cm³的酸置于锥形瓶中，加入几滴酚酞指示剂。用滴定管逐滴加入浓度为 0.100 mol/dm³ 的氢氧化钠溶液，直到溶液变为持久的粉红色。获得了三个滴定读数：23.80 cm³、23.70 cm³ 和 23.90 cm³。
Calculate the mean titre, ignoring any anomalous results (none in this case). The mean volume is (23.80 + 23.70 + 23.90) / 3 = 23.80 cm³. The balanced equation is: HCl + NaOH → NaCl + H₂O. This shows a 1:1 mole ratio. Using the formula: concentration of acid = (concentration of alkali × volume of alkali) / volume of acid. So, c₁ = (0.100 × 23.80) / 25.0 = 0.0952 mol/dm³. Always express the answer to three significant figures where appropriate.
计算平均滴定体积，忽略任何异常结果（此处无异常）。平均体积为 (23.80 + 23.70 + 23.90) / 3 = 23.80 cm³。反应方程式为：HCl + NaOH → NaCl + H₂O。这表明1:1摩尔比。使用公式：酸的浓度 = (碱的浓度 × 碱的体积) / 酸的体积。因此，c₁ = (0.100 × 23.80) / 25.0 = 0.0952 mol/dm³。适当情况下，答案应表达为三位有效数字。
When evaluating the method, consider possible sources of error: the indicator may have been added in excess, the burette may not have been rinsed with the alkali before use, or the conical flask might have been rinsed with acid, which would change the number of moles of acid. Improvements include ensuring all equipment is rinsed with the correct solution and reading the meniscus at eye level to avoid parallax errors.
评估实验方法时，考虑可能的误差来源：指示剂可能添加过量，滴定管在使用前可能未用碱液润洗，或者锥形瓶可能用酸润洗过，这会改变酸的物质的量。改进措施包括确保所有设备用正确的溶液润洗，并在读取弯月面时保持视线水平以避免视差错误。
4. Case Study 3: Verifying Newton’s Second Law with an Air Track | 案例3：用气垫导轨验证牛顿第二定律
A class investigated Newton’s Second Law, F = m × a, using an air track to minimise friction. A glider of mass M was attached to a light string passing over a pulley, with a small mass m hanging at the end to provide the accelerating force. The acceleration was measured using light gates, and the experiment was repeated with different hanging masses. The results for one run: M = 0.400 kg, m = 0.050 kg, acceleration a = 1.05 m/s². The theoretical acceleration is calculated from the force (weight of the hanging mass) divided by the total mass of the system: a = (m × g) / (M + m), where g = 9.81 m/s².
一个班级利用气垫导轨来减小摩擦，研究牛顿第二定律 F = m × a。一个质量为 M 的滑块系在一根跨过滑轮的轻质绳子上，绳子末端悬挂一个小质量 m 以提供加速力。使用光闸测量加速度，并用不同的悬挂质量重复实验。其中一次实验的结果为：M = 0.400 kg，m = 0.050 kg，加速度 a = 1.05 m/s²。理论加速度由力（悬挂重物的重量）除以系统的总质量计算得出：a = (m × g) / (M + m)，其中 g = 9.81 m/s²。
Calculate the expected acceleration: a = (0.050 × 9.81) / (0.400 + 0.050) = 0.4905 / 0.450 = 1.09 m/s² (to 3 s.f.). Compare this with the measured value of 1.05 m/s². A small discrepancy may arise from friction at the pulley or slight misalignment of the track. The independent variable was the net force (varied by changing m), the dependent variable was the acceleration, and the glider mass M was kept constant for this part of the experiment. To improve accuracy, the string should be very light and inelastic, and the hanging mass should not swing.
计算预期加速度：a = (0.050 × 9.81) / (0.400 + 0.050) = 0.4905 / 0.450 = 1.09 m/s²（保留三位有效数字）。将其与测量值 1.05 m/s² 进行比较。微小的偏差可能来自滑轮处的摩擦或轨道的轻微未对齐。自变量是净力（通过改变 m 来变化），因变量是加速度，而滑块质量 M 在这部分实验中保持恒定。为提高准确性，绳子应非常轻且不可伸长，悬挂物不应摆动。
This case study encourages you to relate the formula to a practical setup, calculate percentage difference, and discuss the validity of conclusions. For instance, percentage difference = |1.09 – 1.05| / 1.09 × 100% ≈ 3.7%, which is acceptable within experimental error. You can also plot a graph of force vs acceleration to see if the gradient equals the total mass of the system.
这个案例鼓励你将公式与实际装置联系起来，计算百分差异，并讨论结论的有效性。例如，百分差异 = |1.09 – 1.05| / 1.09 × 100% ≈ 3.7%，这在实验误差许可范围内。你还可以绘制力与加速度的关系图，看看斜率是否等于系统的总质量。
5. Case Study 4: Analysing Population Growth and Biodiversity Data | 案例4：分析人口增长与生物多样性数据
Ecologists surveyed the population of a species of butterfly in a nature reserve over five years. They used mark-release-recapture (Lincoln Index) to estimate the population size. The data collected each year included the number of individuals caught in the first sample (n₁), the second sample (n₂), and the number of marked individuals in the second sample (m₂). For Year 3, n₁ = 40, n₂ = 50, m₂ = 10. The population estimate N is given by: N = (n₁ × n₂) / m₂.
生态学家调查了一个自然保护区中一种蝴蝶的种群数量，历时五年。他们使用标记重捕法（林肯指数）来估算种群大小。每年收集的数据包括第一次样本捕获的个体数（n₁）、第二次样本捕获的个体数（n₂）以及第二次样本中带标记的个体数（m₂）。对于第三年，n₁ = 40，n₂ = 50，m₂ = 10。种群估算值 N 由公式 N = (n₁ × n₂) / m₂ 给出。
Applying the formula: N = (40 × 50) / 10 = 2000 / 10 = 200. So the estimated population in Year 3 is 200 individuals. However, this method assumes that the marks are not lost, the population is closed (no immigration or emigration), and that marking does not affect the butterfly’s survival or chance of recapture. Evaluate how these assumptions might be violated: if a predator selectively eats marked butterflies because they are more visible, the estimate of N would be too high.
应用公式：N = (40 × 50) / 10 = 2000 / 10 = 200。因此，第三年的估计种群数量为200只。但是，该方法假设标记不会丢失、种群是封闭的（没有迁入或迁出），并且标记不影响蝴蝶的存活或再捕获机会。评估这些假设可能被违反的情况：如果捕食者因为标记更显眼而选择性地捕食带标记的蝴蝶，那么 N 的估计值会偏高。
To build a full case study, you could be given a table with n₁, n₂, and m₂ for all five years and asked to graph the population trend, identify reasons for any decline (e.g., habitat loss, climate change), and suggest conservation strategies. This exercise connects mathematical skills with ecological understanding, a common feature of Eduqas science papers.
为了构建一个完整的案例研究，你可能会得到一个包含所有五年 n₁、n₂ 和 m₂ 的表格，并被要求绘制种群趋势图，找出任何下降的原因（例如栖息地丧失、气候变化），并提出保护策略。这个练习将数学技能与生态学理解联系起来，这是Eduqas科学试卷的一个常见特征。
6. Case Study 5: How Does Concentration Affect the Rate of Reaction? | 案例5：浓度如何影响反应速率？
The reaction between magnesium ribbon and dilute hydrochloric acid was studied: Mg + 2HCl → MgCl₂ + H₂. The volume of hydrogen gas produced was measured every 10 seconds using a gas syringe, for three different concentrations of acid: 0.5 mol/dm³, 1.0 mol/dm³, and 1.5 mol/dm³. The mass of magnesium and the volume of acid were kept constant. The results at 1.0 mol/dm³ are shown below.
研究了镁条与稀盐酸之间的反应：Mg + 2HCl → MgCl₂ + H₂。使用气体注射器，针对三种不同的酸浓度（0.5 mol/dm³、1.0 mol/dm³ 和 1.5 mol/dm³），每10秒测量一次产生的氢气体积。镁的质量和酸的体积保持不变。在 1.0 mol/dm³ 浓度下的结果如下所示。
| Time / s | Volume of H₂ / cm³ |
|---|---|
| 0 | 0 |
| 10 | 15 |
| 20 | 28 |
| 30 | 37 |
| 40 | 42 |
| 50 | 45 |
| 60 | 45 |
The reaction stops by 50 seconds because all the magnesium has reacted. To compare rates, you can calculate the average rate up to a certain time, such as the first 30 s: rate = 37 cm³ / 30 s ≈ 1.23 cm³/s. For 1.5 mol/dm³, you would expect a faster initial rate and the same final volume if the same amount of magnesium is used but the acid is in excess. Plotting all three concentrations on one graph allows you to describe the trend: higher concentration leads to more frequent collisions and a steeper initial slope.
反应在50秒左右停止，因为所有镁都已反应完。要比较速率，你可以计算达到某个时间点的平均速率，例如前30秒：速率 = 37 cm³ / 30 s ≈ 1.23 cm³/s。对于 1.5 mol/dm³，如果使用等量的镁且酸过量，你会预期初始速率更快而最终体积相同。将所有三种浓度的数据绘制在一张图上，你可以描述趋势：浓度越高，碰撞越频繁，初始斜率越陡。
Evaluating the experiment, consider why the gas syringe must be clamped firmly, why the magnesium ribbon should not be heavily oxide-coated (it would slow the reaction), and why repeating the trial and averaging volumes at each time point improves reliability. This mirrors a typical analysis question in the exam.
评估实验时，思考为什么气体注射器必须夹紧、为什么镁条不应有过厚的氧化物涂层（它会减慢反应），以及为什么对每个时间点重复实验并取平均值能提高可靠性。这反映了考试中典型的分析问题。
7. Case Study 6: Investigating Ohm’s Law and Resistance of a Wire | 案例6：探究欧姆定律与导线电阻
A student connected a circuit with a power supply, an ammeter, a voltmeter, and a length of nichrome wire. She varied the potential difference (p.d.) across the wire in steps of 0.5 V and recorded the current. The resistance R was calculated for each pair of readings using R = V / I. The data for a wire of length 0.500 m and diameter 0.25 mm is given below.
一名学生连接了一个包含电源、电流表、电压表和一段镍铬合金线的电路。她以0.5 V的步长改变导线两端的电位差（p.d.），并记录电流。对于每一对读数，使用 R = V / I 计算出电阻 R。一根长度为0.500 m、直径为0.25 mm的导线的数据如下。
| Potential difference V / V | Current I / A | Resistance R / Ω |
|---|---|---|
| 0.50 | 0.10 | 5.0 |
| 1.00 | 0.19 | 5.3 |
| 1.50 | 0.29 | 5.2 |
| 2.00 | 0.38 | 5.3 |
The resistance stays approximately constant at around 5.2 Ω, confirming that the nichrome wire obeys Ohm’s law at constant temperature. If the student had taken readings at higher currents, the wire would heat up, causing the resistance to increase because metal ions vibrate more, impeding electron flow. A graph of V against I would be a straight line passing through the origin for an ohmic conductor. Calculate the resistivity ρ using the formula R = ρL / A. The cross-sectional area A of the wire is π × (d/2)² = π × (0.125 × 10⁻³)² = 4.91 × 10⁻⁸ m², and L = 0.500 m, so ρ = R × A / L ≈ 5.2 × 4.91×10⁻⁸ / 0.500 = 5.1×10⁻⁷ Ω m.
电阻大致恒定在5.2 Ω左右，证实镍铬合金线在恒温下遵循欧姆定律。如果学生在更高电流下记录读数，导线会发热，导致电阻增加，因为金属离子振动更剧烈，阻碍电子流动。对于一个欧姆导体，V 与 I 的关系图是一条通过原点的直线。使用公式 R = ρL / A 计算电阻率 ρ。导线的横截面积 A = π × (d/2)² = π × (0.125×10⁻³)² = 4.91×10⁻⁸ m²，L = 0.500 m，因此 ρ = R × A / L ≈ 5.2 × 4.91×10⁻⁸ / 0.500 = 5.1×10⁻⁷ Ω m。
8. Handling Anomalous Results and Calculating Uncertainties | 处理异常结果与计算不确定度
When you are given a set of repeated measurements, you may need to identify an outlier and exclude it before averaging. An outlier is a value that does not fit the pattern and may have been caused by a mistake in reading an instrument or a sudden change in conditions. Consider a set of five replicate measurements for the mass of a precipitate: 2.34 g, 2.36 g, 2.38 g, 2.31 g, and 2.59 g. The value 2.59 g is clearly inconsistent; if you calculate the mean without it, you get 2.35 g (to 2 d.p.), which is far more representative.
当你得到一组重复测量数据时，你可能需要识别出异常值并在求平均值前将其排除。异常值是一个不符合整体规律的值，可能由读取仪器时的错误或条件的突然变化引起。考虑一组沉淀物质量的五次重复测量值：2.34 g、2.36 g、2.38 g、2.31 g 和 2.59 g。2.59 g明显不一致；如果去掉它计算平均值，得到 2.35 g（保留两位小数），这更具代表性。
Uncertainty in a single measurement can often be taken as half the smallest scale division of the measuring instrument. For example, a 100 cm³ measuring cylinder has 2 cm³ divisions, so the uncertainty is ±1 cm³. In titrations, the uncertainty of a burette reading is ±0.05 cm³ per reading, so two readings (initial and final) give an overall uncertainty of ±0.10 cm³. When expressing results, you should consider percentage uncertainties: for the mean titre of 23.80 cm³, the percentage uncertainty = (0.10 / 23.80) × 100% ≈ 0.42%. Case studies often ask you to compare the reliability of two methods based on such calculations.
单次测量的不确定度通常可认为是测量仪器最小刻度值的一半。例如，一个 100 cm³ 的量筒，刻度为 2 cm³ 一格，其不确定度为 ±1 cm³。在滴定中，滴定管读数的每次不确定度为 ±0.05 cm³，因此两次读数（初始和最终）的总不确定度为 ±0.10 cm³。在表达结果时，你应考虑百分不确定度：对于平均滴定体积 23.80 cm³，百分不确定度 = (0.10 / 23.80) × 100% ≈ 0.42%。案例研究常要求你根据此类计算比较两种方法的可靠性。
9. Common Pitfalls When Answering Case Study Questions | 回答案例分析问题时常见陷阱
A frequent mistake is failing to specify the correct variables clearly. Students often confuse the independent variable with the dependent variable, or forget to mention key control variables that should have been kept constant. Another pitfall is writing an evaluation that merely states ‘we could do more repeats’ without explaining why that improves reliability (by reducing the effect of random errors and allowing a more trustworthy mean to be calculated). Examiners also expect you to link your suggestions to the specific context, not just give generic improvements.
一个常见错误是未能清晰说明正确的变量。学生们经常混淆自变量和因变量，或者忘记提及本应保持不变的关键控制变量。另一个陷阱是写出的评估仅仅说“我们可以多做几次重复”，而没有解释为什么这样能提高可靠性（通过减少随机误差的影响，从而计算更可信的平均值）。考官还期望你将建议与具体情境联系起来，而不仅仅是给出泛泛的改进。
In graphical work, remember to label axes with quantity and unit, choose appropriate scales, and plot points with small crosses. Drawing a line of best fit (which may be a straight line or a smooth curve) is essential; never simply join points dot-to-dot unless instructed. If you are asked to calculate a gradient, show the triangle on the graph and give the final value with the correct units. In Maths for Science questions, always round your final answer to the appropriate number of significant figures, usually matching the least precise piece of data given.
在绘图工作中，记得标注轴的名称和单位，选择合适的刻度，并用小十字标记点。画一条最佳拟合线（可能是直线或平滑曲线）是必不可少的；除非另有指示，切勿简单地将点连成折线。如果要求你计算斜率，要在图上画出三角形，并给出具有正确单位的最终值。在科学数学题中，始终将最终答案四舍五入到适当的有效数字位数，通常与所给数据中最不精确的一个相匹配。
10. Practising with a Mark Scheme Mindset | 带着评分方案思维进行练习
To excel in Eduqas Science case studies, you should practise writing answers that hit the mark scheme points. For a question asking you to ‘Describe a method to investigate…’, structure your answer with a clear list of steps: gather equipment, state how the independent variable will be changed, how the dependent variable measured, what will be controlled, and how the results will be recorded. Use precise scientific vocabulary such as ‘repeat and calculate mean’ rather than ‘do it again’.
要精通Eduqas科学的案例分析，你应该练习写出能击中评分方案要点的答案。对于要求“描述一个探究……的方法”的问题，用清晰的步骤列表来组织你的回答：收集设备，说明如何改变自变量，如何测量因变量，将控制什么，以及如何记录结果。使用精确的科学词汇，如“重复并计算平均值”，而不是“再做一次”。
When evaluating a conclusion, always refer back to the evidence. For example, if the data shows a decrease in the dependent variable after a certain point, state the numerical evidence (e.g., ‘the rate dropped from 0.022 s⁻¹ to 0.007 s⁻¹’), and then explain why this supports or contradicts the hypothesis. This demonstrates your ability to analyse, not just describe. Finally, keep an eye on the question command words: ‘evaluate’ means give advantages and disadvantages with a supported judgement; ‘explain’ requires a reason linked to scientific theory; ‘suggest’ invites you to apply your knowledge to a new context.
在评估结论时，始终回顾证据。例如，如果数据显示因变量在某一点后下降，说明数值证据（例如，“速率从 0.022 s⁻¹ 降至 0.007 s⁻¹”），然后解释这如何支持或反驳假设。这表明你分析的能力，而不仅仅是描述。最后，注意问题中的指令词：“评估”意味着给出优点和缺点，并附带一个有依据的判断；“解释”要求一个与科学理论相联系的理由；“建议”邀请你将知识应用于新情境。
11. Applying Knowledge to Real-World Contexts | 将知识应用于现实世界情境
Modern Eduqas papers increasingly embed scientific concepts in authentic scenarios: bioremediation of oil spills, the physics of a car’s braking system, or the chemistry behind food preservatives. When you encounter such a case, first identify the core scientific principles at play, even if the setting feels unfamiliar. For example, a question on hand warmers might ask why supersaturated sodium acetate crystallises upon clicking a metal disc—relating to activation energy and exothermic processes. Your job is to translate the scenario back into the lab-based knowledge you have practised.
现代的Eduqas试卷越来越多地将科学概念嵌入真实情境：石油泄漏的生物修复、汽车制动系统的物理学，或者食品防腐剂背后的化学原理。当你遇到这种案例时，首先识别起作用的的核心科学原理，即使场景感觉陌生。例如，一个关于暖手宝的问题可能会问为什么过饱和醋酸钠在按压金属片时会结晶——这与活化能和放热过程有关。你的任务是将场景转换回你练习过的基于实验室的知识。
In such questions, you may be given a graph of data that you have never seen before, perhaps showing the change in pH during a fermentation or the force-extension curve for a polymer. Do not panic: draw on your understanding of variables and trends. Read the axis labels carefully, describe the relationship (e.g., ‘as the load increases, the extension increases linearly until the elastic limit’), and then use your scientific knowledge to explain the shape. Practice with past papers and specimen assessments provided by Eduqas to gain familiarity with these applied contexts.
在这类问题中，你可能会得到一张你从未见过的数据图，可能展示发酵过程中的pH变化或聚合物的力-伸长曲线。不要慌张：运用你对变量和趋势的理解。仔细阅读坐标轴标签，描述关系（例如，“随着载荷增加，伸长量线性增加，直到弹性极限”），然后用你的科学知识解释曲线形状。通过练习Eduqas提供的过往试卷和样卷，熟悉
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