Case Study Practice: KS3 Cambridge Science | KS3剑桥科学：案例分析实战演练

📚 Case Study Practice: KS3 Cambridge Science | KS3剑桥科学：案例分析实战演练

Welcome to a dedicated case study workout for KS3 Cambridge Science. In the sections that follow, you will work through eight carefully crafted scenarios that mirror the kind of extended questions you might meet in your topic tests and end-of-stage assessments. Each case pushes you to connect your scientific knowledge to a practical situation, interpret data and explain ideas in clear, accurate language. Use this resource to sharpen your thinking and build confidence before your next exam.

欢迎参加KS3剑桥科学案例分析的专项训练。接下来的内容为你准备了八个精心设计的真实情境问题，模拟你在单元测验或阶段末考试中可能遇到的综合应用题。每一个案例都要求你把科学知识与实际场景联系起来、解释数据并用准确清晰的语言表达自己的想法。请用本次实战演练打磨你的解题思路，为下一次考试积累信心。

1. How to Tackle Case Studies | 如何应对案例分析

Effective case-study answering starts with careful reading. Underline the command words—such as ‘describe’, ‘explain’, ‘suggest’ or ‘calculate’—and note the science topic the question explores. Next, identify the variables or organisms involved. Before you begin writing, sketch a mental checklist: what do I already know about this topic, which keywords (e.g. independent variable, neutralisation, particle, energy transfer) must I use, and is there a graph, table or diagram I need to refer to? Always support an explanation with scientific reasoning, not just opinion.

高效的案例分析答题从仔细读题开始。圈出指令词——例如“描述”“解释”“建议”或“计算”——并看清题目涉及的科学话题。接着，找出其中包含的变量或生物。动笔之前先在脑中列一个清单：关于这个话题我已经掌握哪些知识，必须使用哪些关键词（如自变量、中和反应、粒子、能量转移），以及是否需要参考图表或示意图。一定要用科学原理解释，而不能只凭感觉回答。

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2. Case 1: Food Chains and Ecosystems | 案例1：食物链与生态系统

A farmer notices that the barn owl population on her land has fallen sharply over two years. At the same time, she observes more field mice and finds increased damage to her grass and crops. She asks you to explain the possible links between these observations and to draw a simple food chain.

一位农民注意到，过去两年她土地上的仓鸮数量大幅下降。与此同时，田鼠数量增多，草地和庄稼的受损也更严重。她请你解释这些现象之间可能存在的联系，并画出一条简单的食物链。

Start by thinking about feeding relationships. Barn owls are predators of field mice; mice are herbivores that eat grass and cereal crops. A decline in owls means less predation pressure on mice, so the mouse population rises. With more mice feeding, grass and crops are consumed faster, leading to visible damage. A suitable food chain is: grass → field mouse → barn owl. The arrows show the direction of energy transfer. This is an example of how removing a top predator can disturb the balance of an ecosystem.

先从食物关系入手思考。仓鸮是田鼠的捕食者；田鼠是植食动物，以青草和谷物为食。鸮的数量下降意味着田鼠被捕食的压力减小，田鼠种群数量随之上升。更多田鼠取食导致草地和庄稼快速消耗，出现明显损害。合适的食物链是：青草→田鼠→仓鸮。箭头表示能量传递的方向。这是一个说明移除顶级捕食者会破坏生态系统平衡的例子。

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3. Case 2: Neutralisation in Action | 案例2：中和反应的应用

A lake near a factory has become acidic due to rainwater washing acidic gases into the water. Environmental officers decide to add slaked lime, calcium hydroxide, to the lake. Explain the type of reaction that occurs and write a word equation for the process. If the lake contains hydrochloric acid, suggest the salt produced.

工厂附近的湖泊因为雨水将酸性气体带入水体而变酸。环保官员决定向湖中加入熟石灰（氢氧化钙）。请解释发生的反应类型，并写出该过程的文字表达式。如果湖水中含有盐酸，推测会生成哪种盐。

Slaked lime is an alkali. Mixing an acid with an alkali results in a neutralisation reaction. The general pattern is: acid + alkali → salt + water. When calcium hydroxide neutralises hydrochloric acid (HCl), the products are calcium chloride (CaCl₂) and water (H₂O). The word equation is: hydrochloric acid + calcium hydroxide → calcium chloride + water. The calcium chloride is a salt that dissolves and poses far less harm to aquatic life than the original acid.

熟石灰是一种碱。酸与碱混合会发生中和反应。一般规律为：酸 + 碱 → 盐 + 水。当氢氧化钙中和盐酸（HCl）时，产物是氯化钙（CaCl₂）和水（H₂O）。文字表达式为：盐酸 + 氢氧化钙 → 氯化钙 + 水。生成的氯化钙是一种溶解在水中的盐，对水生生物的危害远小于原来的酸。

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4. Case 3: Investigating Friction | 案例3：探究摩擦力

A student wants to find out how the surface type affects the force needed to pull a wooden block at a steady speed. She attaches a newton meter to the block and pulls it across a carpet, a wooden bench and a sheet of plastic. The newton meter readings are 4.2 N, 2.8 N and 1.9 N respectively.

一位学生想探究不同接触面对匀速拉动木块所需力的影响。她把弹簧测力计连接在木块上，分别在地毯、木制桌面和塑料板上拉动木块。弹簧测力计的读数依次为4.2 N、2.8 N和1.9 N。

Identify the independent variable (surface type), the dependent variable (force needed to pull the block, measured in newtons) and two control variables (the mass of the block, the speed of pulling, the area of the block in contact). Explain why pulling at a constant speed is essential: it ensures the pulling force exactly balances the friction force, so the newton meter reading equals the kinetic friction. A fair test requires keeping the block’s mass and the pulling speed the same each time. The results show that carpet produces the greatest friction, while smooth plastic gives the least.

请指出自变量（接触面类型）、因变量（拉动木块所需的力，单位牛顿）以及两个控制变量（木块质量、拉动速度、接触面积等）。解释为什么匀速拉动十分关键：这样可保证拉力恰好与摩擦力平衡，弹簧测力计读数就等于动摩擦力的大小。公平测试要求每次保持木块质量和拉动速度不变。实验结果表明，地毯产生的摩擦力最大，光滑塑料板产生的摩擦力最小。

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5. Case 4: Microscopy and Cells | 案例4：显微镜与细胞

A pupil prepares a temporary slide of onion epidermis and adds a drop of iodine solution before placing the cover slip. Using a ×10 eyepiece and a ×40 objective, she observes rectangular shapes with clearly stained nuclei and cell walls. She sketches one cell and labels the nucleus, cytoplasm and cell wall. Her teacher asks her to calculate the actual length of the cell if it occupies one quarter of the field of view diameter (0.4 mm).

一名学生制作了洋葱表皮的临时装片，盖上盖玻片前滴加了一滴碘液。她使用10倍目镜和40倍物镜观察，看到了许多长方形结构，细胞核和细胞壁清晰着色。她画下了一个细胞并标注了细胞核、细胞质和细胞壁。老师要求她计算细胞的真实长度，已知该细胞占视野直径（0.4 mm）的四分之一。

Iodine solution is a stain that makes the nucleus and cell wall more visible by adding contrast. The total magnification is 10 × 40 = ×400. To find the actual cell length, multiply the field of view diameter by the fraction: 0.4 mm × 1/4 = 0.1 mm. Convert to micrometres: 0.1 mm = 100 µm. A labelled plant cell diagram should also note the absence of chloroplasts in onion epidermis cells, because they grow underground and do not photosynthesise.

碘液是一种染色剂，通过增加对比度使细胞核和细胞壁更加清晰。总放大倍数为10 × 40 = 400倍。计算细胞真实长度：视野直径乘以所占比例，即0.4 mm × 1/4 = 0.1 mm。换算为微米：0.1 mm = 100 µm。标注植物细胞图时还应注意，洋葱表皮细胞不含叶绿体，因为它们长在地下，不进行光合作用。

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6. Case 5: The Particle Model of Matter | 案例5：物质的粒子模型

On a sunny but windy day, a puddle of water disappears much faster than a puddle in a sheltered spot. Using the particle model, explain why the rate of evaporation increases under these conditions. Also, why do wet clothes dry more quickly when spread out rather than left in a crumpled heap?

在一个晴朗又有风的日子，一滩积水比背风处的积水消失得快得多。请运用粒子模型解释为什么在这种条件下蒸发速率会加快。另外，为什么湿衣服摊开晾比揉成一团干得更快？

In the particle model, liquid water consists of particles that move at different speeds. The fastest particles at the surface have enough energy to overcome attractive forces and escape as gas. Wind carries away the water vapour just above the surface, reducing the likelihood of particles returning to the liquid; thus evaporation proceeds more quickly. Sunlight transfers thermal energy to the water, raising the average particle kinetic energy so more particles can escape. Spreading clothes increases the exposed surface area, allowing more liquid particles to be near the air interface at any moment, which boosts the evaporation rate.

在粒子模型中，液态水由运动速度不同的粒子组成。液面处速度最快的粒子有足够能量克服吸引力逸出成为气体。风会带走液面上方刚形成的水蒸气，降低粒子重新返回液体的概率，因此蒸发进行得更快。阳光将热能传递给水，提高了粒子的平均动能，使更多粒子能够逃逸。摊开衣服则增大了暴露的表面积，任何时候都有更多液体粒子靠近空气界面，从而加快蒸发速率。

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7. Case 6: Electrical Circuits | 案例7? Wait, indexing: after 6 we have 7. Let’s continue 7,8,9,10. So 7. Case 6: Electrical Circuits. Yes.

But need to keep numbering consistent. Already used 5 cases, next 6,7,8. So: 7. Case 6: Electrical Circuits. Then 8. Case 7: Photosynthesis Experiment Design. Then 9. Case 8: Earth and the Solar System. Then 10. Key Takeaways. That’s 10 subsections. Good.

Now 7.

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7. Case 6: Electrical Circuits | 案例6：电路

A pupil builds a series circuit with two identical lamps connected to a 6 V battery. Both lamps glow with normal brightness. She then adds a third identical lamp in series. Predict what happens to the brightness of all three lamps. Explain your prediction in terms of current, resistance and energy transfer. Next, she re-wires the three lamps in parallel. How will the brightness change, and why?

一个学生用两个相同灯泡与一个6 V电池连成串联电路，两盏灯都正常发光。随后她又串入第三个相同灯泡。请预测三盏灯亮度会怎样变化，并从电流、电阻和能量转移的角度解释你的预测。接着，她把三个灯泡改为并联连接。亮度会如何改变？原因是什么？

In a series circuit, the current is the same everywhere. Adding a third lamp increases the total resistance because the current must pass through three filaments instead of two. Higher total resistance reduces the current from the battery. Since energy transferred to each lamp per second (power) depends on the current, a lower current makes all three lamps dimmer than the original two. In a parallel circuit, each lamp is connected directly to the battery, so each receives the full 6 V. The total current drawn from the battery becomes larger, but each lamp operates at its designed voltage and therefore glows at normal brightness, unaffected by the other lamps.

在串联电路中，电流处处相等。增加第三个灯泡会增大总电阻，因为电流须通过三根灯丝而非两根。总电阻变大导致电池输出的电流减小。由于每盏灯每秒获得的能量（功率）取决于电流，电流减小使三盏灯都比原先的两盏暗。在并联电路中，每盏灯直接与电池两端相连，因此每盏灯都承受满额6 V电压。电池提供的总电流变大，但每盏灯都在其设计电压下工作，所以都能保持正常亮度，互不影响。

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8. Case 7: Photosynthesis Experiment Design | 案例7：光合作用实验设计

A student investigates how light intensity affects the rate of photosynthesis using pondweed (Elodea). She places a piece of pondweed in 0.2% sodium hydrogencarbonate solution and counts oxygen bubbles released per minute when a lamp is placed at 10 cm, 20 cm and 40 cm from the beaker. The bubble counts are 45, 21 and 6 respectively. Identify the independent, dependent and at least three control variables. Explain the pattern in the results using scientific ideas about photosynthesis.

一位学生利用水草（伊乐藻）探究光照强度对光合作用速率的影响。她把一段水草放入0.2%碳酸氢钠溶液中，并在距离烧杯10 cm、20 cm和40 cm处放置台灯，记录每分钟释放的氧气泡数。结果分别为45、21和6。请指出自变量、因变量以及至少三个控制变量。运用有关光合作用的科学知识解释数据所呈现的规律。

Independent variable: light intensity (varied by changing the lamp distance). Dependent variable: rate of photosynthesis (measured as bubbles per minute). Control variables: concentration of sodium hydrogencarbonate, water temperature, mass and length of pondweed, time allowed for the plant to adjust. The pattern shows that as the lamp is moved further away, light intensity decreases, providing less energy for the light-dependent reactions of photosynthesis. Fewer oxygen bubbles are produced. The sodium hydrogencarbonate supplies carbon dioxide, which is essential for the light-independent stage. This experiment demonstrates the limiting factor effect: at low light intensity, light becomes the factor holding back photosynthesis, even when CO₂ is plentiful.

自变量：光照强度（通过改变灯距来实现）。因变量：光合作用速率（用每分钟气泡数衡量）。控制变量：碳酸氢钠溶液浓度、水温、水草的质量和长度、植株适应时间等。数据显示，随着灯距增加，光照强度降低，为光合作用的光反应提供的能量减少，因此产生的氧气泡数量也随之减少。碳酸氢钠提供二氧化碳，是暗反应阶段不可或缺的原料。本实验体现了限制因子的效应：在低光照下，即便CO₂充足，光也会成为限制光合作用速率的关键因素。

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9. Case 8: Earth and the Solar System | 案例8：地球与太阳系

Look at the night sky during a month and you will see the Moon change shape from a thin crescent to a full disc and back again. Explain why we observe these phases. Also, use your knowledge of Earth’s tilt to account for the seasons experienced in the northern hemisphere.

在一个月中观察夜空，你会看到月亮从弯弯的蛾眉月变成圆盘状的满月，再变回原来的样子。请解释我们为什么会看到这些月相。另外，运用地球倾斜的知识说明北半球季节变化的原因。

The Moon does not produce its own light; we see it because sunlight reflects off its surface. As the Moon orbits Earth, the relative positions of the Sun, Earth and Moon change, altering the fraction of the sunlit side visible from Earth. This cycle produces the phases—new moon, first quarter, full moon and last quarter—over about 29.5 days. Earth’s seasons arise because the planet’s rotation axis is tilted at approximately 23.5° relative to its orbital plane. During June, the northern hemisphere tilts towards the Sun, receiving more concentrated sunlight and longer daylight hours, resulting in summer. In December, it tilts away, so sunlight is more spread out and days are shorter, producing winter.

月球本身不发光，我们看到的月光是太阳光被月球表面反射而来的。月球绕地球公转时，太阳、地球和月球三者的相对位置不断变化，从地球上可看到的月球亮面比例也随之改变，从而形成新月、上弦月、满月和下弦月的月相循环，周期约为29.5天。季节的产生则是因为地球自转轴相对于公转轨道面倾斜约23.5°。北半球在6月朝向太阳倾斜，接收到更加集中的阳光和更长的白昼，形成夏季；12月时北半球偏离太阳，阳光分散，白昼变短，因此进入冬季。

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10. Key Takeaways from Case Study Practice | 实战演练要点总结

Across these eight cases, a few habits stand out as essential. Always read the stem twice, draw out the scientific keywords, and decide which concepts the examiner is testing. When explaining, use cause-and-effect reasoning: link a change in a variable to an outcome through an accepted scientific principle. Support answers with standard terminology such as ‘energy transfer’, ‘neutralisation’, ‘particle kinetic energy’ and ‘limiting factor’. For calculation questions, show your working clearly and give units. Finally, check that you have addressed the command word—if the question says ‘explain’, you must give reasons, not just state facts.

从这八个案例中可以提炼出几个至关重要的答题习惯。务必把题干读两遍，圈出科学关键词，并判断考官在考查哪些概念。解释时要运用因果推理：把变量的变化通过公认的科学原理与结果联系起来。答案中要使用标准术语，如“能量传递”“中和反应”“粒子动能”“限制因子”等。遇到计算题要清晰展示步骤并注明单位。最后，检查是否回应了指令词——如果题目要求“解释”，就必须给出理由，而不能仅仅罗列事实。

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