📚 Year 9 SQA Science: Case Study Practical Exercises | 九年级SQA科学：案例分析实战演练
In the SQA Year 9 science curriculum, applying knowledge to real-world scenarios is just as important as understanding core concepts. This article presents a series of case study exercises that challenge you to think like a scientist, analyse data, and draw evidence-based conclusions. Each case is designed to mirror the style of assessment you may encounter, blending biology, chemistry and physics contexts. Work through the exercises carefully, then compare your reasoning with the step-by-step explanations provided.
在SQA九年级科学课程中，将知识应用于实际情境与理解核心概念同样重要。本文提供了一系列案例分析练习，旨在让你像科学家一样思考，分析数据并得出基于证据的结论。每个案例都模拟你可能遇到的评估风格，融合了生物、化学和物理背景。请仔细完成练习，然后将你的推理与所提供的逐步解释进行对比。
1. Investigating Friction and Motion | 研究摩擦与运动
A student pushes a wooden block across different surfaces and measures the force needed to keep it moving at a steady speed. The results are shown in the table below.
一名学生在不同表面上推动一个木块，并测量使其保持匀速运动所需的力。结果如下表所示。
| Surface | Force (N) | 表面 | 力（牛） |
|---|---|---|---|
| Smooth wood | 2.5 | 光滑木板 | 2.5 |
| Carpet | 6.0 | 地毯 | 6.0 |
| Sandpaper | 8.5 | 砂纸 | 8.5 |
Explain why the force varies, using your understanding of friction. Predict what would happen if the mass of the block were doubled on the carpet surface.
请用你对摩擦力的理解解释力为何不同。预测如果在地毯表面上将木块的质量加倍会发生什么。
Case analysis: When the block moves at constant speed, the applied force balances the frictional force. Different surfaces have different levels of roughness, which affects the friction between the block and the surface. Smoother surfaces produce less friction, so less force is needed; rougher surfaces create more friction, requiring a greater force. If the mass is doubled, the normal contact force increases, and since friction is generally proportional to the normal force, the frictional force would also increase – roughly doubling. Therefore, the pulling force needed on carpet would be approximately 12.0 N, assuming the mass is the only variable changed.
案例分析：当木块匀速运动时，所施加的力与摩擦力平衡。不同表面粗糙程度不同，影响木块与表面之间的摩擦。光滑表面产生的摩擦力较小，因此所需力较小；粗糙表面则产生较大摩擦力，需要更大的力。如果质量加倍，正压力增大，而摩擦力通常与正压力成正比，因此摩擦力也会增大——大约翻倍。所以，在地毯上所需的拉力约为12.0牛，假设仅改变质量这一变量。
2. Interpreting a Cooling Curve | 解读冷却曲线
A beaker of hot water was left to cool in a room. A temperature sensor recorded the data below.
一个盛有热水的烧杯放在房间里冷却。温度传感器记录了以下数据。
| Time (min) | Temperature (°C) | 时间（分钟） | 温度（°C） |
|---|---|---|---|
| 0 | 80 | 0 | 80 |
| 2 | 68 | 2 | 68 |
| 4 | 58 | 4 | 58 |
| 6 | 50 | 6 | 50 |
| 8 | 44 | 8 | 44 |
| 10 | 40 | 10 | 40 |
| 12 | 37 | 12 | 37 |
| 14 | 35 | 14 | 35 |
Describe the trend shown by the data and suggest the likely room temperature. Explain why the rate of cooling decreases over time.
描述数据所显示的趋势，并推测可能的室温。解释为什么冷却速率随时间降低。
Case analysis: The temperature drops rapidly at first, then more slowly. The difference between the water temperature and the surroundings drives heat transfer – the larger the temperature difference, the faster the energy flow. As the water cools, the gradient decreases, so the rate of heat loss falls. The curve appears to be levelling off near 35 °C, suggesting the room temperature is around 20 °C (since cooling would continue until thermal equilibrium is reached). Simple extrapolation would not give exactly room temperature from this limited data, but we can infer it is below 35 °C. In practice, the room might be 20–22 °C. Rate of cooling is proportional to temperature difference, which explains the curve shape.
案例分析：温度起初下降很快，随后逐渐变慢。水与周围环境的温差驱动热传递——温差越大，能量流动越快。随着水温降低，温差减小，热损失速率下降。曲线似乎在35°C附近趋于平缓，表明室温约为20°C（因为冷却持续到热平衡）。从有限数据外推无法精确得到室温，但可推测低于35°C。实际室温可能在20–22°C。冷却速率与温差成正比，这解释了曲线形状。
3. Acid–Metal Reaction and Gas Identification | 酸与金属的反应及气体鉴定
A student adds magnesium ribbon to dilute hydrochloric acid. Bubbles form rapidly, and the solution becomes warm. The gas produced is collected and tested with a burning splint – a squeaky pop is heard.
一名学生将镁条加入稀盐酸中。迅速产生气泡，溶液变热。将产生的气体收集并用燃着的木条检验——听到尖锐的爆鸣声。
Write a word equation for the reaction and name the gas. Explain the energy change and the safety precaution needed when using acid.
写出反应的文字方程式并命名该气体。解释能量变化以及使用酸时所需的安全预防措施。
Case analysis: esium chloride + hydrogen. The gas is hydrogen (the pop test confirms it). The reaction is exothermic; the beaker feels warm because energy is released to the surroundings. When working with acids, wear safety goggles and a lab coat to protect eyes and skin from splashes. The reaction produces a flammable gas, so no naked flames should be nearby during the experiment. The squeaky pop is the hallmark test for hydrogen.
案例分析：镁 + 盐酸 → 氯化镁 + 氢气。气体是氢气（爆鸣实验证实）。反应放热；烧杯变暖是因为能量释放到周围环境中。使用酸时，应佩戴护目镜和实验服，保护眼睛和皮肤免受溅射伤害。反应产生易燃气体，因此实验附近不应有明火。尖锐的爆鸣声是氢气的特征检验现象。
4. Ecosystem Changes and Predator–Prey Relationships | 生态系统变化与捕食-被捕食关系
A woodland food web includes plants, caterpillars, blue tits, and sparrowhawks. Over a year, the caterpillar population crashes due to a disease. Predict the short-term and long-term effects on the other organisms. Explain your reasoning using ecological ideas.
一个林地食物网包括植物、毛虫、蓝山雀和雀鹰。一年里，毛虫种群因疾病而锐减。预测对其他生物的短期和长期影响，并运用生态学概念解释你的推理。
Case analysis: In the short term, blue tits lose a major food source, so their population may decline due to starvation or reduced breeding success. Sparrowhawks that prey on blue tits may also experience a drop in food supply, affecting their numbers. Meanwhile, plants may flourish because fewer caterpillars are eating leaves. In the long term, the ecosystem may rebalance: with fewer blue tits, caterpillar numbers could recover, but if the disease persists, the balance will depend on alternative prey or food sources. This demonstrates interdependence and the ripple effect of population change.
案例分析：短期来看，蓝山雀失去了主要食物来源，其种群数量可能因饥饿或繁殖成功率下降而减少。以蓝山雀为食的雀鹰也会面临食物短缺，数量受影响。同时，植物可能繁茂起来，因为吃叶子的毛虫减少了。长期来看，生态系统可能重新平衡：蓝山雀减少，毛虫数量可能恢复，但如果疾病持续存在，平衡将取决于替代猎物或食物来源。这展示了相互依存关系以及种群变化的连锁效应。
5. Electrical Circuit Fault-Finding | 电路故障排查
A pupil builds a series circuit with two bulbs and a switch. When the switch is closed, neither bulb lights. The ammeter reads zero, but the voltmeter across the battery shows 6 V. Suggest two possible faults and describe how to test for each.
一名学生搭建了一个由两个灯泡和一个开关组成的串联电路。合上开关后，两个灯泡都不亮。电流表读数为零，但跨接在电池两端的电压表显示6 V。请提出两种可能的故障，并描述如何检测每种故障。
Case analysis: Fault 1: One of the bulbs is blown (filament broken), causing a break in the series circuit. Testing: remove each bulb and test with a multimeter set to resistance; a broken filament shows infinite resistance. Fault 2: A loose connection or broken wire interrupts current flow. Testing: systematically check each connection and wiggle wires while observing the ammeter; if current briefly appears, the loose point is identified. The voltmeter across the battery reading normal voltage confirms the battery is not dead; the problem lies in the circuit continuity.
案例分析：故障一：其中一个灯泡烧坏（灯丝断裂），导致串联电路断路。测试：取下每个灯泡，用万用表电阻档检测；断开的灯丝显示无限大电阻。故障二：接线松动或导线断裂，中断电流。测试：逐一检查每个连接点，轻轻晃动导线并观察电流表；如果电流瞬间出现，则找到了松动处。电池端电压正常说明电池没问题，问题出在电路连续性上。
6. Photosynthesis Rate and Light Intensity | 光合作用速率与光照强度
Elodea pondweed was placed in a beaker of water, and the number of oxygen bubbles released per minute was counted at different distances from a lamp.
将伊乐藻水草放在水烧杯中，在距离灯不同距离处记录每分钟释放的氧气气泡数。
| Distance (cm) | Bubbles per min | 距离（厘米） | 每分钟气泡数 |
|---|---|---|---|
| 10 | 45 | 10 | 45 |
| 20 | 23 | 20 | 23 |
| 30 | 12 | 30 | 12 |
| 40 | 5 | 40 | 5 |
Explain the relationship between distance and bubble rate. Why does doubling the distance not exactly halve the bubble rate? (Hint: light intensity follows an inverse square law.)
解释距离与气泡速率之间的关系。为什么距离加倍并不使气泡速率正好减半？（提示：光强遵循平方反比定律。）
Case analysis: As distance increases, light intensity decreases, so the rate of photosynthesis (shown by oxygen production) drops. However, light intensity is inversely proportional to the square of the distance (Intensity ∝ 1/d²). So, doubling the distance reduces the intensity to one-fourth, not one-half. This explains why bubble rate falls more sharply than a simple inverse relationship. At 20 cm, intensity is a quarter of that at 10 cm, yet bubble rate drops from 45 to 23 – not exactly to 11.25 because other limiting factors (like CO₂ or temperature) also influence photosynthesis.
案例分析：距离增加，光强减弱，因此光合作用速率（以氧气产量表示）下降。然而，光强与距离的平方成反比（光强 ∝ 1/d²）。所以，距离加倍，光强降为原来的四分之一，而不是一半。这解释了为何气泡速率的下降比简单反比关系更剧烈。距离20 cm时，光强是10 cm时的四分之一，但气泡数从45降至23，而非正好降至11.25，因为其他限制因素（如CO₂浓度或温度）也影响光合作用。
7. Neutralisation and Soil pH Adjustment | 中和反应与土壤pH调节
A gardener finds that the soil in a flower bed is too acidic (pH 4.5) for growing roses, which prefer pH 6.5. She adds crushed limestone (calcium carbonate) to the soil. Explain how this works, using a balanced chemical equation. Why might adding too much limestone be harmful?
园丁发现花坛土壤酸性过强（pH 4.5），不适合玫瑰生长（喜pH 6.5）。她将碎石灰石（碳酸钙）加入土壤中。请用配平的化学方程式解释其原理。为什么过量添加石灰石可能有害？
Case analysis: : CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂. The carbonate reacts, removing acidity and raising the pH. Over-application would push the pH too high (alkaline), making nutrients like iron and manganese less available to plants, causing yellowing and poor growth. The balanced pH is crucial for nutrient solubility.
案例分析：石灰石是碳酸钙，属于碱。它与土壤中的酸（如雨水带来的氢离子）发生中和反应：CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂。碳酸盐反应，除去酸性，提高pH值。过量施用会使pH过高（碱性），导致铁、锰等微量元素溶解度下降，植物无法吸收，出现黄化、生长不良。平衡的pH对营养溶解至关重要。
8. Force Extension and Hooke's Law | 力的伸长与胡克定律
A spring is loaded with increasing weights, and its length is measured. The data are plotted and show a straight line up to a certain point, then the graph curves. Use Hooke's Law to explain the initial part, and identify what the curved part represents. Calculate the spring constant if a 4 N load causes an extension of 10 cm.
弹簧加载逐渐增加的砝码，测量其长度。所绘制的数据图在一定范围内呈直线，随后曲线弯曲。请用胡克定律解释初始阶段，并说明弯曲部分代表什么。若4 N的载荷产生10 cm的伸长量，计算弹簧常数。
Case analysis: Hooke' up to the limit of proportionality: F = kx. The straight line shows this elastic behaviour – rs, and the spring may not recover its shape. For the calculation: x = 10 cm = 0.10 m. k = F / x = 4 N / 0.10 m = 40 N/m. Always convert units to SI.
案例分析：胡克定律指出，在比例极限内，伸长量与力成正比：F = kx。直线部分表现弹性行为——卸载后弹簧恢复原长。曲线部分表明已超过弹性极限，发生永久变形，弹簧可能无法恢复形状。计算：x = 10 cm = 0.10 m。k = F / x = 4 N / 0.10 m = 40 N/m。务必转换为国际单位。
9. Diffusion of Coloured Dye in Water | 有色染料在水中的扩散
A crystal of purple potassium permanganate is placed at the bottom of a beaker of cold water. Over time, the purple colour spreads throughout the water without stirring. Explain what is happening at the particle level and describe two ways to speed up the process.
将一粒紫色高锰酸钾晶体放在冷水烧杯底部。随着时间推移，紫色在没有搅拌的情况下弥漫至整杯水中。从粒子层面解释发生了什么，并描述两种加速该过程的方法。
Case analysis: The potassium permanganate particles are in constant random motion. They collide with water molecules and gradually move from a region of high concentration to low concentration – this is diffusion. Speeding up: (1) Increase temperature – heating gives particles more kinetic energy, so they move faster and diffuse more quickly. (2) Stirring – although the question states 'without stirring', as a method to accelerate, stirring creates convection currents that distribute the dye faster.
案例分析：高锰酸钾粒子处于持续的无规则运动中。它们与水分子碰撞，逐渐从高浓度区域向低浓度区域移动——这就是扩散。加速方法：（1）提高温度——加热使粒子获得更大动能，运动加快，扩散更快。（2）搅拌——尽管题目说无搅拌，但作为加速手段，搅拌产生对流，能更快分散染料。
10. Interpreting Data on Global Warming | 解读全球变暖数据
A graph shows average global temperature and CO₂ concentration from 1880 to 2020. Both lines trend upward, with CO₂ rising sharply after 1950. A student claims that the temperature increase is caused solely by CO₂. Critically evaluate this claim, suggesting other possible factors and any limitations of the data.
一张图表显示了从 1880 年到 2020 年的全球平均温度和 CO₂ 浓度。两条曲线均呈上升趋势，CO₂ 在 1950 年后急剧上升。一名学生声称温度升高完全由 CO₂ 引起。请批判性地评价这一说法，提出其他可能因素及数据的局限性。
Case analysis: While the correlation between CO₂ and temperature is strong, correlation does not prove causation. Other greenhouse gases (methane, water vapour), changes in solar radiation, volcanic activity, and deforestation can also affect climate. The data show a clear link, but scientists use climate models to separate natural variations from human influence. A limitation is that the early temperature record may be less accurate and global coverage sparse. Also, time lags exist in the climate system. Nonetheless, the overwhelming scientific evidence supports that increased CO₂ from burning fossil fuels is the primary driver of recent warming.
案例分析：尽管CO₂与温度之间存在强相关性，但相关性不能证明因果关系。其他温室气体（甲烷、水蒸气）、太阳辐射变化、火山活动和森林砍伐也可影响气候。数据显示明显关联，但科学家使用气候模型区分自然变化和人类影响。一个局限性是早期温度记录可能不太准确且全球覆盖稀疏。此外，气候系统中存在时间滞后。然而，压倒性的科学证据支持，化石燃料燃烧产生增加的CO₂是近期变暖的主要驱动力。
Published by TutorHao | SQA Science Revision Series | aleveler.com
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