Case Study Practice in Year 9 CIE Science | 科学案例分析实战演练

📚 Case Study Practice in Year 9 CIE Science | 科学案例分析实战演练

Welcome to our case study practice designed for Year 9 CIE Science. This article provides real-world scenarios across physics, chemistry, and biology, helping you apply scientific knowledge to analyse data, design experiments, and solve problems. Case study questions are common in your assessments, and practising them will boost your confidence and skills.

欢迎来到专为Year 9 CIE科学设计的案例分析实战演练。本文提供物理、化学和生物学科的现实情景，帮助你运用科学知识分析数据、设计实验和解决问题。案例分析题在你的测评中很常见，通过练习能增强你的信心和能力。

1. Case Study 1: Temperature Data and the Greenhouse Effect | 案例1：温度数据与温室效应

Scientists have collected data on global average temperature and atmospheric carbon dioxide (CO₂) concentration over the last 100 years. The table below shows measurements at 20-year intervals.

科学家收集了过去100年全球平均温度和大气中二氧化碳（CO₂）浓度的数据。下表显示了每20年间隔的测量值。

Table: Global average temperature and CO₂ concentration | 表格：全球平均温度和CO₂浓度

Year	Temperature (°C)	CO₂ (ppm)
1920	13.8	300
1940	13.9	310
1960	14.0	320
1980	14.2	340
2000	14.6	370
2020	15.0	415

From the table, we can observe that both temperature and CO₂ concentration have increased significantly since 1920. The rise is especially steep after 1980.

从表中我们可以观察到，自1920年以来温度和CO₂浓度都显著增加，尤其是在1980年之后上升更为急剧。

Question 1: Describe the trend in global average temperature between 1920 and 2020.

问题1：描述1920年至2020年间全球平均温度的变化趋势。

Answer: The temperature increased gradually from 13.8°C in 1920 to 14.2°C in 1980, and then rose more sharply to 15.0°C in 2020.

回答：气温从1920年的13.8°C缓慢升高到1980年的14.2°C，随后更急剧地上升到2020年的15.0°C。

Question 2: Explain why many scientists think the increase in CO₂ is causing the temperature rise.

问题2：解释为什么许多科学家认为CO₂的增加导致了气温升高。

Answer: CO₂ is a greenhouse gas. It traps infrared radiation (heat) that would otherwise escape into space. More CO₂ means more heat is trapped, which warms the atmosphere. This is the enhanced greenhouse effect.

回答：CO₂是一种温室气体。它会捕获本应逃逸到太空的红外辐射（热量）。更多的CO₂意味着更多热量被锁定，从而使大气变暖。这就是增强的温室效应。

Data also show that the rise in CO₂ from burning fossil fuels matches the timing of industrial growth, supporting the link.

数据还显示，燃烧化石燃料导致的CO₂上升与工业发展的时间相吻合，进一步支持了这种联系。

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2. Case Study 2: Plant Growth and Light Intensity | 案例2：植物生长与光照强度

A student investigated how light intensity affects the growth of bean seedlings. Two groups of seedlings were used: one placed under bright light, the other under dim light. Both groups received the same amount of water and were kept at the same temperature. After three weeks, the average height of the bright-light seedlings was 25 cm, while the dim-light seedlings averaged only 10 cm.

一位学生研究了光照强度如何影响豆苗的生长。两组幼苗分别被放在强光和弱光下。两组获得相同的水量和相同温度。三周后，强光下的幼苗平均高度为25厘米，而弱光下的平均仅10厘米。

Question: Identify and explain the independent variable, dependent variable, and two control variables in this experiment.

问题：指出并解释本实验中的自变量、因变量和两个控制变量。

Answer: Independent variable – light intensity (changed by the student); Dependent variable – height of seedlings (measured); Control variables – amount of water, temperature, type of soil, number of seeds (kept the same).

回答：自变量——光照强度（由学生改变）；因变量——幼苗高度（测量）；控制变量——水量、温度、土壤类型、种子数量（保持不变）。

Question: Suggest why the seedlings in bright light grew taller. Use your knowledge of photosynthesis.

问题：利用光合作用的知识解释为什么强光下的幼苗长得更高。

Answer: Light provides the energy for photosynthesis, the process by which plants make glucose (food). More light means more photosynthesis, producing more glucose for growth and energy. The dim-light plants had less energy for growth.

回答：光为光合作用提供能量，光合作用是植物制造葡萄糖（食物）的过程。光线更强意味着光合作用更多，产生更多葡萄糖用于生长和能量。弱光下的植物生长能量较少。

To make the results more reliable, the student could repeat the experiment several times and calculate averages, or use more seedlings in each group.

为了使结果更可靠，学生可以重复实验多次并计算平均值，或在每组使用更多幼苗。

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3. Case Study 3: Reaction Rate and Concentration | 案例3：反应速率与浓度

Marble chips (calcium carbonate) were reacted with hydrochloric acid (HCl) of different concentrations. In each trial, 1.0 g of marble chips was added to 50 cm³ of acid, and the time to collect 40 cm³ of carbon dioxide gas was recorded.

大理石碎片（碳酸钙）与不同浓度的盐酸（HCl）反应。每次试验中，1.0克大理石碎片加入50 cm³酸中，记录收集40 cm³二氧化碳所需的时间。

Concentration of HCl (mol/dm³)	Time to collect 40 cm³ CO₂ (s)
0.5	80
1.0	40
1.5	27

Calculate the average rate of gas production for each concentration in cm³/s.

计算每种浓度下气体产生的平均速率，以cm³/s为单位。

Rate = Volume of gas / Time

速率 = 气体体积 / 时间

0.5 mol/dm³: Rate = 40 / 80 = 0.5 cm³/s. 1.0 mol/dm³: 40 / 40 = 1.0 cm³/s. 1.5 mol/dm³: 40 / 27 ≈ 1.48 cm³/s.

0.5 mol/dm³：速率 = 40 / 80 = 0.5 cm³/s。1.0 mol/dm³：40 / 40 = 1.0 cm³/s。1.5 mol/dm³：40 / 27 ≈ 1.48 cm³/s。

Explain why increasing concentration increases the rate of reaction, using the particle collision theory.

用粒子碰撞理论解释为什么增加浓度会提高反应速率。

Answer: Higher concentration means there are more HCl particles per unit volume. This increases the frequency of successful collisions between HCl particles and the marble surface, leading to a faster reaction.

回答：浓度更高意味着单位体积内有更多的HCl粒子。这增加了HCl粒子与大理石表面成功碰撞的频率，导致反应更快。

A control variable would be the surface area of the marble chips (particle size) and the temperature. These must be kept constant to ensure a fair test.

控制变量包括大理石碎片的大小（表面积）和温度。必须保持这些不变以确保公平测试。

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4. Case Study 4: Balanced and Unbalanced Forces in Sports | 案例4：体育中的平衡与不平衡力

A cyclist with a total mass of 80 kg is moving along a straight road. When she pedals, she produces a forward driving force of 300 N. Air resistance and friction total 180 N.

一位总质量为80 kg的自行车运动员正在笔直的道路上行驶。当她踩踏板时，产生300 N的前进驱动力。空气阻力和摩擦力总和为180 N。

Calculate the resultant force acting on the cyclist and her acceleration.

计算作用在运动员身上的合力以及她的加速度。

Resultant force = Driving force − Total resistive force = 300 N − 180 N = 120 N

Acceleration = Resultant force / mass = 120 N / 80 kg = 1.5 m/s²

合力 = 驱动力 − 总阻力 = 300 N − 180 N = 120 N。加速度 = 合力 / 质量 = 120 N / 80 kg = 1.5 m/s²。

When the cyclist reaches a steady speed, the driving force equals the resistive forces. Explain what the resultant force and acceleration are at this stage.

当运动员达到稳定速度时，驱动力等于阻力。解释此时合力和加速度的情况。

Answer: The resultant force is zero because driving force = resistive force, so the forces are balanced. Acceleration is also zero, meaning the cyclist moves at constant speed (Newton’s first law).

回答：合力为零，因为驱动力等于阻力，力达到平衡。加速度也为零，意味着运动员以恒定速度运动（牛顿第一定律）。

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