📚 Typical Worked Examples for CCEA IGCSE Science | IGCSE CCEA 科学：典型例题详解
In CCEA IGCSE Science, the Double Award specification covers a wide range of topics in Biology, Chemistry and Physics. Mastering worked examples is essential for success, as it helps students apply concepts, practise calculations and develop problem‑solving skills for exam‑style questions. This article provides detailed step‑by‑step solutions to typical examples from each major topic area.
在 CCEA IGCSE 科学双奖课程中，涵盖生物、化学和物理的广泛主题。掌握典型例题对于成功至关重要，因为它帮助学生应用概念、练习计算并培养解决考试风格问题的能力。本文针对每个主要主题领域提供了详细的逐步解答。
1. Kinematics – Equations of Motion | 运动学 – 运动方程
Question: A train moving at 30 m/s decelerates uniformly at 2 m/s² until it stops. Calculate the distance covered during braking.
题目：一列以 30 m/s 运动的火车以 2 m/s² 匀减速直至停止。计算制动过程中行驶的距离。
Identify the known quantities: initial velocity u = 30 m/s, final velocity v = 0 m/s, acceleration a = −2 m/s² (negative because it is decelerating). The unknown is displacement s.
识别已知量：初速度 u = 30 m/s，末速度 v = 0 m/s，加速度 a = −2 m/s²（负值表示减速）。未知量为位移 s。
The appropriate equation of motion linking v, u, a and s without time is:
不需要时间的运动学方程为：
v² = u² + 2as
Substitute the values: 0² = (30)² + 2 × (−2) × s → 0 = 900 − 4s → 4s = 900 → s = 225 m.
代入数值：0² = (30)² + 2 × (−2) × s → 0 = 900 − 4s → 4s = 900 → s = 225 m。
The braking distance is 225 metres.
制动距离为 225 米。
2. Forces and Newton’s Laws | 力与牛顿定律
Question: A block of mass 5 kg is pulled along a smooth horizontal surface by a horizontal force of 20 N. Calculate the acceleration of the block.
题目：一个质量为 5 kg 的物块在光滑水平面上受到 20 N 的水平拉力。计算物块的加速度。
The surface is smooth, so friction is negligible. Apply Newton’s second law: resultant force F = m × a.
表面光滑，摩擦力可忽略。应用牛顿第二定律：合力 F = m × a。
F = ma
Rearrange to find acceleration: a = F / m = 20 N / 5 kg = 4 m/s².
变形求加速度：a = F / m = 20 N / 5 kg = 4 m/s²。
The acceleration of the block is 4 m/s² in the direction of the applied force.
物块的加速度为 4 m/s²，方向与施加的力一致。
3. Energy, Work and Power | 能量、功与功率
Question: A student of mass 50 kg climbs a flight of stairs of vertical height 12 m in 15 s. Calculate the work done against gravity and the power developed. (g = 10 m/s²)
题目：一名质量为 50 kg 的学生用 15 s 爬上一段垂直高度为 12 m 的楼梯。计算克服重力做的功和产生的功率。(g = 10 m/s²)
Work done against gravity (W) is equal to the gain in gravitational potential energy: W = mgh.
克服重力做的功 (W) 等于增加的重力势能：W = mgh。
W = mgh
W = 50 kg × 10 m/s² × 12 m = 6000 J.
W = 50 kg × 10 m/s² × 12 m = 6000 J。
Power is the rate of doing work: P = W / t = 6000 J / 15 s = 400 W.
功率是做功的速率：P = W / t = 6000 J / 15 s = 400 W。
Therefore, the work done is 6000 J and the power is 400 W.
因此，做功为 6000 J，功率为 400 W。
4. Electric Circuits and Ohm’s Law | 电路与欧姆定律
Question: A resistor of 15 Ω is connected across a battery of 6 V. Calculate the current in the circuit and the charge passing through the resistor in 5 minutes.
题目：一个 15 Ω 的电阻器接在 6 V 的电池两端。计算电路中的电流及 5 分钟内通过电阻器的电荷量。
Using Ohm’s law: V = IR, rearranging gives I = V / R.
使用欧姆定律：V = IR，变形得 I = V / R。
V = IR
I = 6 V / 15 Ω = 0.4 A.
I = 6 V / 15 Ω = 0.4 A。
Charge Q is given by Q = I × t. Time t = 5 min = 300 s.
电荷量 Q = I × t。时间 t = 5 min = 300 s。
Q = 0.4 A × 300 s = 120 C.
Q = 0.4 A × 300 s = 120 C。
The current is 0.4 A and the charge is 120 coulombs.
电流为 0.4 A，电荷量为 120 库仑。
5. Atomic Structure and Radioactivity | 原子结构与放射性
Question: Uranium‑238 undergoes alpha decay to form thorium. Write the nuclear equation and determine the atomic number and mass number of the daughter nucleus.
题目：铀‑238 发生 α 衰变生成钍。写出核反应方程，并确定子核的原子序数和质量数。
An alpha particle is a helium nucleus with 2 protons and 2 neutrons, symbol ⁴₂He.
α 粒子是一个氦核，含有 2 个质子和 2 个中子，符号为 ⁴₂He。
The general equation for alpha decay: ²³⁸₉₂U → ⁴₂He + ²³⁴₉₀Th.
α 衰变的一般方程：²³⁸₉₂U → ⁴₂He + ²³⁴₉₀Th。
Checking: mass number 238 = 4 + 234, atomic number 92 = 2 + 90.
验证：质量数 238 = 4 + 234，原子序数 92 = 2 + 90。
The daughter nucleus thorium has an atomic number of 90 and a mass number of 234.
子核钍的原子序数为 90，质量数为 234。
6. Chemical Bonding and Structure | 化学键与结构
Question: Describe the bonding in a crystal of sodium chloride and explain why it has a high melting point.
题目：描述氯化钠晶体中的键合，并解释为何它具有高熔点。
Sodium chloride is an ionic compound. Sodium atoms lose one electron to form Na⁺ ions, while chlorine atoms gain one electron to form Cl⁻ ions.
氯化钠是离子化合物。钠原子失去一个电子形成 Na⁺ 离子，氯原子得到一个电子形成 Cl⁻ 离子。
The oppositely charged ions are held together by strong electrostatic forces of attraction, forming a giant ionic lattice. A large amount of energy is required to overcome these forces, hence the high melting point.
带相反电荷的离子通过强大的静电吸引力结合在一起，形成巨大的离子晶格。需要大量能量来克服这些力，因此熔点高。
In the lattice, each Na⁺ is surrounded by six Cl⁻ ions and vice versa, giving a cubic arrangement typical of NaCl.
在晶格中，每个 Na⁺ 被六个 Cl⁻ 包围，反之亦然，形成典型的 NaCl 立方排列。
7. Moles, Mass and Gas Volumes | 摩尔、质量与气体体积
Question: Calculate the number of moles in 4.9 g of sulfuric acid, H₂SO₄. (H = 1, S = 32, O = 16)
题目：计算 4.9 g 硫酸 (H₂SO₄) 的物质的量。(H = 1, S = 32, O = 16)
First find the molar mass of H₂SO₄: (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g/mol.
首先求 H₂SO₄ 的摩尔质量：(2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g/mol。
The number of moles n = mass / molar mass = 4.9 g / 98 g/mol.
物质的量 n = 质量 / 摩尔质量 = 4.9 g / 98 g/mol。
n = m / M
n = 4.9 / 98 = 0.05 mol.
n = 4.9 / 98 = 0.05 mol。
If this sample were a gas at r.t.p., its volume would be 0.05 mol × 24 dm³/mol = 1.2 dm³. (1 mol of any gas at r.t.p. occupies 24 dm³)
如果该样品在常温常压下为气体，其体积为 0.05 mol × 24 dm³/mol = 1.2 dm³。（常温常压下 1 mol 任何气体体积为 24 dm³）
8. Rates of Reaction | 反应速率
Question: Marble chips (calcium carbonate) react with dilute hydrochloric acid to produce carbon dioxide gas. Explain two ways to increase the rate of this reaction, using collision theory.
题目：大理石碎片（碳酸钙）与稀盐酸反应生成二氧化碳气体。根据碰撞理论，解释两种加快该反应速率的方法。
Increasing the concentration of the acid provides more H⁺ ions per unit volume, increasing the frequency of successful collisions between reactant particles, thus raising the rate.
增加酸的浓度使单位体积内的 H⁺ 离子增多，提高了反应物粒子间成功碰撞的频率，从而加快反应速率。
Using powdered marble instead of large chips increases the surface area of the solid reactant, allowing more collisions to occur at the same time, which also speeds up the reaction.
使用粉末状大理石代替大块碎片增加了固体反应物的表面积，使同时发生的碰撞更多，这也加快了反应速率。
The reaction equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Monitoring the volume of CO₂ collected over time can be used to measure the rate.
反应方程式：CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。通过测量不同时间收集到的 CO₂ 体积可以监测反应速率。
9. Electrolysis and Redox | 电解与氧化还原
Question: Predict the products of the electrolysis of molten sodium chloride using inert electrodes. Write the half‑equations.
题目：预测用惰性电极电解熔融氯化钠的产物，并写出半反应方程式。
Molten NaCl contains Na⁺ and Cl⁻ ions. At the cathode (negative electrode), reduction occurs: Na⁺ + e⁻ → Na (liquid sodium metal).
熔融 NaCl 含有 Na⁺ 和 Cl⁻ 离子。在阴极（负极），发生还原反应：Na⁺ + e⁻ → Na（液态金属钠）。
At the anode (positive electrode), oxidation occurs: 2Cl⁻ → Cl₂ + 2e⁻ (chlorine gas is produced).
在阳极（正极），发生氧化反应：2Cl⁻ → Cl₂ + 2e⁻（生成氯气）。
The overall reaction is: 2NaCl → 2Na + Cl₂. Inert electrodes (e.g., graphite) do not take part in the reaction.
总反应为：2NaCl → 2Na + Cl₂。惰性电极（如石墨）不参与反应。
10. Cell Structure and Microscopy | 细胞结构与显微镜
Question: A plant cell has an actual diameter of 0.04 mm. Under a light microscope it appears to have a diameter of 16 mm. Calculate the magnification.
题目：一个植物细胞的实际直径为 0.04 mm。在光学显微镜下观察到的直径为 16 mm。计算放大倍数。
Magnification = size of image / actual size of the object. Both measurements must be in the same units.
放大倍数 = 图像大小 / 物体的实际大小。两者单位必须一致。
Image size = 16 mm, actual size = 0.04 mm.
图像大小 = 16 mm，实际大小 = 0.04 mm。
Magnification = 16 mm / 0.04 mm = 400
放大倍数 = 16 / 0.04 = 400 倍。
The micrograph is 400 times larger than the real cell. The formula can also be written as M = I / A.
这张显微图是真实细胞的 400 倍大。该公式也可写成 M = I / A。
11. Photosynthesis and Plant Nutrition | 光合作用与植物营养
Question: Write the word and balanced chemical equation for photosynthesis. State the conditions required and explain how leaves are adapted for this process.
题目：写出光合作用的文字表达式和配平的化学方程式，说明所需条件，并解释叶片如何适应这一过程。
Word equation: carbon dioxide + water → glucose + oxygen, using light energy and chlorophyll.
文字表达式：二氧化碳 + 水 → 葡萄糖 + 氧气，需要光能和叶绿素。
Balanced chemical equation:
配平的化学方程式：
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Leaves are adapted by having a large surface area to absorb light, thin structure for short diffusion paths, stomata for gas exchange, and chloroplasts containing chlorophyll to trap light energy.
叶片的适应性包括：较大的表面积以吸收光线，薄的结构使扩散路径短，气孔用于气体交换，以及含有叶绿素的叶绿体捕获光能。
A deficiency in magnesium or light can limit the rate of photosynthesis.
缺乏镁或光照会限制光合作用的速率。
12. Digestion and Enzymes | 消化与酶
Question: Describe the digestion of starch from the mouth to the small intestine, naming the enzymes involved and their products.
题目：描述淀粉从口腔到小肠的消化过程，指出所涉及的酶及其产物。
Digestion of starch begins in the mouth, where salivary amylase breaks starch into maltose. The food is then swallowed and passes into the stomach, but amylase is denatured by stomach acid.
淀粉的消化从口腔开始，唾液淀粉酶将淀粉分解为麦芽糖。然后食物被吞咽进入胃，但淀粉酶会被胃酸变性。
In the small intestine, pancreatic amylase continues breaking down starch to maltose. Finally, maltase, a membrane‑bound enzyme on the intestinal lining, breaks maltose into glucose, which is absorbed into the blood.
在小肠中，胰淀粉酶继续将淀粉分解为麦芽糖。最后，肠壁上的膜结合酶麦芽糖酶将麦芽糖分解为葡萄糖，被吸收进入血液。
The overall conversion: starch → maltose → glucose. Enzymes are specific and work at optimum pH and temperature.
总体转化过程：淀粉 → 麦芽糖 → 葡萄糖。酶具有专一性，并在最适 pH 和温度下发挥作用。
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