📚 IGCSE Edexcel Science: Detailed Explanation of Typical Example Questions | IGCSE Edexcel 科学：典型例题详解
Typical example questions are a powerful way to deepen your understanding of the IGCSE Edexcel Science syllabus. By working through carefully selected problems from Physics, Chemistry and Biology, you can strengthen your conceptual knowledge and learn how to apply principles to unfamiliar situations. This article provides detailed, step-by-step solutions to common question types, with each explanation presented in both English and Chinese to support dual-language learners. The focus is on clear reasoning, correct use of scientific vocabulary, and examination-style approaches that will help boost your confidence and performance.
典型例题是加深理解 IGCSE Edexcel 科学大纲的有效方法。通过练习精心挑选的物理、化学和生物题目，你可以强化概念知识，并学会将原理应用于陌生情境。本文对常见题型进行详细的分步讲解，每个解释都用中英双语呈现，以支持双语学习者。重点在于清晰的推理、正确使用科学词汇以及考试风格的解题方法，这将帮助你提升信心和成绩。
1. Kinematics: Distance from a Velocity-Time Graph | 运动学：从速度-时间图求距离
Question: A car accelerates uniformly from rest to 20 m/s in 10 s. It then travels at constant velocity for 20 s before decelerating uniformly to rest in a further 10 s. Calculate the total distance travelled using a velocity-time graph.
问题：一辆汽车从静止匀加速至20 m/s，用时10 s。然后以恒定速度行驶20 s，之后匀减速至静止，再花10 s。利用速度-时间图计算总行驶距离。
To solve this, we first sketch the velocity-time graph. The graph consists of three sections: an increasing straight line from (0,0) to (10,20), a horizontal line at 20 m/s from t=10 s to t=30 s, and a decreasing straight line from (30,20) to (40,0).
为了解决这个问题，我们首先画出速度-时间图。图形由三段组成：一条从(0,0)到(10,20)的上升直线，一条在20 m/s从 t=10 s 到 t=30 s 的水平线，以及一条从(30,20)到(40,0)的下降直线。
The total area under a velocity-time graph gives the distance travelled. The shape formed is a trapezium with parallel sides of length 20 s (the constant-velocity section) and 40 s (the total time base), and a height of 20 m/s.
速度-时间图下方的总面积代表行驶距离。形成的形状是一个梯形，其平行边长度为20 s（匀速段对应的时间）和40 s（总时间），高度为20 m/s。
Distance = Area of trapezium = ½ × (a + b) × h
Substituting the values: a = 20 s, b = 40 s, h = 20 m/s.
代入数值：a = 20 s，b = 40 s，h = 20 m/s。
Distance = ½ × (20 s + 40 s) × 20 m/s = ½ × 60 × 20 = 600 m
Alternatively, calculate the area of each section: a triangle (0.5×10×20 = 100 m), a rectangle (20×20 = 400 m), and another triangle (0.5×10×20 = 100 m). The sum is 100 + 400 + 100 = 600 m. This confirms the result.
另一种方法是分别计算每段面积：三角形(0.5×10×20 = 100 m)、矩形(20×20 = 400 m)和另一个三角形(0.5×10×20 = 100 m)。总和为100 + 400 + 100 = 600 m。这验证了结果。
2. Energy: Efficiency of an Electric Motor | 能量：电动机的效率
Question: A motor lifts a 5 kg mass through a vertical height of 2 m in 3 s. The motor has a power rating of 50 W. Calculate the efficiency of the motor. (Use g = 10 N/kg)
问题：一台电动机在3 s内将一个5 kg的物体竖直提升2 m。电动机的额定功率为50 W。计算电动机的效率。（取g = 10 N/kg）
First, find the useful work output. The work done against gravity is the gain in gravitational potential energy.
首先，求出有用功输出。克服重力所做的功等于重力势能的增加量。
Useful work = m × g × h = 5 kg × 10 N/kg × 2 m = 100 J
Next, calculate the total electrical energy input. The motor uses 50 W for 3 s, so the energy supplied is power × time.
接下来，计算总输入电能。电动机功率50 W，运行3 s，因此提供能量为功率 × 时间。
Energy input = Power × time = 50 W × 3 s = 150 J
Efficiency is the ratio of useful output energy to total input energy, often expressed as a percentage.
效率是有用输出能量与总输入能量之比，通常以百分数表示。
Efficiency = (Useful energy output / Total energy input) × 100%
Efficiency = (100 J / 150 J) × 100% = 66.7%
The remaining energy is transferred to the surroundings, mainly as thermal energy due to friction and electrical heating. This is a typical single-stage efficiency calculation common in IGCSE Energy topics.
剩余的能量传递到周围环境，主要以摩擦和电热形式散失。这是 IGCSE 能量专题中典型的单级效率计算。
3. Electricity: Series Circuit Calculations | 电学：串联电路计算
Question: A 12 V battery is connected to two resistors in series: 4 Ω and 6 Ω. Calculate the total current flowing and the voltage across each resistor.
问题：一个12 V的电池与两个电阻串联：4 Ω 和 6 Ω。计算电路中的总电流以及每个电阻两端的电压。
For components in series, the total resistance is the sum of individual resistances.
对于串联元件，总电阻等于各个电阻之和。
Rtotal = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω
Apply Ohm’s Law to find the current I. The same current flows through all series components.
应用欧姆定律求电流 I。同一电流流过所有串联元件。
I = V / Rtotal = 12 V / 10 Ω = 1.2 A
Now determine the potential difference across each resistor using V = I × R.
现在使用 V = I × R 确定每个电阻两端的电势差。
Voltage across 4 Ω: V₁ = 1.2 A × 4 Ω = 4.8 V
Voltage across 6 Ω: V₂ = 1.2 A × 6 Ω = 7.2 V
Check that the sum of the voltages equals the supply voltage: 4.8 V + 7.2 V = 12 V. This confirms the conservation of energy in the circuit. In exam answers, always show the addition of voltmeter readings as a verification step.
检查电压之和是否等于电源电压：4.8 V + 7.2 V = 12 V。这验证了电路中的能量守恒。在考试答案中，始终要展示将电压表读数相加作为验证步骤。
4. Radioactivity: Half-life Calculation | 放射性：半衰期计算
Question: A sample of radioactive iodine-131 has an initial count rate of 200 counts per second. After 24 days, its count rate falls to 25 counts per second. Determine the half-life of iodine-131.
问题：一份放射性碘-131样品最初的计数率为200次/秒。24天后，计数率降至25次/秒。求碘-131的半衰期。
We first find how many times the activity has halved. The final count rate is 25, and the initial is 200. The ratio is 25/200 = 1/8.
我们首先找出活度减半的次数。最终计数率为25，初始为200。比值为25/200 = 1/8。
Each half-life reduces the count rate by half. After n half-lives, the fraction remaining is (½)ⁿ.
每个半衰期使计数率减半。经过 n 个半衰期后，剩余比例为 (½)ⁿ。
(½)ⁿ = 1/8 → n = 3
The sample has gone through three half-lives in 24 days. Therefore, one half-life is:
样品在24天内经历了三个半衰期。因此，一个半衰期为：
Half-life = 24 days ÷ 3 = 8 days
A quick tabular check: start 200; after 8 days: 100; after 16 days: 50; after 24 days: 25. This matches the data. This method is essential for all half-life problems, and you should clearly show the halving steps.
快速表格检验：初始200；8天后：100；16天后：50；24天后：25。这与数据相符。这种方法对所有半衰期问题都至关重要，你应该清晰地展示对半步骤。
5. Moles: Mass of Product from a Reacting Mass | 摩尔：由反应物质量求产物质量
Question: What mass of magnesium oxide (MgO) is formed when 2.4 g of magnesium reacts completely with oxygen? (Relative atomic masses: Mg = 24, O = 16)
问题：当2.4 g镁与氧气完全反应时，生成多少质量的氧化镁（MgO）？（相对原子质量：Mg = 24，O = 16）
Write the balanced equation: 2Mg + O₂ → 2MgO. The mole ratio between Mg and MgO is 1 : 1 (2:2).
写出配平的化学方程式：2Mg + O₂ → 2MgO。镁与氧化镁的摩尔比为1 : 1（2:2）。
Calculate moles of magnesium used: n = mass / Mᵣ.
计算所用镁的摩尔数：n = 质量 / 相对原子质量。
Moles of Mg = 2.4 g / 24 g/mol = 0.10 mol
From the equation, 0.10 mol Mg produces 0.10 mol MgO.
根据方程式，0.10 mol 镁生成0.10 mol 氧化镁。
Molar mass of MgO = 24 + 16 = 40 g/mol. Now find the mass.
氧化镁的摩尔质量 = 24 + 16 = 40 g/mol。现在求出质量。
Mass of MgO = 0.10 mol × 40 g/mol = 4.0 g
Always confirm that the mass of product is greater than the reactant mass, since oxygen atoms have been added. This stoichiometric approach is a fundamental skill in IGCSE Chemistry.
始终确认产物质量大于反应物质量，因为加入了氧原子。这种化学计量方法是 IGCSE 化学中的基本技能。
6. Electrolysis of Molten Lead(II) Bromide | 熔融溴化铅的电解
Question: When molten lead(II) bromide (PbBr₂) is electrolysed using inert electrodes, state the products at the anode and cathode, and write the half-equations.
问题：用惰性电极电解熔融溴化铅（PbBr₂）时，写出阳极和阴极的产物，并写出半反应方程式。
In molten lead(II) bromide, the ions present are Pb²⁺ and Br⁻. The cathode attracts positive ions (cations) and the anode attracts negative ions (anions).
在熔融溴化铅中，存在 Pb²⁺ 和 Br⁻ 离子。阴极吸引阳离子，阳极吸引阴离子。
At the cathode, Pb²⁺ ions gain electrons and are reduced to lead metal. This appears as a grey solid.
在阴极，Pb²⁺ 离子得到电子，被还原成金属铅，呈灰色固体。
Cathode half-equation: Pb²⁺ + 2e⁻ → Pb
At the anode, Br⁻ ions lose electrons and are oxidised to bromine gas. Bubbles of reddish-brown gas are observed.
在阳极，Br⁻ 离子失去电子，被氧化成溴气，可观察到红棕色气泡。
Anode half-equation: 2Br⁻ → Br₂ + 2e⁻
The overall reaction is: PbBr₂ → Pb + Br₂. Because the electrodes are inert, they do not take part in the reaction. This question tests knowledge of redox definitions and ion discharge in molten compounds.
总反应为：PbBr₂ → Pb + Br₂。由于电极是惰性的，它们不参与反应。此题考查氧化还原定义和熔融化合物中离子放电的知识。
7. Titration: Concentration of Sodium Hydroxide | 滴定：氢氧化钠的浓度
Question: 25.0 cm³ of sodium hydroxide (NaOH) solution is completely neutralised by 20.0 cm³ of 0.10 mol/dm³ sulfuric acid (H₂SO₄). Calculate the concentration of the NaOH solution.
问题：25.0 cm³ 的氢氧化钠（NaOH）溶液被20.0 cm³ 0.10 mol/dm³的硫酸（H₂SO₄）完全中和。计算氢氧化钠溶液的浓度。
First, write the balanced equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. The mole ratio of NaOH to H₂SO₄ is 2 : 1.
首先，写出配平的化学方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。NaOH 与 H₂SO₄ 的摩尔比为2 : 1。
Calculate the moles of sulfuric acid used: n = c × V (in dm³). Convert 20.0 cm³ to 0.0200 dm³.
计算所用硫酸的摩尔数：n = c × V（单位 dm³）。将20.0 cm³转换为0.0200 dm³。
Moles of H₂SO₄ = 0.10 mol/dm³ × 0.0200 dm³ = 0.0020 mol
Using the 2:1 ratio, the moles of NaOH that reacted are twice this value.
根据2:1的比例，反应的 NaOH 摩尔数是此值的两倍。
Moles of NaOH = 2 × 0.0020 mol = 0.0040 mol
Now, calculate the concentration of NaOH. The volume of NaOH is 25.0 cm³ = 0.0250 dm³.
现在计算 NaOH 的浓度。NaOH 的体积为25.0 cm³ = 0.0250 dm³。
Concentration of NaOH = 0.0040 mol / 0.0250 dm³ = 0.16 mol/dm³
Always give the final answer to an appropriate number of significant figures, here 0.16 mol/dm³. This titration calculation is a classic IGCSE Chemistry problem; remember to convert cm³ to dm³ by dividing by 1000.
最终答案应保留合适数量的有效数字，此处为0.16 mol/dm³。这个滴定计算是经典的 IGCSE 化学问题；记得将 cm³ 除以1000转换为 dm³。
8. Enzyme Activity: Effect of Temperature | 酶活性：温度的影响
Question: An experiment on catalase activity produces a graph showing the volume of oxygen produced over time at 20°C and 40°C. At 40°C, the initial rate is higher but the total volume of oxygen collected is less than at 20°C. Explain these observations.
问题：一项关于过氧化氢酶活性的实验得到一张图表，显示了在20°C和40°C下随时间产生的氧气体积。在40°C时，初始速率更高，但收集到的氧气总体积比20°C时少。请解释这些观察结果。
At the higher temperature, particles have more kinetic energy, so enzyme and substrate molecules move faster. This leads to more frequent successful collisions and a higher initial rate of reaction.
在较高温度下，粒子具有更高的动能，因此酶和底物分子运动更快。这导致更频繁的有效碰撞，初始反应速率更高。
However, enzymes are proteins with a specific three-dimensional shape. At temperatures above their optimum (often around 37°C for many enzymes), the bonds maintaining the active site begin to break. This causes denaturation.
然而，酶是具有特定三维形状的蛋白质。在高于最适温度（许多酶约为37°C）时，维持活性位点的键开始断裂，导致变性。
Denaturation is irreversible and changes the shape of the active site, so the substrate no longer fits. This means the enzyme loses its catalytic function, and the reaction stops earlier. Therefore, less total product is formed.
变性是不可逆的，它改变了活性位点的形状，使得底物无法再与之契合。这意味着酶失去了催化功能，反应提前停止。因此，形成的产物总量较少。
Always link your explanation to the lock-and-key model or the induced-fit model, and mention denaturation specifically. This is a high-mark explanation question common in Biology papers.
你的解释一定要联系锁钥模型或诱导契合模型，并专门提及变性。这是生物试卷中常见的高分解释题。
9. Photosynthesis: Limiting Factors | 光合作用：限制因素
Question: A graph showing the rate of photosynthesis against light intensity rises linearly at first and then levels off at high light intensity. Explain what limits the rate at the plateau and how the rate could be further increased.
问题：一张显示光合作用速率与光照强度关系的图表开始呈线性上升，然后在较高光照强度下趋于平稳。解释在平台期限制速率的因素，以及如何进一步提高速率。
At low light intensity, light is the limiting factor because there is insufficient energy to drive the light-dependent reactions. As light increases, the rate increases proportionally.
在低光照强度下，光是限制因素，因为没有足够的能量驱动光反应。随着光照增加，速率成比例增加。
When the graph plateaus, light is no longer limiting; something else has become the limiting factor. This could be the concentration of carbon dioxide (CO₂) or temperature, depending on the experimental conditions.
当图形达到
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