📚 Typical Exam Questions Explained for IGCSE WJEC Science | IGCSE WJEC 科学：典型例题详解
Welcome to this comprehensive walkthrough of typical IGCSE WJEC Science exam questions. Understanding how to approach and answer these questions is key to achieving a high grade. In this article, we will cover worked examples from Physics, Chemistry, and Biology, highlighting the reasoning, formulas, and presentation expected by examiners. Each section is followed by a detailed explanation in both English and Chinese to support bilingual learners.
欢迎阅读这份全面的 IGCSE WJEC 科学典型例题解析。掌握如何应对并解答这些题目是取得高分的关键。在本文中，我们将涵盖物理、化学和生物的典型示例，重点展示解题思路、公式以及阅卷人期望的作答格式。每个部分都配有详细的中英文解释，以帮助双语学习者。
1. Physics: Speed, Acceleration and Distance | 物理：速度、加速度与路程
Question: A car accelerates from rest to 20 m/s in 5 seconds. Calculate its acceleration. The car then continues at a constant speed of 20 m/s for a further 10 seconds. What is the total distance travelled during the whole 15 seconds?
题目：一辆汽车从静止加速到 20 m/s，用时 5 秒。计算它的加速度。随后汽车以 20 m/s 的恒定速度再行驶 10 秒。在整个 15 秒内，汽车行驶的总路程是多少？
Step 1 – Acceleration: Acceleration is the rate of change of velocity. Use a = (v – u) / t, where v = 20 m/s, u = 0 m/s, t = 5 s.
步骤1 – 加速度：加速度是速度的变化率。使用公式 a = (v – u) / t，其中 v = 20 m/s，u = 0 m/s，t = 5 s。
a = (20 – 0) / 5 = 4 m/s²
Step 2 – Distance during acceleration: Use s = u t + ½ a t² = 0 × 5 + ½ × 4 × 5² = 0 + ½ × 4 × 25 = 50 m.
步骤2 – 加速阶段的路程：使用 s = u t + ½ a t² = 0 × 5 + ½ × 4 × 5² = 0 + ½ × 4 × 25 = 50 m。
Step 3 – Distance at constant speed: In the next 10 s, speed is constant so distance = speed × time = 20 × 10 = 200 m.
步骤3 – 匀速阶段的路程：在接下来的 10 秒内，速度恒定，因此路程 = 速度 × 时间 = 20 × 10 = 200 m。
Total distance = 50 m + 200 m = 250 m. Always include units.
总路程 = 50 m + 200 m = 250 m。务必带上单位。
2. Physics: Ohm’s Law in Circuits | 物理：电路中的欧姆定律
Question: A resistor has a potential difference of 6 V across it and a current of 0.5 A flowing through it. Calculate its resistance. If the voltage is increased to 12 V and the resistance stays the same, what will be the new current?
题目：一个电阻两端的电压为 6 V，流过的电流为 0.5 A。计算它的电阻值。如果电压升高到 12 V 而电阻保持不变，新的电流是多少？
Ohm’s law states that V = I × R. Rearranging gives R = V / I. Here R = 6 V / 0.5 A = 12 Ω.
欧姆定律指出 V = I × R。变形后得 R = V / I。这里 R = 6 V / 0.5 A = 12 Ω。
For the second part, using I = V / R: I = 12 V / 12 Ω = 1 A. Doubling the voltage doubles the current when resistance is constant.
第二部分使用 I = V / R：I = 12 V / 12 Ω = 1 A。当电阻不变时，电压翻倍，电流也翻倍。
Remember to check units: voltage in volts (V), current in amperes (A), resistance in ohms (Ω).
请记住检查单位：电压用伏特 (V)，电流用安培 (A)，电阻用欧姆 (Ω)。
3. Chemistry: Moles and Mass Calculations | 化学：摩尔与质量计算
Question: Calculate the mass of 0.5 moles of calcium carbonate, CaCO₃. Relative atomic masses: Ca = 40, C = 12, O = 16.
题目：计算 0.5 摩尔碳酸钙 (CaCO₃) 的质量。相对原子质量：Ca = 40，C = 12，O = 16。
First, find the molar mass (Mᵣ) of CaCO₃. Molar mass = 40 (Ca) + 12 (C) + (3 × 16) (O₃) = 40 + 12 + 48 = 100 g/mol.
首先，计算 CaCO₃ 的摩尔质量 (Mᵣ)。摩尔质量 = 40 (Ca) + 12 (C) + (3 × 16) (O₃) = 40 + 12 + 48 = 100 g/mol。
Mass = moles × molar mass = 0.5 × 100 = 50 g
A common error is forgetting to multiply oxygen’s mass by three. Always write the formula correctly to count all atoms.
一个常见错误是忘记将氧的质量乘以三。始终正确写出化学式，数清所有原子。
4. Chemistry: Collision Theory and Reaction Rate | 化学：碰撞理论与反应速率
Question: Explain, using collision theory, why increasing the temperature increases the rate of a chemical reaction.
题目：用碰撞理论解释为什么升高温度会加快化学反应速率。
Collision theory states that for a reaction to occur, particles must collide with sufficient energy (greater than or equal to the activation energy) and with the correct orientation.
碰撞理论指出，要发生反应，粒子必须发生碰撞，且碰撞能量必须足够（大于或等于活化能），同时取向必须正确。
When temperature rises, particles gain more kinetic energy and move faster. This leads to two effects: more frequent collisions per second, and a much greater proportion of collisions having energy ≥ activation energy. The second factor is more significant, greatly increasing the number of successful collisions per unit time.
当温度升高时，粒子获得更多动能，运动加快。这会带来两个效应：每秒碰撞次数增加，以及具有活化能以上能量的碰撞比例大幅上升。第二个因素更为重要，它显著增加了单位时间内有效碰撞的次数。
5. Biology: Enzyme Activity and Graphs | 生物：酶活性与图表
Question: The graph below shows how the activity of a human enzyme changes with temperature. Describe and explain the shape of the graph.
题目：下图显示了人体内某种酶的活性随温度变化的情况。描述并解释该图形的形状。
As temperature increases from 0°C, the enzyme activity increases because the enzyme and substrate molecules have more kinetic energy, move faster, and collide more frequently. The rate of formation of enzyme–substrate complexes rises.
当温度从 0°C 开始升高时，酶活性增加，这是因为酶和底物分子获得更多动能，运动更快，碰撞更频繁，酶-底物复合物的形成速率上升。
The activity reaches a maximum at the optimum temperature (around 37°C for many human enzymes). Beyond this point, the enzyme begins to denature: the weak bonds holding the three-dimensional shape of the active site break, the active site changes shape, and the substrate no longer fits. Activity falls sharply.
活性在最适温度（对许多人体酶而言约为 37°C）达到最大值。超过该点后，酶开始变性：维持活性部位三维形状的弱键断裂，活性部位形状改变，底物不再契合，活性急剧下降。
6. Biology: Genetics and Punnett Squares | 生物：遗传与庞纳特方格
Question: In pea plants, the tall allele (T) is dominant over the dwarf allele (t). A heterozygous tall plant is crossed with a dwarf plant. Determine the expected genotypes and phenotypes of the offspring.
题目：在豌豆植株中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。将一株杂合高茎植株与一株矮茎植株杂交，确定后代的预期基因型与表现型。
Parent genotypes: Tt (heterozygous tall) × tt (dwarf). The Tt parent produces gametes T and t; the tt parent produces only t.
亲本基因型：Tt（杂合高茎） × tt（矮茎）。Tt 亲本产生配子 T 和 t；tt 亲本只产生 t。
The Punnett square shows the combinations:
庞纳特方格展示了所有组合：
| t | t | |
| T | Tt | Tt |
| t | tt | tt |
Genotypic ratio: 1 Tt : 1 tt (50% heterozygous tall, 50% homozygous dwarf). Phenotypic ratio: 1 tall : 1 dwarf.
基因型比例：1 Tt : 1 tt（50% 杂合高茎，50% 纯合矮茎）。表现型比例：1 高茎 : 1 矮茎。
7. Scientific Enquiry: Variables and Method Design | 科学探究：变量与方法设计
Context: A student investigates how the concentration of hydrochloric acid affects the rate of its reaction with magnesium ribbon. She measures the volume of hydrogen gas produced over time.
背景：一名学生研究盐酸浓度对镁条反应速率的影响。她测量了随时间产生的氢气体积。
The independent variable is the concentration of hydrochloric acid (e.g., 0.5, 1.0, 1.5 mol/dm³). The dependent variable is the volume of gas collected in a fixed time, or the time taken to produce a certain volume. Control variables include the mass/length of magnesium, volume of acid, temperature, and surface area of magnesium.
自变量是盐酸的浓度（例如 0.5、1.0、1.5 mol/dm³）。因变量是固定时间内收集到的气体体积，或产生一定体积气体所需的时间。控制变量包括镁条的质量/长度、酸的体积、温度以及镁条的表面积。
A valid method: Place a known length of magnesium ribbon into a flask with a set volume of acid. Immediately attach a gas syringe or invert a measuring cylinder filled with water. Record the volume of hydrogen at regular intervals (e.g., every 10 seconds). Repeat with each acid concentration and calculate the mean initial rate.
有效的方法：将已知长度的镁条放入装有固定体积酸的烧瓶中，立即接上气体注射器或倒扣一个装满水的量筒。每隔一定时间（如每 10 秒）记录氢气体积。对每种酸浓度重复实验，并计算平均初始速率。
8. Data Analysis: Interpreting Distance–Time Graphs | 数据分析：解读距离-时间图
Question: A distance–time graph for a cyclist shows three sections: a straight sloping line from A to B, a horizontal line from B to C, and a steeper straight sloping line from C to D. Describe the motion in each section and explain how you would calculate the speed.
题目：一位自行车骑手的距离-时间图显示了三个部分：从 A 到 B 是一条倾斜的直线，从 B 到 C 是一条水平线，从 C 到 D 是一条更陡的倾斜直线。描述每一段的运动，并解释如何计算速度。
Section A–B: The straight sloping line indicates constant speed. The gradient of the line equals the speed; a steeper gradient means a higher speed.
A–B 段：倾斜直线表示匀速运动。直线的斜率等于速度；斜率越大速度越快。
Section B–C: The horizontal line shows that distance does not change – the cyclist is stationary (speed = 0).
B–C 段：水平线表示距离没有变化——骑手静止不动（速度为 0）。
Section C–D: The steeper straight sloping line shows that the cyclist is moving again at a higher constant speed than in section A–B. To find the speed in each moving section, divide the change in distance (rise) by the change in time (run).
C–D 段：更陡的倾斜直线表示骑手再次运动，且速度比 A–B 段更高。要计算任一运动段的速度，用距离的变化量（垂直增量）除以时间的变化量（水平增量）。
9. Chemistry: Electrolysis of Molten Lead Bromide | 化学：熔融溴化铅的电解
Question: Describe what happens at the cathode and anode during the electrolysis of molten lead bromide (PbBr₂), and write the half-equations.
题目：描述熔融溴化铅 (PbBr₂) 电解时阴极和阳极各自发生的现象，并写出半反应方程式。
Molten lead bromide contains Pb²⁺ and Br⁻ ions. During electrolysis, cations (Pb²⁺) move to the negative cathode, where they gain electrons (reduction). At the cathode: silvery liquid lead metal forms.
熔融溴化铅中含有 Pb²⁺ 和 Br⁻ 离子。电解时，阳离子 (Pb²⁺) 移向负极（阴极），在那里得到电子（还原）。在阴极：形成银白色液态金属铅。
Cathode half-equation: Pb²⁺ + 2e⁻ → Pb
At the positive anode, anions (Br⁻) are attracted and lose electrons (oxidation). Brown fumes of bromine gas are observed.
在正极（阳极），阴离子 (Br⁻) 被吸引并失去电子（氧化）。可观察到红棕色的溴蒸气。
Anode half-equation: 2Br⁻ → Br₂ + 2e⁻
Remember that oxidation occurs at the anode and reduction at the cathode (OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons).
记住，氧化发生在阳极，还原发生在阴极（OIL RIG：氧化失电子，还原得电子）。
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