📚 IGCSE Science: Typical Worked Examples | IGCSE 科学：典型例题详解
This article presents a selection of typical IGCSE Science questions drawn from Physics, Chemistry and Biology, each followed by a step‑by‑step solution. The worked examples are designed to help you master key concepts, build confidence in applying knowledge, and practise the reasoning skills required for the exam. By studying these model answers, you will see how marks are awarded and learn to avoid common pitfalls.
本文精选了物理、化学和生物三科的 IGCSE 科学典型例题，并配有详细的逐步解答。这些例题旨在帮助你掌握核心概念，建立运用知识的信心，并训练考试所需的推理能力。通过分析这些模范答案，你将了解阅卷的给分点，学会避开常见错误。
1. Physics: Motion Graphs – Worked Example | 物理：运动图像典型例题
A car accelerates from rest at 2 m/s² for 5 seconds, then travels at constant speed for 10 seconds, and finally decelerates uniformly to stop in 4 seconds. The student is asked to sketch the velocity–time graph and use it to calculate the total distance travelled.
一辆汽车从静止开始以 2 m/s² 的加速度行驶 5 秒，然后匀速行驶 10 秒，最后匀减速至停止，耗时 4 秒。要求画出速度–时间图像，并利用图像计算行驶的总距离。
Step 1: Determine the velocity at the end of the acceleration phase. Using v = u + at, where u = 0, a = 2 m/s², t = 5 s, we have v = 0 + 2 × 5 = 10 m/s. So the graph rises from (0,0) to (5,10) with a straight line.
步骤一：计算加速阶段结束时的速度。使用 v = u + at，其中 u = 0，a = 2 m/s²，t = 5 s，得到 v = 0 + 2 × 5 = 10 m/s。因此图像从 (0,0) 直线上升至 (5,10)。
Step 2: The constant‑speed section is a horizontal line at 10 m/s from t = 5 s to t = 15 s. The deceleration phase: from 10 m/s to 0 in 4 s, giving a straight line from (15,10) to (19,0).
步骤二：匀速阶段为一条位于 10 m/s 的水平线，时间从 5 s 到 15 s。减速阶段：从 10 m/s 到 0，耗时 4 s，是一条从 (15,10) 到 (19,0) 的直线。
Step 3: The total distance is the area under the v–t graph. Divide the area into a triangle (0‑5 s), a rectangle (5‑15 s) and a second triangle (15‑19 s). Area = (½ × 5 × 10) + (10 × 10) + (½ × 4 × 10) = 25 + 100 + 20 = 145 m.
步骤三：总距离为 v–t 图像下的面积。将图形分割为一个三角形 (0‑5 s)、一个矩形 (5‑15 s) 和另一个三角形 (15‑19 s)。面积 = (½ × 5 × 10) + (10 × 10) + (½ × 4 × 10) = 25 + 100 + 20 = 145 m。
Key point: The slope of a v–t graph gives acceleration; the area gives displacement. Always check that the units are consistent and show your working clearly.
关键点：v–t 图像的斜率表示加速度；面积表示位移。务必检查单位的一致性，并清晰展示解题步骤。
2. Physics: Circuits and Ohm’s Law – Worked Example | 物理：电路与欧姆定律典型例题
A circuit consists of a 12 V battery and two resistors in series: R₁ = 4 Ω and R₂ = 6 Ω. Calculate (a) the total resistance, (b) the current flowing in the circuit, and (c) the potential difference across each resistor.
一个电路由 12 V 电池和两个串联电阻组成：R₁ = 4 Ω，R₂ = 6 Ω。计算：(a) 总电阻，(b) 电路中的电流，(c) 每个电阻两端的电压。
For series resistors, total resistance Rₜ = R₁ + R₂ = 4 + 6 = 10 Ω. This is simply the sum.
对于串联电阻，总电阻 Rₜ = R₁ + R₂ = 4 + 6 = 10 Ω。这就是直接相加。
Using Ohm’s law (V = IR), the current I = V / Rₜ = 12 V / 10 Ω = 1.2 A. The current is the same through all components in a series circuit.
使用欧姆定律 (V = IR)，电流 I = V / Rₜ = 12 V / 10 Ω = 1.2 A。在串联电路中，通过所有元件的电流相同。
Potential difference across R₁: V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V. Across R₂: V₂ = I × R₂ = 1.2 A × 6 Ω = 7.2 V. As a check, V₁ + V₂ = 4.8 + 7.2 = 12 V, which equals the battery voltage.
R₁ 两端的电压：V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V。R₂ 两端的电压：V₂ = I × R₂ = 1.2 A × 6 Ω = 7.2 V。检验：V₁ + V₂ = 4.8 + 7.2 = 12 V，等于电池电压。
Common mistake: forgetting that current is constant in series. Always redraw the circuit if you need to visualise the pathway.
常见错误：忘记串联电路中电流处处相等。如需直观理解电流路径，可以重新绘制电路图。
3. Physics: Energy and Work – Worked Example | 物理：能量与功典型例题
A ball of mass 0.5 kg is dropped from a height of 10 m. Taking g = 10 m/s², calculate (a) the gravitational potential energy (GPE) at the start, (b) the velocity of the ball just before it hits the ground, assuming no air resistance.
一个质量为 0.5 kg 的小球从 10 m 高度落下。取 g = 10 m/s²，计算：(a) 起始时的重力势能 (GPE)，(b) 小球落地前的速度，假设没有空气阻力。
GPE = m × g × h = 0.5 × 10 × 10 = 50 J. All of this energy is converted to kinetic energy (KE) at ground level, assuming no energy losses.
重力势能 GPE = m × g × h = 0.5 × 10 × 10 = 50 J。假设无能量损失，这些能量在落地时全部转化为动能 (KE)。
Using KE = ½ m v² and GPE = KE, we have 50 = ½ × 0.5 × v² → 50 = 0.25 v² → v² = 200 → v = √200 ≈ 14.1 m/s. The velocity can also be found using the equation v² = u² + 2gh, with u = 0.
利用 KE = ½ m v² 且 GPE = KE，得到 50 = ½ × 0.5 × v² → 50 = 0.25 v² → v² = 200 → v = √200 ≈ 14.1 m/s。也可用运动学公式 v² = u² + 2gh 求解，其中 u = 0。
Always state the principle of conservation of energy when appropriate. In exam questions, show the substitution step clearly to gain full marks.
在适当的时候务必说明能量守恒定律。考试答题时，清楚地写出代入步骤才能获得全部分数。
4. Chemistry: Balancing Chemical Equations – Worked Example | 化学：配平化学方程式典型例题
Balance the equation for the reaction of iron with oxygen to form iron(III) oxide: Fe + O₂ → Fe₂O₃.
配平铁与氧气反应生成氧化铁的化学方程式：Fe + O₂ → Fe₂O₃。
Step 1: Count atoms on each side. Left: Fe = 1, O = 2. Right: Fe = 2, O = 3. The numbers are unbalanced.
步骤一：数清两边各元素的原子数。左侧：Fe = 1，O = 2。右侧：Fe = 2，O = 3。原子数不相等。
Step 2: Balance Fe by placing a coefficient 2 in front of Fe on the left: 2Fe + O₂ → Fe₂O₃. Now Fe is balanced (2 on each side).
步骤二：在左侧 Fe 前放置系数 2 以平衡铁原子：2Fe + O₂ → Fe₂O₃。此时铁原子两边均为 2，已平衡。
Step 3: Balance oxygen. Left: O = 2. Right: O = 3. To balance, we can use fractions: O₂ needs to provide 3 oxygen atoms, so we use 3/2 O₂, giving 2Fe + (3/2)O₂ → Fe₂O₃. Then multiply all coefficients by 2 to clear the fraction: 4Fe + 3O₂ → 2Fe₂O₃.
步骤三：平衡氧原子。左侧：O = 2，右侧：O = 3。可先用分数：使 O₂ 提供 3 个氧原子，采用 3/2 O₂，得到 2Fe + (3/2)O₂ → Fe₂O₃。然后将所有系数乘以 2 消除分数：4Fe + 3O₂ → 2Fe₂O₃。
Final check: Fe: 4 on left, 4 on right; O: 6 on left, 6 on right. The equation is balanced.
最终检查：铁原子左右均为 4；氧原子左右均为 6。方程式已配平。
5. Chemistry: Moles and Mass Calculations – Worked Example | 化学：摩尔与质量计算典型例题
When calcium carbonate is heated, it decomposes: CaCO₃ → CaO + CO₂. Calculate the mass of carbon dioxide produced when 10 g of CaCO₃ is completely decomposed. (Relative atomic masses: Ca = 40, C = 12, O = 16)
加热碳酸钙时发生分解反应：CaCO₃ → CaO + CO₂。计算 10 g 碳酸钙完全分解时生成的二氧化碳质量。(相对原子质量：Ca = 40，C = 12，O = 16)
First, find the molar mass of CaCO₃: 40 + 12 + (3 × 16) = 100 g/mol. Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol.
首先计算 CaCO₃ 的摩尔质量：40 + 12 + (3 × 16) = 100 g/mol。CO₂ 的摩尔质量 = 12 + (2 × 16) = 44 g/mol。
Number of moles of CaCO₃ used = mass / molar mass = 10 g / 100 g/mol = 0.10 mol. From the equation, 1 mol CaCO₃ produces 1 mol CO₂, so 0.10 mol CaCO₃ produces 0.10 mol CO₂.
所用 CaCO₃ 的摩尔数 = 质量 / 摩尔质量 = 10 g / 100 g/mol = 0.10 mol。由方程式可知，1 mol CaCO₃ 生成 1 mol CO₂，因此 0.10 mol CaCO₃ 生成 0.10 mol CO₂。
Mass of CO₂ = moles × molar mass = 0.10 mol × 44 g/mol = 4.4 g. The answer is 4.4 g.
CO₂ 的质量 = 摩尔数 × 摩尔质量 = 0.10 mol × 44 g/mol = 4.4 g。答案为 4.4 g。
Always write the balanced equation and show the molar ratio clearly. Many errors arise from misreading the stoichiometry.
一定要写出配平的方程式，并清晰标明摩尔比。许多错误源于对反应计量比的误读。
6. Chemistry: Rates of Reaction – Worked Example | 化学：反应速率典型例题
A student investigates the rate of reaction between marble chips (calcium carbonate) and dilute hydrochloric acid. Concentrated acid is available. Explain how increasing the concentration of the acid affects the reaction rate, and sketch the expected volume‑of‑gas‑against‑time graph for two different concentrations.
某学生研究大理石碎片（碳酸钙）与稀盐酸的反应速率，实验室有浓盐酸可供使用。解释增大酸浓度如何影响反应速率，并绘制两种不同浓度下预期气体体积–时间图。
Increasing the concentration of the acid means there are more H⁺ ions per unit volume. This leads to a greater frequency of successful collisions between reactant particles, so the rate of reaction increases. The same total volume of gas is eventually produced because the mass of marble chips is unchanged.
增大酸浓度意味着单位体积内有更多的 H⁺ 离子。这导致反应物粒子之间成功碰撞的频率增加，因此反应速率加快。由于大理石碎片的质量不变，最终产生的气体总体积相同。
On the graph, the curve for the higher concentration rises more steeply from the start and levels off earlier, while the lower concentration curve rises more slowly. Both curves reach the same final volume. The independent variable is acid concentration; the volume of gas produced is the dependent variable. Control variables: mass and surface area of marble chips, temperature, volume of acid.
在图像上，较高浓度对应的曲线从一开始就上升得更陡，并更早趋于平缓；较低浓度对应的曲线上升较慢。两条曲线最终达到相同的气体体积。自变量是酸的浓度；因变量是产生的气体体积。控制变量：大理石碎片的质量和表面积、温度、酸的体积。
In exam descriptions, always link the change to collision theory: more particles in a given volume → more frequent successful collisions → faster rate.
在考试描述中，务必将变化与碰撞理论联系起来：给定体积内粒子增多 → 成功碰撞更频繁 → 速率加快。
7. Biology: Cells and Microscopy – Worked Example | 生物：细胞与显微镜典型例题
An animal cell is viewed under a light microscope at a magnification of ×400. The image of the cell has a diameter of 2.0 mm when measured on the micrograph. Calculate the actual diameter of the cell in micrometres (µm).
在光学显微镜下以 400 倍的放大倍数观察一个动物细胞。显微照片上测量出的细胞图像直径为 2.0 mm。计算该细胞的实际直径，以微米 (µm) 表示。
Actual size = image size / magnification. First, convert 2.0 mm to micrometres: 1 mm = 1000 µm, so 2.0 mm = 2000 µm.
实际大小 = 图像大小 / 放大倍数。先将 2.0 mm 换算为微米：1 mm = 1000 µm，因此 2.0 mm = 2000 µm。
Then actual diameter = 2000 µm / 400 = 5 µm. This is a typical size for many animal cells, such as a cheek cell.
于是实际直径 = 2000 µm / 400 = 5 µm。这是许多动物细胞（如口腔上皮细胞）的典型尺寸。
You may also be asked to label the nucleus, cytoplasm and cell membrane on a diagram. Remember that animal cells do not have a cell wall, chloroplasts or a large permanent vacuole.
你也可能被要求在图上标出细胞核、细胞质和细胞膜。请记住，动物细胞没有细胞壁、叶绿体和大型中央液泡。
When using the formula, ensure the units of image size and actual size are the same. Magnification has no units.
使用公式时，确保图像大小与实际大小的单位一致。放大倍数没有单位。
8. Biology: Photosynthesis Experiment – Worked Example | 生物：光合作用实验典型例题
Pondweed (Elodea) is often used to study the rate of photosynthesis by counting the number of oxygen bubbles produced per minute. A student wants to investigate the effect of light intensity. Describe how to set up a fair test and identify the key variables.
经常使用水蕴草（伊乐藻）通过计数每分钟产生的氧气气泡数来研究光合作用的速率。某学生想探究光照强度的影响。描述如何设计公平实验并确定关键变量。
Place a piece of pondweed in a beaker of water with a source of carbon dioxide (such as sodium hydrogencarbonate solution). Position a lamp at a measured distance from the beaker. Count the number of bubbles released over a fixed time, e.g., 1 minute, and repeat for each distance. To change light intensity, vary the distance of the lamp (closer = higher intensity).
将一段水蕴草放入盛有水和二氧化碳源（如碳酸氢钠溶液）的烧杯中。将一盏灯放置在距离烧杯一定距离的位置。计数固定时间（如 1 分钟）内释放的气泡数，并在每个距离下重复实验。通过改变灯的距离来改变光照强度（越近强度越高）。
Independent variable: light intensity (distance of lamp). Dependent variable: number of bubbles per minute (rate of photosynthesis). Control variables: temperature (use a water bath if needed), concentration of CO₂ solution, same piece of pondweed or same mass, same volume of water, and time allowed for the plant to acclimatise.
自变量：光照强度（灯的距离）。因变量：每分钟气泡数（光合作用速率）。控制变量：温度（必要时使用水浴）、CO₂ 溶液浓度、同一段水蕴草或相同质量、相同体积的水，以及植株适应环境的时间。
A common improvement is to collect the gas in a syringe or inverted measuring cylinder for more accurate volume measurement, rather than counting bubbles, because bubbles can vary in size.
常见的改进方案是用注射器或倒置量筒收集气体以更精确地测量体积，而不是计数气泡，因为气泡大小可能不一致。
9. Biology: Genetics and Punnett Squares – Worked Example | 生物：遗传与旁氏表典型例题
In pea plants, the allele for tall stems (T) is dominant over the allele for dwarf stems (t). Two heterozygous tall plants (Tt) are crossed. Predict the possible genotypes and phenotypes of the offspring and their expected ratios.
在豌豆植株中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。将两株杂合高茎植株 (Tt) 杂交。预测后代可能的基因型和表现型及其预期比例。
First, construct a Punnett square. The gametes from each parent are T and t. Combine them: top row T, t; left column T, t. The resulting offspring genotypes are: TT, Tt, Tt, tt.
首先构建旁氏表。每个亲本产生的配子是 T 和 t。将其组合：上行 T、t；左列 T、t。后代基因型为：TT、Tt、Tt、tt。
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
Genotypic ratio: 1 TT : 2 Tt : 1 tt. Phenotypic ratio: Tall plants (TT and Tt) : dwarf (tt) = 3 : 1. There is a 75% probability of a tall offspring and 25% dwarf.
基因型比例：1 TT : 2 Tt : 1 tt。表现型比例：高茎 (TT 和 Tt) : 矮茎 (tt) = 3 : 1。后代为高茎的概率为 75%，矮茎为 25%。
Remember, dominant alleles mask the effect of recessive alleles. Always define the symbols clearly and show the Punnett square step by step to earn full marks.
请记住，显性等位基因掩盖隐性等位基因的效应。始终明确定义所用符号，并逐步展示旁氏表以获得满分。
10. Data Analysis and Experimental Design – Worked Example | 数据分析与实验设计典型例题
A student cools a sample of octadecanoic acid and records the temperature every 30 seconds until it solidifies. The data table shows a plateau region around 69 °C. Explain the shape of the cooling curve and how it can be used to determine the melting point.
某学生冷却硬脂酸样品，每隔 30 秒记录一次温度，直到其凝固。数据表显示在约 69 °C 处有一段平台。解释冷却曲线的形状，以及如何利用该曲线确定熔点。
In a cooling curve for a pure substance, the temperature falls steadily as the liquid cools. When the substance begins to freeze, the temperature remains constant for a period even though heat is still being lost. This happens because the energy released by the forming intermolecular bonds compensates for the energy lost to the surroundings. The constant temperature during the plateau is the freezing point (which equals the melting point for a pure substance).
在纯物质的冷却曲线中，随着液体冷却，温度稳步下降。当物质开始凝固时，尽管仍在散热，温度会在一段时间内保持不变。这是因为分子间键形成时释放的能量补偿了散失到环境中的能量。平台阶段的恒定温度即为凝固点（对于纯物质，凝固点等于熔点）。
To determine the melting point from the data, identify the temperature at which the plateau occurs. Plot the data points and draw two best‑fit straight lines: one through the falling portion and one through the flat portion. The intersection of these lines gives an accurate value of the melting point. Repeated readings improve reliability.
根据数据确定熔点的方法是找出平台出现的温度。描出数据点，画两条最佳拟合直线：一条穿过下降部分，一条穿过平台部分。两条线的交点即为精确的熔点值。重复测量可以提高可靠性。
Controlled variables in such an experiment include the mass of the substance, the volume of air around the tube, and the starting temperature. Always comment on the precision of the thermometer used and suggest improvements, such as using a data logger.
此类实验的控制变量包括物质的质量、试管周围的空气体积以及起始温度。一定要对所用温度计的精确度予以评价，并提出改进建议，例如使用数据记录仪。
This type of question often requires you to sketch the curve and label the axes correctly: time on the x‑axis, temperature on the y‑axis, with units.
这类题目通常要求你绘制曲线并正确标注坐标轴：x 轴为时间，y 轴为温度，并标明单位。
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