GCSE CIE Science: Typical Worked Examples | GCSE CIE 科学：典型例题详解

📚 GCSE CIE Science: Typical Worked Examples | GCSE CIE 科学：典型例题详解

Mastering GCSE CIE Science requires not only understanding core concepts but also applying them to exam-style questions. This article walks you through carefully selected typical problems from Physics, Chemistry and Biology, with step-by-step explanations in both English and Chinese. Each example highlights common assessment objectives, including knowledge recall, data analysis and experimental design. Use these worked examples to strengthen your problem-solving skills and avoid common pitfalls.

掌握 GCSE CIE 科学不仅需要理解核心概念，更要能将它们应用到考试题型中。本文精选了物理、化学和生物的典型例题，用中英双语逐步解析。每个例子都突出了常见的考核目标，包括知识回忆、数据分析和实验设计。通过这些例题详解，你可以强化解题技巧，避免常见错误。

1. Physics: Kinematics and Motion Graphs | 物理：运动学与图像题

Example: A car accelerates uniformly from rest and reaches a velocity of 20 m/s in 10 seconds. It then travels at constant velocity for 30 seconds before decelerating uniformly to rest in 5 seconds. (a) Sketch a velocity-time graph for the motion. (b) Calculate the acceleration during the first 10 s. (c) Determine the total distance travelled.

例题：一辆汽车从静止开始匀加速，10 秒内速度达到 20 m/s。然后匀速行驶 30 秒，最后匀减速，5 秒内停下来。(a) 画出该运动的速度-时间图像。(b) 计算前 10 秒的加速度。(c) 求行驶的总距离。

The velocity-time graph consists of three straight-line segments: a rising line from (0,0) to (10,20), a horizontal line from (10,20) to (40,20), and a falling line from (40,20) to (45,0).

速度-时间图由三条直线段组成：从 (0,0) 到 (10,20) 的上升线段，从 (10,20) 到 (40,20) 的水平线段，以及从 (40,20) 到 (45,0) 的下降线段。

(b) Acceleration a = Δv / Δt = (20 – 0) / 10 = 2.0 m/s².

(b) 加速度 a = Δv / Δt = (20 – 0) / 10 = 2.0 m/s²。

(c) Total distance is the area under the graph. Area = (½ × 10 × 20) + (30 × 20) + (½ × 5 × 20) = 100 + 600 + 50 = 750 m.

Always check that your graph axes are labelled with correct units (time in s, velocity in m/s). CIE examiners often award marks for plotting key points accurately, so use a ruler and sharp pencil.

务必检查坐标轴标签及单位（时间用 s，速度用 m/s）。CIE 考官通常会给准确描点打得分，所以务必使用直尺和尖铅笔作画。

2. Physics: Electrical Circuits and Resistance | 物理：电路与电阻

Example: Two resistors, 4 Ω and 6 Ω, are connected in parallel. This parallel combination is then connected in series with a 2 Ω resistor and a 12 V battery of negligible internal resistance. Calculate: (a) the total resistance of the circuit; (b) the current supplied by the battery; (c) the potential difference across the 6 Ω resistor.

例题：两个电阻 4 Ω 和 6 Ω 并联。该并联组合再与一个 2 Ω 电阻和一个内阻可忽略的 12 V 电池串联。计算：(a) 电路的总电阻；(b) 电池供应的电流；(c) 6 Ω 电阻两端的电势差。

Step 1: For parallel resistors, 1/R_parallel = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, so R_parallel = 12/5 = 2.4 Ω.

步骤一：并联电阻，1/R_并联 = 1/4 + 1/6 = 3/12 + 2/12 = 5/12，因此 R_并联 = 12/5 = 2.4 Ω。

Step 2: Total resistance R_total = R_parallel + 2 = 2.4 + 2 = 4.4 Ω.

步骤二：总电阻 R_总 = R_并联 + 2 = 2.4 + 2 = 4.4 Ω。

(b) Current I = V / R_total = 12 / 4.4 ≈ 2.73 A.

(b) 电流 I = V / R_总 = 12 / 4.4 ≈ 2.73 A。

(c) The p.d. across the parallel block is V_parallel = I × R_parallel = 2.73 × 2.4 ≈ 6.55 V. Since resistors in parallel share the same p.d., the voltage across the 6 Ω resistor is also 6.55 V. You could also verify using current division, but recognising equal voltage in parallel is quicker.

Be careful to combine series and parallel sections stepwise. Always leave your working so that method marks can be awarded even if a numerical slip occurs.

要逐步合并串联和并联部分。务必保留计算过程，这样即使出现数字错误也能拿到方法分。

3. Chemistry: Mole Calculations and Stoichiometry | 化学：摩尔计算与化学计量

Example: Calcium carbonate, CaCO₃, decomposes on heating: CaCO₃ → CaO + CO₂. A student heats 10.0 g of pure CaCO₃ until no further mass change. (a) Calculate the number of moles of CaCO₃ used. (b) Determine the mass of calcium oxide, CaO, produced. (c) What volume of CO₂ gas is released at room temperature and pressure (r.t.p., 24 dm³/mol)? (Aᵣ: Ca = 40, C = 12, O = 16)

例题：碳酸钙 CaCO₃ 受热分解：CaCO₃ → CaO + CO₂。某学生加热 10.0 g 纯净 CaCO₃ 至不再发生质量变化。(a) 计算所用 CaCO₃ 的物质的量 (mol)。(b) 求生成的氧化钙 CaO 的质量。(c) 在室温常压 (r.t.p., 24 dm³/mol) 下，释放的 CO₂ 体积是多少？(Aᵣ: Ca = 40, C = 12, O = 16)

(a) Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 100 g/mol. Moles = mass / Mᵣ = 10.0 / 100 = 0.100 mol.

(a) CaCO₃ 摩尔质量 = 40 + 12 + (3 × 16) = 100 g/mol。物质的量 = 质量 / Mᵣ = 10.0 / 100 = 0.100 mol。

(b) From the equation, 1 mol CaCO₃ produces 1 mol CaO. Moles of CaO = 0.100 mol. Mᵣ of CaO = 40 + 16 = 56 g/mol. Mass of CaO = 0.100 × 56 = 5.60 g.

(b) 根据方程式，1 mol CaCO₃ 生成 1 mol CaO。CaO 的物质的量 = 0.100 mol。CaO 的 Mᵣ = 40 + 16 = 56 g/mol。CaO 的质量 = 0.100 × 56 = 5.60 g。

Stoichiometry questions often involve a clear mole ratio. Always write the balanced equation and triple-check relative atomic masses from the Periodic Table. Keep to 3 significant figures if the given data suggests it.

化学计量题常涉及明确的摩尔比。务必写出配平的方程式，并反复核对周期表中的相对原子质量。若题目数据暗示，保留三位有效数字。

4. Chemistry: Rates of Reaction and Energy Changes | 化学：反应速率与能量变化

Example: Marble chips (CaCO₃) are added to excess hydrochloric acid. The volume of CO₂ gas collected is recorded every 30 seconds. A table of results is given. (a) Plot a graph of volume (cm³) against time (s). (b) Use the graph to determine the initial rate of reaction in cm³/s. (c) Explain why the curve levels off. (d) On the same axes, sketch the curve expected if the experiment were repeated with smaller marble chips at the same temperature.

例题：将大理石碎片 (CaCO₃) 加入过量盐酸中。每 30 秒记录收集到的 CO₂ 体积。数据表已给出。(a) 绘制体积 (cm³) 对时间 (s) 的图像。(b) 利用图像确定初始反应速率，单位 cm³/s。(c) 解释曲线为何趋于水平。(d) 在同一坐标轴上画出若用更小的大理石碎片在其他条件不变时预期的曲线。

(a) Plot a smooth curve through the points; it should start steep and gradually flatten. (b) The initial rate is the slope of the tangent at t = 0. Choose two points on the tangent, e.g., (0,0) and (20, 48). Rate = 48/20 = 2.4 cm³/s.

(a) 描点并画出平滑曲线，曲线先陡后趋平。(b) 初始速率是 t = 0 处切线的斜率。在切线上取两点，如 (0,0) 和 (20, 48)。速率 = 48/20 = 2.4 cm³/s。

(c) The reaction slows down because the acid is used up (although excess here, the marble chips are the limiting reactant) – actually here the marble chips are the solid and they become smaller, but the primary reason is the decrease in surface area as marble chips are consumed, and also the concentration of acid drops slightly; typically, the levelling off occurs when one reactant is used up. Since acid is in excess, the reaction stops when all CaCO₃ has reacted.

(d) Smaller chips have a larger total surface area, increasing collision frequency. The curve should start more steeply (higher initial rate) but level off at the same final volume. Label it ‘smaller chips’.

(d) 更小的大理石碎片具有更大的总表面积，增加了碰撞频率。曲线会开始得更陡（更高的初始速率），但最终体积相同。记得标出 ‘更小碎片’。

For rates questions, always mention collision theory: surface area, concentration, temperature or catalyst. Examiners expect clear sketches with labels.

关于速率的题目，一定要提到碰撞理论：表面积、浓度、温度或催化剂。考官期望清晰的草图并带有标签。

5. Biology: Cell Structure and Magnification | 生物：细胞结构与放大倍数

Example: A student views a plant cell under a light microscope using an eyepiece graticule. The cell image measures 18 graticule divisions. With a stage micrometer, 20 divisions on the graticule correspond to 0.1 mm. (a) Calculate the actual size of the cell in micrometres (µm). (b) If the drawing magnification is ×400, what is the length of the cell in the student’s drawing? (c) Name two organelles that would not be visible with a light microscope.

例题：一学生使用目镜测微尺在光学显微镜下观察一个植物细胞。细胞图像占 18 个测微尺分度。借助镜台测微尺得知，20 个测微尺分度相当于 0.1 mm。(a) 计算该细胞的实际大小，以微米 (µm) 表示。(b) 如果绘图放大倍率为 ×400，学生绘图中该细胞多长？(c) 举出两个在光学显微镜下无法看到的细胞器。

(a) First, find the length of one eyepiece graticule division: 0.1 mm / 20 = 0.005 mm. Convert to µm: 0.005 mm = 5 µm. Actual size = 18 × 5 = 90 µm.

(a) 首先，计算每个测微尺分度的实际长度：0.1 mm / 20 = 0.005 mm。转换为微米：0.005 mm = 5 µm。实际大小 = 18 × 5 = 90 µm。

(b) Drawing magnification = image size / actual size. Rearranged: drawing size = actual size × magnification = 90 µm × 400 = 36,000 µm = 36 mm. (Always convert to a sensible unit like mm for a drawing.)

(b) 绘图放大倍率 = 图像大小 / 实际大小。变形得：绘图大小 = 实际大小 × 放大倍率 = 90 µm × 400 = 36,000 µm = 36 mm。(通常绘图用 mm 表示更合理。)

(c) Ribosomes and mitochondria are too small to be resolved clearly by a light microscope; electron microscopes are needed. (Accept also chloroplast details, endoplasmic reticulum, etc.)

Magnification calculations often trip up students with unit conversions. Remember: 1 mm = 1000 µm. Always show the conversion step to gain full marks.

放大倍数计算常因单位换算而丢分。记住：1 mm = 1000 µm。一定要写出换算步骤以获得满分。

6. Biology: Inheritance and Genetic Crosses | 生物：遗传与基因杂交

Example: In pea plants, the allele for tall stems (T) is dominant to the allele for short stems (t). A heterozygous tall plant is crossed with a short plant. (a) State the genotypes of the parents. (b) Draw a Punnett square to show the cross. (c) Determine the probability that an offspring will be tall. (d) Explain what is meant by ‘heterozygous’.

例题：在豌豆中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。一株杂合高茎豌豆与一株矮茎豌豆杂交。(a) 写出亲本的基因型。(b) 绘制旁氏方格表示该杂交。(c) 求出后代为高茎的概率。(d) 解释 “杂合” 的含义。

(a) Heterozygous tall: Tt. Short plant: tt (since recessive trait only appears when homozygous).

(a) 杂合高茎：Tt。矮茎植株：tt（因为隐性性状只在纯合时表现）。

(b) Punnett square: gametes from Tt are T and t; from tt are t and t. Offspring: Tt, Tt, tt, tt.

(b) 旁氏方格：Tt 个体产生的配子为 T 和 t；tt 个体产生的配子均为 t。子代基因型：Tt, Tt, tt, tt。

(c) Probability of tall = 2/4 = 1/2 or 50%. (d) Heterozygous means having two different alleles for a particular gene, e.g., Tt.

Genetic diagrams must include parental phenotypes, genotypes, gametes and offspring genotypes. Use a ruler to draw the square and label clearly.

遗传图解须包含亲代表现型、基因型、配子和子代基因型。用直尺画格子并清晰标注。

7. Experimental Skills: Planning and Data Analysis | 实验技能：实验设计与数据分析

Example: A student investigates how the concentration of sugar solution affects the mass of potato cylinders. She places potato pieces of equal size into different sugar concentrations for 30 minutes and records the change in mass. Describe how she should ensure the investigation is a fair test. State the independent, dependent and control variables. Explain how she could process the data to conclude.

例题：某学生探究糖溶液浓度对土豆条质量的影响。她把大小相同的土豆条浸入不同浓度的糖溶液中 30 分钟，记录质量变化。描述她如何确保实验是公平测试。指出自变量、因变量和控制变量。解释她可以如何处理数据得出结论。

A fair test requires that only the sugar concentration is changed. Control variables: temperature, volume of solution, incubation time, surface area of potato pieces, blotting method. Independent variable: sugar concentration. Dependent variable: change in mass (or percentage change).

公平测试要求只有糖浓度发生变化。控制变量：温度、溶液体积、浸泡时间、土豆条表面积、吸干方法。自变量：糖浓度。因变量：质量变化（或质量变化百分比）。

She should calculate percentage change in mass to normalise for slightly different starting masses: % change = (final mass – initial mass)/initial mass × 100%. Plot a graph of percentage change against concentration. A line of best fit helps identify the isotonic point where mass does not change.

她应计算质量变化百分比以标准化不同起始质量的影响：变化% = (最终质量 – 初始质量)/初始质量 × 100%。绘制百分比变化对浓度的图像。通过最佳拟合线可以得出质量不发生变化的等渗点。

Always describe how to measure and control each variable. Suggest repeats and calculation of a mean. Mention safety precautions if relevant, e.g., careful handling of glassware.

务必描述如何测量和控制每个变量。建议重复实验并计算平均值。如有必要，提一下安全注意事项，如小心处理玻璃器皿。

8. Common Mistakes and Exam Tips | 常见错误与应试技巧

Many marks are lost through careless errors rather than lack of knowledge. Always read the question stem carefully – look for command words such as ‘describe’, ‘explain’ or ‘calculate’. In calculations, show your formula, substitution and final answer with correct units and significant figures. For graph questions, use a sharp pencil, label axes with quantities and units, and draw a smooth line or curve of best fit unless asked for a dot-to-dot line.

很多失分是由于粗心错而非知识缺乏。仔细审题——注意 ‘描述’、’解释’ 或 ‘计算’ 等指令词。计算题要写出公式、代入数据和最终结果，并带上正确的单位与有效数字。作图题须用尖铅笔，坐标轴标出物理量和单位，绘制平滑最佳拟合线（除非要求折线）。

In biology and chemistry, precise scientific vocabulary is rewarded. Instead of ‘it goes up’, write ‘the rate of reaction increases’. Use chemical equations with state symbols where required. For ‘explain’ questions, always link to a scientific principle, such as particles, collisions, diffusion or osmosis.

在生物和化学中，精确的科学术语能得分。不要写 ‘它上升了’，而应写 ‘反应速率增加’。必要时写出带状态符号的化学方程式。回答 ‘解释’ 类问题时，一定要关联科学原理，如粒子、碰撞、扩散或渗透。

Time management is crucial. If you are stuck on a calculation, write down any formula or relationship you know – you might earn a method mark. Leave space and move on, then return later. Finally, check the back page and ensure you have attempted all parts.

时间管理至关重要。如果卡在计算题，写下任何你知道的公式或关系式——你可能拿到方法分。留出空白做下一题，稍后再回看。最后，检查背面页，确保答完了所有部分。

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GCSE CIE Science: Typical Worked Examples | GCSE CIE 科学：典型例题详解