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

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

In the OCR GCSE Science exams, applying knowledge to solve problems is crucial. Worked examples help you understand the step-by-step reasoning required to tackle questions on Physics, Chemistry, and Biology. This article presents ten typical worked examples that cover key areas, boosting your confidence and exam technique.

在 OCR GCSE 科学考试中，运用知识解决问题至关重要。典型例题能帮助你掌握解答物理、化学和生物问题所需的逐步推理过程。本文精选了十个涵盖关键领域的典型例题详解，以增强你的信心和应试技巧。

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1. Speed from a Distance-Time Graph | 速度与距离-时间图

Example: A cyclist travels 450 m in 25 seconds. Calculate the average speed and explain what a flat horizontal line on a distance-time graph represents.

例题：一位骑车人在25秒内行驶了450米。计算平均速度，并解释距离-时间图上的水平直线代表什么。

Step 1: Use the formula speed = distance ÷ time. The equation is centred below.

步骤1：使用公式 速度 = 距离 ÷ 时间。其方程居中显示如下。

v = d ÷ t

Step 2: Substitute the values: v = 450 m ÷ 25 s = 18 m/s.

步骤2：代入数值：v = 450 m ÷ 25 s = 18 m/s。

Step 3: A flat horizontal line on a distance-time graph indicates that the object is stationary because distance does not change with time.

步骤3：距离-时间图上的水平直线表示物体处于静止状态，因为距离不随时间变化。

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2. Energy Transfers and Efficiency | 能量转换与效率

Example: An electric motor uses 800 J of electrical energy to lift a weight, but only 600 J is transferred as useful gravitational potential energy. Calculate the percentage efficiency and identify where the wasted energy goes.

例题：一台电动机用800焦耳的电能提起重物，但只有600焦耳转化为有用的重力势能。计算百分比效率，并指出浪费的能量去了哪里。

Step 1: Efficiency = (useful output energy ÷ total input energy) × 100%.

步骤1：效率 =（有用的输出能量 ÷ 输入总能量）× 100%。

Step 2: Substitute: Efficiency = (600 J ÷ 800 J) × 100% = 75%.

步骤2：代入：效率 = (600 J ÷ 800 J) × 100% = 75%。

Step 3: The wasted energy (200 J) is dissipated as thermal energy (heat) to the surroundings and as sound energy from the motor.

步骤3：浪费的能量（200 J）以热能和电动机发出的声能形式散失到周围环境中。

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3. Balancing Chemical Equations | 配平化学方程式

Example: Balance the equation for the combustion of methane: __ CH₄ + __ O₂ → __ CO₂ + __ H₂O.

例题：配平甲烷燃烧的方程式：__ CH₄ + __ O₂ → __ CO₂ + __ H₂O。

the number of each atom on both sides. Initially: left C=1, H=4, O=2; right C=1, H=2, O=3.

步骤1：列出两边每种原子的数目。初始时：左边 C=1, H=4, O=2；右边 C=1, H=2, O=3。

in front of H₂O: CH₄ + __ O₂ → CO₂ + 2H₂O. Now left: C=1, H=4, O=2; right: C=1, H=4, O=4.

步骤2：通过在水分子前加系数2来平衡氢原子：CH₄ + __ O₂ → CO₂ + 2H₂O。现在左边：C=1, H=4, O=2；右边：C=1, H=4, O=4。

y putting a coefficient 2 before O₂: CH₄ + 2O₂ → CO₂ + 2H₂O. The equation is now balanced.

步骤3：通过在 O₂ 前加系数2来平衡氧原子：CH₄ + 2O₂ → CO₂ + 2H₂O。方程式现已配平。

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4. Mole Calculations | 摩尔计算

Example: : Calculate the number of moles in 11 g of carbon dioxide (CO₂). (Relative atomic masses: C = 12, O = 16.)

例题：计算11克二氧化碳（CO₂）的摩尔数。（相对原子质量：C = 12，O = 16。）

Step 1: Determine the relative formula mass (Mᵣ) of CO₂: 12 + (2 × 16) = 44.

步骤1：确定 CO₂ 的相对分子质量（Mᵣ）：12 + (2 × 16) = 44。

Step 2: Use the formula moles = mass ÷ Mᵣ.

步骤2：使用公式 摩尔数 = 质量 ÷ Mᵣ。

Step 3: Substitute: moles = 11 g ÷ 44 g/mol = 0.25 mol.

步骤3：代入：摩尔数 = 11 g ÷ 44 g/mol = 0.25 mol。

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5. Electrolysis of Molten Compounds | 熔融化合物的电解

Example: Explain what happens during the electrolysis of molten lead(II) bromide (PbBr₂). Include the half-equations at each electrode.

例题：解释在电解熔融溴化铅（PbBr₂）时会发生什么。写出每个电极的半反应方程式。

Step 1: Molten PbBr₂ contains Pb²⁺ and Br⁻ ions that are free to move.

步骤1：熔融 PbBr₂ 中含有可自由移动的 Pb²⁺ 和 Br⁻ 离子。

electrode), lead ions gain electrons and are reduced to form molten lead: Pb²⁺ + 2e⁻ → Pb.

步骤2：在阴极（负极），铅离子得到电子被还原为液态铅：Pb²⁺ + 2e⁻ → Pb。

bromide ions lose electrons and are oxidised to form brown bromine gas: 2Br⁻ → Br₂ + 2e⁻.

步骤3：在阳极（正极），溴离子失去电子被氧化生成棕色的溴气：2Br⁻ → Br₂ + 2e⁻。

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6. Microscopy and Magnification | 显微镜与放大倍数

Example: 0. The image diameter is measured as 18 mm, while the actual specimen diameter is 0.0045 mm. Verify the total magnification using both methods.

例题：用10倍目镜和40倍物镜观察标本。图像直径实测为18毫米，标本实际直径为0.0045毫米。用两种方法验证总放大倍数。

1: Total magnification = eyepiece magnification × objective magnification = 10 × 40 = 400.

方法1：总放大倍数 = 目镜放大倍数 × 物镜放大倍数 = 10 × 40 = 400。

m is 18 mm, 0.0045 mm is correct. Wait, 18 ÷ 0.0045 = 4000 not 400. This is a mistake. Let's correct: actual size should be 0.045 mm to give 400. I' 18 ÷ 0.045 = 400. Or use 0.0045 and get 4000. Need to match the eyepiece magnification. I' : actual diameter = 0.045 mm. Then 18 ÷ 0.045 = 400. So example: actual size 0.045 mm. Let's adjust accordingly.)

我们使用实际直径0.045毫米。那么18 ÷ 0.045 = 400。所以例题数据：实际直径0.045 mm。

2: Magnification = image size ÷ actual size = 18 mm ÷ 0.045 mm = 400. Both methods agree.

步骤2：放大倍数 = 图像大小 ÷ 实际大小 = 18 mm ÷ 0.045 mm = 400。两种方法结果一致。

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7. Genetics: Punnett Squares | 遗传学：庞尼特方格

Example: In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). Predict the outcome of a cross between a heterozygous tall plant and a homozygous short plant. Give the phenotypic ratio.

例题：在豌豆植株中，高茎等位基因（T）对矮茎等位基因（t）为显性。预测杂合高茎植株与纯合矮茎植株杂交的结果。给出表现型比例。

1: Determine the genotypes of the parents: heterozygous tall = Tt, homozygous short = tt.

步骤1：确定亲本的基因型：杂合高茎 = Tt，纯合矮茎 = tt。

Step 2: Set up a Punnett square. Male gametes (t, t) across the top, female gametes (T, t) down the side. The possible combinations are Tt and tt.

步骤2：绘制庞尼特方格。雄配子（t, t）置于顶部，雌配子（T, t）置于侧边。可能的组合为 Tt 和 tt。

Step 3: Offspring genotypes: 50% Tt (tall) and 50% tt (short). Phenotypic ratio is 1 tall : 1 short.

步骤3：后代基因型：50% Tt（高茎）和50% tt（矮茎）。表现型比例为 1 高 : 1 矮。

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8. Enzyme Activity and Temperature | 酶活性与温度

Example: Describe how increasing temperature affects the rate of an enzyme-catalysed reaction, and explain why the rate eventually falls.

例题：描述升高温度如何影响酶催化反应的速率，并解释为什么速率最终会下降。

Step 1: As temperature increases, the kinetic energy of enzyme and substrate molecules increases, causing more frequent and successful collisions. This raises the reaction rate.

步骤1：随着温度升高，酶和底物分子的动能增加，导致更频繁有效的碰撞，从而提高反应速率。

Step 2: Once the temperature exceeds the enzyme's optimum (usually around 37–40 °C for human enzymes), the enzyme's active site begins to lose its specific shape. The enzyme denatures.

步骤2：一旦温度超过酶的最适温度（人体酶通常约37–40 °C），酶的活性位点开始失去其特定形状，酶变性。

Step 3: The substrate no longer fits the active site, so the rate of reaction decreases rapidly and cannot be restored.

步骤3：底物无法再与活性位点结合，因此反应速率迅速下降且不可恢复。

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9. Ohm's Law and Resistance | 欧姆定律与电阻

Example: A resistor has a potential difference of 4 V across it and a current of 0.2 A flowing through it. Calculate the resistance. What current would flow if the voltage is doubled?

例题：某电阻两端的电压为4伏特，流过的电流为0.2安培。计算其电阻。如果电压加倍，电流会变成多少？

Step 1: Use Ohm's law R = V ÷ I. Substitute: R = 4 V ÷ 0.2 A = 20 Ω.

步骤1：使用欧姆定律 R = V ÷ I。代入：R = 4 V ÷ 0.2 A = 20 Ω。

f the voltage is doubled to 8 V, I = V ÷ R = 8 V ÷ 20 Ω = 0.4 A. The current also doubles.

步骤2：该电阻器的电阻是恒定的（假设温度不变）。如果电压加倍至8 V，I = V ÷ R = 8 V ÷ 20 Ω = 0.4 A。电流也加倍。

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10. Titration and Concentration | 滴定与浓度

Example: the concentration of the NaOH solution. The balanced equation is NaOH + HCl → NaCl + H₂O.

例题：25.0 cm³ 浓度为0.100 mol/dm³ 的盐酸（HCl）恰好被28.0 cm³ 的氢氧化钠（NaOH）溶液中和。计算 NaOH 溶液的浓度。配平方程式为 NaOH + HCl → NaCl + H₂O。

: moles = concentration × volume (in dm³) = 0.100 mol/dm³ × (25.0/1000) dm³ = 0.00250 mol.

步骤1：计算所用 HCl 的摩尔数：摩尔数 = 浓度 × 体积（以 dm³ 计）= 0.100 mol/dm³ × (25.0/1000) dm³ = 0.00250 mol。

: From the equation, the mole ratio of HCl to NaOH is 1:1, so moles of NaOH = 0.00250 mol.

步骤2：由方程式可知，HCl 与 NaOH 的摩尔比为1:1，因此 NaOH 的摩尔数 = 0.00250 mol。

lume (in dm³) = 0.00250 mol ÷ (28.0/1000) dm³ = 0.0893 mol/dm³ (to 3 significant figures).

步骤3：NaOH 的浓度 = 摩尔数 ÷ 体积（dm³）= 0.00250 mol ÷ (28.0/1000) dm³ = 0.0893 mol/dm³（保留三位有效数字）。

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