IB CCEA Science: Worked Examples Explained | IB CCEA 科学：典型例题详解

📚 IB CCEA Science: Worked Examples Explained | IB CCEA 科学：典型例题详解

In IB CCEA Science, applied problem-solving is at the heart of the assessment. This article walks through worked examples from physics, chemistry and biology, showing clear, step-by-step logic that you can replicate in your own exams. Each section pairs an English explanation with a matching Chinese version, ensuring you grasp both the scientific reasoning and the technical language required for top marks.

在 IB CCEA 科学课程中，应用型解题是考评的核心。本文通过物理、化学和生物的典型例题，展示清晰、分步的逻辑，帮助你在自己的考试中照此推理。每个要点均配有中英文对照解释，确保你同时掌握科学推理和拿高分所需的专业表达。

1. Kinematics – Motion with Constant Acceleration | 运动学 – 匀加速直线运动

A car accelerates uniformly from rest at 2.5 m s⁻² for 8.0 s. Calculate its final velocity and the distance travelled during this time.

一辆汽车从静止开始以 2.5 m s⁻² 的加速度匀加速运动 8.0 s。计算其末速度及这段时间内的位移。

We list the known quantities: initial velocity u = 0, acceleration a = 2.5 m s⁻², time t = 8.0 s. The relevant SUVAT equations are v = u + at and s = ut + ½at².

列出已知量：初速度 u = 0，加速度 a = 2.5 m s⁻²，时间 t = 8.0 s。适用的匀加速方程是 v = u + at 和 s = ut + ½at²。

v = 0 + (2.5)(8.0) = 20 m s⁻¹

s = 0 × 8.0 + ½ × 2.5 × (8.0)² = 80 m

Thus the final velocity is 20 m s⁻¹ and the distance covered is 80 m. Always check that the units are consistent and that the direction of acceleration matches the increase in speed.

因此末速度为 20 m s⁻¹，位移为 80 m。务必检查单位是否一致，以及加速度方向与速度增加的方向是否匹配。

2. Mole Calculations & Stoichiometry | 摩尔计算与化学计量

Calcium carbonate decomposes on heating: CaCO₃ → CaO + CO₂. What mass of carbon dioxide is produced when 10.0 g of pure CaCO₃ is completely decomposed? (Mᵣ: CaCO₃ = 100.1, CO₂ = 44.0)

碳酸钙受热分解：CaCO₃ → CaO + CO₂。当 10.0 g 纯 CaCO₃ 完全分解时，产生多少质量的二氧化碳？(相对分子质量：CaCO₃ = 100.1，CO₂ = 44.0)

First calculate the number of moles of CaCO₃: n = mass / Mᵣ = 10.0 / 100.1 ≈ 0.0999 mol. The stoichiometric ratio between CaCO₃ and CO₂ is 1 : 1, so the moles of CO₂ produced are also 0.0999 mol.

首先计算 CaCO₃ 的物质的量：n = 质量 / 相对分子质量 = 10.0 / 100.1 ≈ 0.0999 mol。CaCO₃ 与 CO₂ 的化学计量比为 1 : 1，因此生成的 CO₂ 物质的量也是 0.0999 mol。

Mass of CO₂ = moles × Mᵣ = 0.0999 × 44.0 = 4.40 g (to three significant figures).

CO₂ 的质量 = 物质的量 × 相对分子质量 = 0.0999 × 44.0 = 4.40 g（保留三位有效数字）。

Always show the balanced equation and check the molar ratio before doing any mass-mole conversions. Avoid rounding intermediate values too early.

进行质量-物质的量换算前一定要写出配平的方程式并检查摩尔比。避免过早对中间值进行舍入。

3. Monohybrid Cross – Dominant & Recessive Alleles | 单基因杂交 – 显性与隐性等位基因

In garden peas, tall stem (T) is dominant over short stem (t). A heterozygous tall plant (Tt) is crossed with a short plant (tt). Predict the genotypic ratio and phenotypic ratio of the offspring using a Punnett square.

在豌豆中，高茎 (T) 对矮茎 (t) 为显性。一株杂合高茎植株 (Tt) 与一株矮茎植株 (tt) 杂交。请用旁氏表预测后代的基因型比和表型比。

The cross is Tt × tt. Gametes from the heterozygous parent are T and t; the short parent produces only t. Construct a Punnett square:

杂交组合为 Tt × tt。杂合亲本产生的配子为 T 和 t；矮茎亲本只产生 t。构建旁氏表：

	t	t
T	Tt	Tt
t	tt	tt

Offspring genotypes: 2 Tt : 2 tt, which simplifies to a genotypic ratio of 1 Tt : 1 tt. The phenotypes are tall (Tt) and short (tt), giving a phenotypic ratio of 1 tall : 1 short.

后代基因型：2 Tt : 2 tt，简化为基因型比 1 Tt : 1 tt。表型为高茎 (Tt) 和矮茎 (tt)，表型比为 1 高 : 1 矮。

This illustrates Mendel’s law of segregation – the two alleles for a trait separate during gamete formation so that each gamete carries only one allele.

这体现了孟德尔的分离定律——一对等位基因在配子形成时彼此分离，每个配子只携带其中一个等位基因。

4. Energy & Specific Heat Capacity | 能量与比热容

A 250 g aluminium block is heated from 22 °C to 95 °C. The specific heat capacity of aluminium is 0.897 J g⁻¹ °C⁻¹. Calculate the thermal energy absorbed by the block.

一块 250 g 的铝块从 22 °C 加热至 95 °C。铝的比热容为 0.897 J g⁻¹ °C⁻¹。计算铝块吸收的热能。

Use the formula Q = m c Δθ, where m is mass, c is specific heat capacity, and Δθ is the temperature change. Δθ = 95 – 22 = 73 °C.

使用公式 Q = m c Δθ，其中 m 为质量，c 为比热容，Δθ 为温度变化。Δθ = 95 – 22 = 73 °C。

Q = 250 g × 0.897 J g⁻¹ °C⁻¹ × 73 °C = 250 × 0.897 × 73 = 16 370.25 J ≈ 16.4 kJ

Always ensure mass is in grams if c is given per gram, or convert to kilograms for specific heat capacity in J kg⁻¹ °C⁻¹. The final answer is often expressed in kilojoules for convenience.

如果比热容的单位是每克每度，质量就用克；若为每千克每度，则需换算。最终答案通常以千焦表示更为方便。

5. Acid-Base Titration – Determining Concentration | 酸碱滴定 – 测定浓度

25.0 cm³ of sulfuric acid (H₂SO₄) of unknown concentration is neutralised by 23.5 cm³ of 0.100 mol dm⁻³ sodium hydroxide (NaOH). Find the concentration of the sulfuric acid. The neutralisation reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

25.0 cm³ 未知浓度的硫酸 (H₂SO₄) 被 23.5 cm³ 0.100 mol dm⁻³ 的氢氧化钠 (NaOH) 中和。求硫酸的浓度。中和反应：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。

Moles of NaOH used = concentration × volume = 0.100 mol dm⁻³ × (23.5 / 1000) dm³ = 0.00235 mol. According to the equation, 1 mol of H₂SO₄ reacts with 2 mol of NaOH, so moles of H₂SO₄ = 0.00235 ÷ 2 = 0.001175 mol.

所用 NaOH 的物质的量 = 浓度 × 体积 = 0.100 mol dm⁻³ × (23.5 / 1000) dm³ = 0.00235 mol。根据方程式，1 mol H₂SO₄ 与 2 mol NaOH 反应，因此 H₂SO₄ 的物质的量 = 0.00235 ÷ 2 = 0.001175 mol。

Concentration of H₂SO₄ = moles / volume = 0.001175 mol ÷ (25.0 / 1000) dm³ = 0.0470 mol dm⁻³. Remember to convert cm³ to dm³ by dividing by 1000.

H₂SO₄ 的浓度 = 物质的量 / 体积 = 0.001175 mol ÷ (25.0 / 1000) dm³ = 0.0470 mol dm⁻³。务必记得将 cm³ 除以 1000 换算为 dm³。

6. Newton’s Second Law – Force, Mass, Acceleration | 牛顿第二定律 – 力、质量、加速度

A 12 kg crate is pulled along a smooth horizontal floor with a horizontal force of 36 N. Calculate the acceleration of the crate.

一个 12 kg 的板条箱在光滑水平地面上受到 36 N 的水平拉力。求板条箱的加速度。

Newton’s second law states F = m a, so a = F / m. Substituting the values: a = 36 N / 12 kg = 3.0 m s⁻². The direction of acceleration is the same as the applied force.

牛顿第二定律 F = m a，因此 a = F / m。代入数值：a = 36 N / 12 kg = 3.0 m s⁻²。加速度方向与施加力的方向相同。

If friction were present, the net force would be (applied force – friction). For example, if a friction force of 6 N opposes the motion, net force = 36 – 6 = 30 N, giving a = 30/12 = 2.5 m s⁻². Always use the resultant force in the direction of motion.

如果存在摩擦力，净力为（施加力 – 摩擦力）。例如若有 6 N 的摩擦力阻碍运动，净力 = 36 – 6 = 30 N，加速度 a = 30/12 = 2.5 m s⁻²。始终要用运动方向上的合力。

7. Osmosis and Water Potential | 渗透作用与水势

A plant cell with a water potential (Ψ) of –650 kPa is immersed in a sucrose solution that has a water potential of –300 kPa. Predict the net movement of water and the likely effect on the cell.

一个水势 (Ψ) 为 –650 kPa 的植物细胞浸入水势为 –300 kPa 的蔗糖溶液中。预测水分的净移动方向及对细胞可能产生的影响。

Water moves from a region of higher water potential (less negative) to a region of lower water potential (more negative). Here –300 kPa is higher than –650 kPa, so water will move out of the cell into the surrounding solution.

水分从水势较高（负值较小）的区域向水势较低（负值较大）的区域移动。此处 –300 kPa 高于 –650 kPa，因此水分将离开细胞，进入周围溶液。

As water leaves, the cell membrane pulls away from the cell wall – a process called plasmolysis. The cell becomes flaccid. If the difference in water potential is large, permanent damage may occur.

随着水分流失，细胞膜会从细胞壁上剥离，这一过程称为质壁分离。细胞变得萎软。若水势差很大，可能造成永久性损伤。

8. Wave Speed, Frequency & Wavelength | 波速、频率与波长

A sound wave in air has a frequency of 256 Hz and a wavelength of 1.34 m. Calculate its speed. Determine how far the wave travels in 2.5 s.

空气中的声波频率为 256 Hz，波长为 1.34 m。计算其波速，并求该波在 2.5 s 内传播的距离。

The wave equation is v = f λ. So v = 256 Hz × 1.34 m = 343 m s⁻¹ (to three significant figures). This matches the typical speed of sound in air at room temperature.

波动方程为 v = f λ。因此 v = 256 Hz × 1.34 m = 343 m s⁻¹（保留三位有效数字）。这与室温下空气中的典型声速一致。

Distance travelled = speed × time = 343 m s⁻¹ × 2.5 s = 857.5 m ≈ 858 m. Always keep the units consistent: frequency in hertz (s⁻¹), wavelength in metres, speed in m s⁻¹.

传播距离 = 速度 × 时间 = 343 m s⁻¹ × 2.5 s = 857.5 m ≈ 858 m。始终保持单位一致：频率用赫兹 (s⁻¹)，波长用米，速度用 m s⁻¹。

9. Redox Reactions & Half-Equations | 氧化还原反应与半反应式

When a piece of zinc metal is placed in copper(II) sulfate solution, a reaction occurs: Zn(s) + CuSO₄(aq) → Cu(s) + ZnSO₄(aq). Write the two half-equations and identify the oxidising agent.

将锌片放入硫酸铜溶液时发生反应：Zn(s) + CuSO₄(aq) → Cu(s) + ZnSO₄(aq)。写出两个半反应式，并指出氧化剂。

Oxidation half-equation (Zn loses electrons): Zn → Zn²⁺ + 2e⁻. Reduction half-equation (Cu²⁺ gains electrons): Cu²⁺ + 2e⁻ → Cu. The electrons lost by zinc are gained by copper ions.

氧化半反应（Zn 失去电子）：Zn → Zn²⁺ + 2e⁻。还原半反应（Cu²⁺ 得到电子）：Cu²⁺ + 2e⁻ → Cu。锌失去的电子被铜离子获得。

The oxidising agent is the species that accepts electrons – here it is Cu²⁺. The reducing agent is Zn. Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.

氧化剂是接受电子的物质——此处是 Cu²⁺。还原剂是 Zn。记住 OIL RIG：氧化是失电子，还原是得电子。

10. Data Analysis – Interpreting a Calibration Curve | 数据分析 – 解读校准曲线

A student measures the absorbance of four standard protein solutions to construct a calibration curve. The results are:

0.0 mg cm⁻³ → absorbance 0.00
0.2 mg cm⁻³ → absorbance 0.18
0.4 mg cm⁻³ → absorbance 0.35
0.6 mg cm⁻³ → absorbance 0.54

An unknown sample gives an absorbance of 0.27. Use the graph to determine its protein concentration.

某学生测量了四种标准蛋白质溶液的吸光度以构建校准曲线。结果如下：

0.0 mg cm⁻³ → 吸光度 0.00
0.2 mg cm⁻³ → 吸光度 0.18
0.4 mg cm⁻³ → 吸光度 0.35
0.6 mg cm⁻³ → 吸光度 0.54

一个未知样品的吸光度为 0.27。利用图像确定其蛋白质浓度。

Plot absorbance (y-axis) against concentration (x-axis). The points show a roughly linear relationship. Draw a best-fit straight line through the origin. For an absorbance of 0.27, find the corresponding concentration on the x-axis. Interpolation gives a value of approximately 0.30 mg cm⁻³.

以吸光度为 y 轴，浓度为 x 轴作图。数据点大致呈线性关系。画一条通过原点的最佳拟合直线。当吸光度为 0.27 时，在 x 轴上找到对应的浓度。内插得到大约 0.30 mg cm⁻³。

If the line equation is determined (e.g., y = 0.90x), you can also calculate: 0.27 = 0.90x → x = 0.30 mg cm⁻³. Always check the correlation coefficient to ensure reliability.

若确定了直线方程（如 y = 0.90x），也可以计算：0.27 = 0.90x → x = 0.30 mg cm⁻³。务必检查相关系数以确保可靠性。

11. Electrolysis Calculations – Faraday’s Laws | 电解计算 – 法拉第定律

Calculate the mass of copper deposited at the cathode when a current of 0.80 A is passed through aqueous CuSO₄ for 1.5 hours. (F = 96 500 C mol⁻¹, Mᵣ of Cu = 63.5)

计算当 0.80 A 的电流通过硫酸铜溶液 1.5 小时后，在阴极上析出的铜的质量。（F = 96 500 C mol⁻¹，Cu 的相对原子质量 = 63.5）

First find the total charge: Q = I × t. Convert time to seconds: 1.5 h = 1.5 × 3600 = 5400 s. So Q = 0.80 A × 5400 s = 4320 C.

首先求总电荷量：Q = I × t。将时间换算为秒：1.5 h = 1.5 × 3600 = 5400 s。因此 Q = 0.80 A × 5400 s = 4320 C。

The cathode half-reaction is Cu²⁺ + 2e⁻ → Cu, so 2 moles of electrons deposit 1 mole of copper. Moles of electrons = Q / F = 4320 / 96 500 ≈ 0.04477 mol. Moles of Cu = 0.04477 / 2 = 0.02238 mol.

阴极半反应为 Cu²⁺ + 2e⁻ → Cu，故 2 mol 电子沉积 1 mol 铜。电子的物质的量 = Q / F = 4320 / 96 500 ≈ 0.04477 mol。Cu 的物质的量 = 0.04477 / 2 = 0.02238 mol。

Mass of Cu = moles × Mᵣ = 0.02238 × 63.5 = 1.42 g. Always check the electrode reaction to determine the correct mole ratio.

Cu 的质量 = 物质的量 × 相对原子质量 = 0.02238 × 63.5 = 1.42 g。务必根据电极反应确定正确的物质的量比。

12. Integrated Problem – Combining Concepts | 综合问题 – 概念融合

A solar panel absorbs 2.50 × 10⁴ J of sunlight and converts 18% of this into electrical energy. The electrical energy is used to electrolyse acidified water: 2H₂O → 2H₂ ↑ + O₂ ↑. Calculate the volume of hydrogen gas produced at room temperature and pressure (molar volume = 24.0 dm³ mol⁻¹). The overall energy required to produce 1 mole of H₂ is 286 kJ.

一块太阳能板吸收了 2.50 × 10⁴ J 的太阳光，并将其中的 18% 转化为电能。该电能用于电解酸化水：2H₂O → 2H₂ ↑ + O₂ ↑。计算在常温常压下产生的氢气体积（摩尔体积 = 24.0 dm³ mol⁻¹）。已知生成 1 mol H₂ 需要能量 286 kJ。

Useful electrical energy = 0.18 × 2.50 × 10⁴ J = 4.50 × 10³ J = 4.50 kJ. Since 286 kJ are needed for 1 mol H₂, the number of moles of H₂ produced = 4.50 kJ / 286 kJ mol⁻¹ = 0.01573 mol.

有用电能 = 0.18 × 2.50 × 10⁴ J = 4.50 × 10³ J = 4.50 kJ。生成 1 mol H₂ 需要 286 kJ，因此产生的 H₂ 物质的量 = 4.50 kJ / 286 kJ mol⁻¹ = 0.01573 mol。

Volume of H₂ = moles × molar volume = 0.01573 mol × 24.0 dm³ mol⁻¹ = 0.378 dm³ (or 378 cm³). This cross-topic problem links energy conversion, electrolysis and molar volume.

H₂ 的体积 = 物质的量 × 摩尔体积 = 0.01573 mol × 24.0 dm³ mol⁻¹ = 0.378 dm³（或 378 cm³）。这道跨章节的题目将能量转换、电解和摩尔体积联系起来。

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IB CCEA Science: Worked Examples Explained | IB CCEA 科学：典型例题详解