📚 IB & Edexcel Science: Typical Worked Examples Explained | IB & Edexcel 科学：典型例题详解
Mastering science in the IB and Edexcel programmes requires a strong command of both conceptual understanding and problem-solving skills. This article presents a series of carefully selected worked examples spanning physics, chemistry, and biology. Each example is broken down into clear logical steps, providing you with a blueprint for tackling assessment-style questions with confidence.
在 IB 和 Edexcel 科学课程中取得优异成绩，既需要扎实的概念理解，也需要过硬的解题技巧。本文精选了一系列涵盖物理、化学和生物的典型例题，并将每道题拆解为清晰的逻辑步骤，为你提供应对考试题型、增强解题信心的蓝本。
1. Kinematics: Using SUVAT Equations | 运动学：SUVAT 方程的应用
Problem: A car accelerates uniformly from rest at 3.0 m s⁻² for 8.0 s. Calculate the distance travelled during this time.
题目：一辆汽车从静止开始以 3.0 m s⁻² 的加速度匀加速运动 8.0 s。计算这段时间内汽车行驶的距离。
Step 1: Identify the known variables. Initial velocity u = 0 m s⁻¹, acceleration a = 3.0 m s⁻², time t = 8.0 s. The unknown is displacement s.
步骤 1：列出已知量。初速度 u = 0 m s⁻¹，加速度 a = 3.0 m s⁻²，时间 t = 8.0 s。未知量为位移 s。
Step 2: Choose the appropriate SUVAT equation. Since u, a, t are known and final velocity v is not required, use s = u t + ½ a t².
步骤 2：选择合适的 SUVAT 方程。已知 u、a、t，且不需要末速度 v，因此使用 s = u t + ½ a t²。
Step 3: Substitute the values. s = (0)(8.0) + ½ (3.0)(8.0)² = 0 + ½ × 3.0 × 64 = 96 m.
步骤 3：代入数值。s = (0)(8.0) + ½ (3.0)(8.0)² = 0 + ½ × 3.0 × 64 = 96 m。
Step 4: Check units and significant figures. The answer is 96 m, given to two significant figures consistent with the data.
步骤 4：检查单位和有效数字。答案为 96 m，与数据一致保留两位有效数字。
2. Forces and Newton’s Laws: Inclined Plane | 力与牛顿定律：斜面问题
Problem: A block of mass 5.0 kg rests on a smooth incline of 30° to the horizontal. Calculate the acceleration of the block down the slope. (Take g = 9.8 m s⁻²)
题目：一质量为 5.0 kg 的物块静止在光滑斜面上，斜面与水平面成 30° 角。计算物块沿斜面下滑的加速度。（取 g = 9.8 m s⁻²）
Step 1: Resolve the weight into components. The weight W = m g = 5.0 × 9.8 = 49 N. The component parallel to the slope is W sin θ = 49 sin 30° = 49 × 0.5 = 24.5 N.
步骤 1：分解重力。重力 W = m g = 5.0 × 9.8 = 49 N。沿斜面方向的分量为 W sin θ = 49 sin 30° = 49 × 0.5 = 24.5 N。
Step 2: Apply Newton’s second law along the slope. The net force F = 24.5 N, so acceleration a = F / m = 24.5 / 5.0 = 4.9 m s⁻².
步骤 2：沿斜面应用牛顿第二定律。合力 F = 24.5 N，因此加速度 a = F / m = 24.5 / 5.0 = 4.9 m s⁻²。
Step 3: Note that the normal reaction and perpendicular component of weight cancel, so they do not affect the motion along the incline.
步骤 3：注意支持力与重力的垂直斜面分量相互抵消，因此不影响沿斜面的运动。
Key insight: The acceleration is independent of mass, a = g sin θ = 9.8 × sin 30° = 4.9 m s⁻², a useful check.
关键点：加速度与质量无关，a = g sin θ = 9.8 × sin 30° = 4.9 m s⁻²，可作为验算。
3. Electrical Circuits: Combining Resistances | 电路：电阻的组合
Problem: A 12 Ω resistor is connected in parallel with a 6.0 Ω resistor. This combination is then connected in series with a 4.0 Ω resistor across a 9.0 V battery of negligible internal resistance. Find the current supplied by the battery.
题目：一个 12 Ω 电阻与一个 6.0 Ω 电阻并联，然后与一个 4.0 Ω 电阻串联，接在 9.0 V 内阻可忽略的电池两端。求电池提供的电流。
Step 1: Calculate the equivalent resistance of the parallel section. 1/R_parallel = 1/12 + 1/6.0 = 1/12 + 2/12 = 3/12, so R_parallel = 4.0 Ω.
步骤 1：计算并联部分的等效电阻。1/R_parallel = 1/12 + 1/6.0 = 1/12 + 2/12 = 3/12，因此 R_parallel = 4.0 Ω。
Step 2: Add the series resistor. Total resistance R_total = R_parallel + 4.0 = 4.0 + 4.0 = 8.0 Ω.
步骤 2：加上串联电阻。总电阻 R_total = R_parallel + 4.0 = 4.0 + 4.0 = 8.0 Ω。
Step 3: Use Ohm’s law. I = V / R_total = 9.0 / 8.0 = 1.125 A, which rounds to 1.1 A (2 significant figures).
步骤 3：应用欧姆定律。I = V / R_total = 9.0 / 8.0 = 1.125 A，约等于 1.1 A（两位有效数字）。
Step 4: Optionally verify potential difference across each part to ensure consistency. Voltage across parallel pair = I × 4.0 = 4.5 V, leaving 4.5 V across the 4.0 Ω series resistor, total 9.0 V.
步骤 4：可选验证电各部分电压以确保一致性。并联部分电压 = I × 4.0 = 4.5 V，剩余 4.5 V 加在 4.0 Ω 串联电阻上，总和 9.0 V。
4. Mole Concept and Stoichiometry: Mass to Moles | 摩尔概念与化学计量：质量与摩尔换算
Problem: How many moles of water are produced when 4.0 g of hydrogen gas (H₂) reacts completely with excess oxygen? (Molar mass of H₂ = 2.0 g mol⁻¹)
题目：当 4.0 g 氢气 (H₂) 与过量的氧气完全反应时，生成多少摩尔水？(H₂ 的摩尔质量 = 2.0 g mol⁻¹)
Step 1: Write the balanced equation: 2 H₂ + O₂ → 2 H₂O.
步骤 1：写出配平的化学方程式：2 H₂ + O₂ → 2 H₂O。
Step 2: Convert the given mass of H₂ to moles. Moles of H₂ = mass / M = 4.0 g / 2.0 g mol⁻¹ = 2.0 mol.
步骤 2：将给定的 H₂ 质量转换为摩尔。H₂ 的摩尔数 = 质量 / 摩尔质量 = 4.0 g / 2.0 g mol⁻¹ = 2.0 mol。
Step 3: Use the mole ratio from the equation. 2 mol H₂ produces 2 mol H₂O, so 2.0 mol H₂ produces 2.0 mol H₂O.
步骤 3：根据方程式中的摩尔比。2 mol H₂ 生成 2 mol H₂O，因此 2.0 mol H₂ 生成 2.0 mol H₂O。
Step 4: The answer is 2.0 mol of water. The fact that oxygen is in excess ensures complete reaction of hydrogen.
步骤 4：答案为 2.0 mol 水。氧气过量确保了氢气完全反应。
5. Titration Calculation: Determining Concentration | 滴定计算：确定浓度
Problem: 25.0 cm³ of sulfuric acid (H₂SO₄) is neutralised by 20.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide (NaOH) solution. Calculate the concentration of the acid.
题目：25.0 cm³ 硫酸 (H₂SO₄) 被 20.0 cm³ 0.100 mol dm⁻³ 氢氧化钠 (NaOH) 溶液中和。计算硫酸的浓度。
Step 1: Write the neutralisation equation: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O.
步骤 1：写出中和反应方程式：H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O。
Step 2: Calculate moles of NaOH used. Moles = concentration × volume (in dm³) = 0.100 × (20.0 / 1000) = 0.00200 mol.
步骤 2：计算所用 NaOH 的摩尔数。摩尔 = 浓度 × 体积 (dm³) = 0.100 × (20.0 / 1000) = 0.00200 mol。
Step 3: From the equation, 2 mol NaOH react with 1 mol H₂SO₄, so moles of H₂SO₄ = 0.00200 / 2 = 0.00100 mol.
步骤 3：根据方程式，2 mol NaOH 与 1 mol H₂SO₄ 反应，因此 H₂SO₄ 的摩尔数 = 0.00200 / 2 = 0.00100 mol。
Step 4: Calculate concentration of H₂SO₄. Volume of acid = 25.0 cm³ = 0.0250 dm³. Concentration = moles / volume = 0.00100 / 0.0250 = 0.0400 mol dm⁻³.
步骤 4：计算 H₂SO₄ 的浓度。酸的体积 = 25.0 cm³ = 0.0250 dm³。浓度 = 摩尔 / 体积 = 0.00100 / 0.0250 = 0.0400 mol dm⁻³。
6. Chemical Equilibrium: Kc and the ICE Table | 化学平衡：Kc 与 ICE 表
Problem: For the reaction H₂(g) + I₂(g) ⇌ 2 HI(g) at a certain temperature, Kc = 50.0. If 1.00 mol of H₂ and 1.00 mol of I₂ are placed in a 1.00 dm³ vessel, calculate the equilibrium concentrations of all species.
题目：对于反应 H₂(g) + I₂(g) ⇌ 2 HI(g)，在一定温度下 Kc = 50.0。若将 1.00 mol H₂ 和 1.00 mol I₂ 放入 1.00 dm³ 容器中，计算各物质的平衡浓度。
Step 1: Set up an ICE (Initial, Change, Equilibrium) table. Initial concentrations: [H₂] = 1.00, [I₂] = 1.00, [HI] = 0. Change: let x be the amount of H₂ that reacts. Then [H₂] changes by -x, [I₂] by -x, and [HI] by +2x.
步骤 1：建立 ICE（初始、变化、平衡）表格。初始浓度：[H₂] = 1.00，[I₂] = 1.00，[HI] = 0。变化量：设 H₂ 反应了 x，则 [H₂] 变化 -x，[I₂] 变化 -x，[HI] 变化 +2x。
| Species | Initial (mol dm⁻³) | Change | Equilibrium |
|---|---|---|---|
| H₂ | 1.00 | -x | 1.00 – x |
| I₂ | 1.00 | -x | 1.00 – x |
| HI | 0 | +2x | 2x |
Step 2: Write the expression for Kc. Kc = [HI]² / ([H₂][I₂]) = (2x)² / ((1.00 – x)(1.00 – x)) = 50.0.
步骤 2：写出 Kc 表达式。Kc = [HI]² / ([H₂][I₂]) = (2x)² / ((1.00 – x)(1.00 – x)) = 50.0。
Step 3: Take square root of both sides. 2x / (1.00 – x) = √50.0 ≈ 7.07. Solve for x: 2x = 7.07(1.00 – x) → 2x = 7.07 – 7.07x → 9.07x = 7.07 → x = 0.779.
步骤 3：两边开平方。2x / (1.00 – x) = √50.0 ≈ 7.07。求解 x：2x = 7.07(1.00 – x) → 2x = 7.07 – 7.07x → 9.07x = 7.07 → x = 0.779。
Step 4: Calculate equilibrium concentrations. [H₂] = 1.00 – 0.779 = 0.221 mol dm⁻³, [I₂] = 0.221 mol dm⁻³, [HI] = 2 × 0.779 = 1.56 mol dm⁻³ (to 3 s.f.).
步骤 4：计算平衡浓度。[H₂] = 1.00 – 0.779 = 0.221 mol dm⁻³，[I₂] = 0.221 mol dm⁻³，[HI] = 2 × 0.779 = 1.56 mol dm⁻³（保留三位有效数字）。
7. Genetics: Monohybrid Cross and Phenotypic Ratios | 遗传学：单基因杂交与表现型比例
Problem: In pea plants, the allele for tall stems (T) is dominant over dwarf stems (t). A heterozygous tall plant is crossed with a dwarf plant. Predict the phenotypic ratio 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。预测后代表现型的比例。
Step 1: Determine the genotypes of parents. Heterozygous tall: Tt. Dwarf: tt. (Dwarf must be homozygous recessive because t is recessive.)
步骤 1：确定亲本基因型。杂合高茎：Tt。矮茎：tt。（矮茎必须是隐性纯合，因为 t 是隐性。）
Step 2: Set up a Punnett square. Gametes from Tt parent: T and t. Gametes from tt parent: t and t.
步骤 2：画出庞纳特方格。Tt 亲本的配子：T 和 t。tt 亲本的配子：t 和 t。
| T | t | |
| t | Tt | tt |
| t | Tt | tt |
Step 3: Offspring genotypes: 2 Tt (heterozygous tall) and 2 tt (dwarf). Phenotypic ratio: 2 tall : 2 dwarf, which simplifies to 1 tall : 1 dwarf.
步骤 3：后代基因型：2 个 Tt（杂合高茎）和 2 个 tt（矮茎）。表现型比例：2 高 : 2 矮，简化为 1 高 : 1 矮。
Step 4: This test cross confirms that the tall parent was heterozygous.
步骤 4：这一测交证实了高茎亲本是杂合子。
8. Enzyme Kinetics: Interpreting the Michaelis-Menten Curve | 酶动力学：解读米氏曲线
Problem: An enzyme-catalysed reaction has a maximum rate (Vmax) of 100 µmol min⁻¹ and a Michaelis constant (Km) of 0.5 mM. At a substrate concentration of 1.0 mM, calculate the reaction rate using the Michaelis-Menten equation.
题目：某酶催化反应的最大速率 (Vmax) 为 100 µmol min⁻¹，米氏常数 (Km) 为 0.5 mM。当底物浓度为 1.0 mM 时，利用米氏方程计算反应速率。
Step 1: Recall the Michaelis-Menten equation: rate v = (Vmax × [S]) / (Km + [S]).
步骤 1：回顾米氏方程：速率 v = (Vmax × [S]) / (Km + [S])。
Step 2: Substitute the given values. Vmax = 100 µmol min⁻¹, Km = 0.5 mM, [S] = 1.0 mM.
步骤 2：代入已知值。Vmax = 100 µmol min⁻¹，Km = 0.5 mM，[S] = 1.0 mM。
Step 3: v = (100 × 1.0) / (0.5 + 1.0) = 100 / 1.5 ≈ 66.7 µmol min⁻¹.
步骤 3：v = (100 × 1.0) / (0.5 + 1.0) = 100 / 1.5 ≈ 66.7 µmol min⁻¹。
Step 4: Check the logic. When [S] equals twice Km, the rate is 2/3 Vmax, consistent with the equation.
步骤 4：检查逻辑。当 [S] 为 Km 的两倍时，速率应为 2/3 Vmax，与方程一致。
Step 5: In IB Biology or Edexcel questions, you may also be asked to sketch the curve and explain the effect of competitive inhibitors on Km and Vmax.
步骤 5：在 IB 生物学或 Edexcel 题目中，也可能要求绘制曲线并解释竞争性抑制剂对 Km 和 Vmax 的影响。
9. Data Analysis: Interpreting a Graph in Required Practicals | 数据分析：解读实验必做中的图表
Problem: In an investigation of the effect of light intensity on photosynthesis, a student measures the volume of oxygen produced per minute by a pondweed. The data are plotted as a graph of oxygen production rate against light intensity. The graph initially rises linearly, then levels off. Explain the shape of the graph.
题目：在研究光照强度对光合作用影响的实验中，学生测量了水草每分钟产生的氧气体积。数据以氧气产生速率对光照强度的关系作图。图像起初线性上升，然后趋于平缓。解释该图像形状的原因。
Step 1: At low light intensity, light is the limiting factor; the rate of photosynthesis increases proportionally with intensity, giving a linear relationship.
步骤 1：在低光照强度下，光是限制因子；光合作用速率与光照强度成正比，呈现线性关系。
Step 2: As intensity further increases, the rate levels off because another factor, such as CO₂ concentration or temperature, becomes limiting. The light saturation point is reached.
步骤 2：当光照强度继续增加时，速率趋于平缓，因为另一个因子如 CO₂ 浓度或温度成为限制因子。达到了光饱和点。
Step 3: If the line plateaus, Vmax for the Calvin cycle enzymes may be reached. In exam answers, students must link the shape to the concept of limiting factors.
步骤 3：如果曲线出现平台，可能是卡尔文循环酶达到了 Vmax。在考试答案中，学生必须将曲线形状与限制因子的概念联系起来。
Step 4: For full marks, mention that beyond the saturation point, increasing light intensity has no effect on the rate because the enzymes are working at maximum capacity.
步骤 4：要拿满分，需要提到超过饱和点后，增加光照强度对速率无影响，因为酶已经以最大能力工作。
10. Stoichiometry in Solutions: Reacting Volumes | 溶液中的化学计量：反应体积
Problem: Calculate the volume of 0.200 mol dm⁻³ HCl required to neutralise 50.0 cm³ of 0.150 mol dm⁻³ Ba(OH)₂ solution.
题目：计算中和 50.0 cm³ 0.150 mol dm⁻³ Ba(OH)₂ 溶液所需的 0.200 mol dm⁻³ HCl 的体积。
Step 1: Balanced equation: Ba(OH)₂ + 2 HCl → BaCl₂ + 2 H₂O.
步骤 1：配平方程式：Ba(OH)₂ + 2 HCl → BaCl₂ + 2 H₂O。
Step 2: Moles of Ba(OH)₂ = 0.150 × (50.0 / 1000) = 0.00750 mol.
步骤 2：Ba(OH)₂ 的摩尔数 = 0.150 × (50.0 / 1000) = 0.00750 mol。
Step 3: Mole ratio: 1 mol Ba(OH)₂ reacts with 2 mol HCl, so moles of HCl needed = 2 × 0.00750 = 0.0150 mol.
步骤 3：摩尔比：1 mol Ba(OH)₂ 与 2 mol HCl 反应，所需 HCl 摩尔数 = 2 × 0.00750 = 0.0150 mol。
Step 4: Volume of HCl = moles / concentration = 0.0150 / 0.200 = 0.0750 dm³ = 75.0 cm³.
步骤 4：HCl 体积 = 摩尔数 / 浓度 = 0.0150 / 0.200 = 0.0750 dm³ = 75.0 cm³。
11. Electrical Power and Energy | 电功率与电能
Problem: A 12 V battery is connected to a 6.0 Ω resistor for 5.0 minutes. Calculate the energy dissipated as heat in the resistor.
题目：一个 12 V 电池连接到一个 6.0 Ω 电阻上，通电 5.0 分钟。计算电阻上以热量形式消耗的能量。
Step 1: Find current using Ohm’s law. I = V / R = 12 / 6.0 = 2.0 A.
步骤 1：用欧姆定律求电流。I = V / R = 12 / 6.0 = 2.0 A。
Step 2: Calculate power using P = I V or P = I² R. P = (2.0)² × 6.0 = 24 W.
步骤 2：计算功率 P = I V 或 P = I² R。P = (2.0)² × 6.0 = 24 W。
Step 3: Convert time to seconds: t = 5.0 min × 60 = 300 s.
步骤 3：将时间转换为秒：t = 5.0 min × 60 = 300 s。
Step 4: Energy E = P × t = 24 × 300 = 7200 J (or 7.2 kJ).
步骤 4：能量 E = P × t = 24 × 300 = 7200 J（或 7.2 kJ）。
12. Osmosis and Water Potential in Biology | 生物学中的渗透与水势
Problem: A plant cell with a water potential (Ψ) of -300 kPa is placed in a sucrose solution of Ψ = -500 kPa. Determine the direction of net water movement and explain.
题目：一个水势 (Ψ) 为 -300 kPa 的植物细胞被放入水势为 -500 kPa 的蔗糖溶液中。判断水的净移动方向并解释。
Step 1: Recall that water moves from a region of higher water potential to lower water potential.
步骤 1：回顾水总是从水势高的区域向水势低的区域移动。
Step 2: Compare the values. -300 kPa is higher (less negative) than -500 kPa.
步骤 2：比较数值。-300 kPa 比 -500 kPa 更高（负值更小）。
Step 3: Therefore, water will move out of the cell into the solution, causing the cell to become flaccid (plasmolysis may occur).
步骤 3：因此，水将从细胞内部移向外部溶液，导致细胞失水变软（可能发生质壁分离）。
Step 4: In IB and Edexcel questions, always remember to state the direction and the consequence for the cell, using correct terminology.
步骤 4：在 IB 和 Edex
Published by TutorHao | IB Science Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导