Worked Examples for GCSE CCEA Science | GCSE CCEA 科学：典型例题详解

📚 Worked Examples for GCSE CCEA Science | GCSE CCEA 科学：典型例题详解

This article provides carefully selected worked examples covering the key topics in GCSE CCEA Science. Each example demonstrates effective problem-solving techniques, essential for tackling exam questions. The step-by-step explanations in English and Chinese will help you master core concepts in Biology, Chemistry and Physics.

本文精选了覆盖GCSE CCEA科学重点主题的典型例题并详细解答。每个例题都展示了应对考试题目的有效解题技巧。中英双语的逐步解析将帮助你掌握生物、化学和物理的核心概念。

1. Enzyme Activity Graph Interpretation | 酶活性图表解读

Example: The table below shows how temperature affects the rate of an enzyme‑controlled reaction. Use the data to determine the optimum temperature and explain the shape of the curve.

例题：下表显示了温度如何影响酶促反应的速率。利用数据确定最适温度并解释曲线形状。

Temperature (°C)	Rate of Reaction (arbitrary units)
10	0.5
20	2.5
30	5.0
40	7.5
50	2.0
60	0.0

The optimum temperature is 40 °C because the reaction rate reaches its maximum value (7.5 units). Below 40 °C, increasing temperature supplies more kinetic energy to molecules, leading to more frequent successful collisions and a faster rate. Above 50 °C, the rate drops sharply because the enzyme denatures – the active site changes shape irreversibly, so the substrate can no longer bind, and the reaction stops.

最适温度是40 °C，因为此时反应速率达到最大值（7.5单位）。40 °C以下，升高温度给分子提供了更多动能，有效碰撞频率增加，速率加快。50 °C以上速率急剧下降是因为酶变性了——活性部位的形状发生不可逆改变，底物无法结合，反应停止。

---

2. Calculating the Average Rate of Reaction | 计算平均反应速率

Example: Magnesium ribbon was added to excess dilute hydrochloric acid. The volume of hydrogen gas produced was recorded every 10 seconds:

例题：将镁条加入过量稀盐酸中。每10秒记录一次产生的氢气体积：

Time (s)	Volume of H₂ (cm³)
0	0
10	24
20	40
30	47
40	48

Calculate the average rate of reaction between 0 s and 20 s, and between 20 s and 40 s. Explain why the rates differ.

计算0秒到20秒之间，以及20秒到40秒之间的平均反应速率。解释为什么速率不同。

Average rate = change in volume / change in time. For the first interval: rate = (40 – 0) cm³ / (20 – 0) s = 40/20 = 2.0 cm³/s. For the second interval: rate = (48 – 40) cm³ / (40 – 20) s = 8/20 = 0.4 cm³/s. The rate decreases because the concentration of hydrochloric acid falls as it is used up. With fewer acid particles per unit volume, the frequency of successful collisions between magnesium and H⁺ ions drops, so the gas production slows down.

平均速率 = 体积变化 / 时间变化。第一个区间：速率 = (40 – 0) cm³ / (20 – 0) s = 40/20 = 2.0 cm³/s。第二个区间：速率 = (48 – 40) cm³ / (40 – 20) s = 8/20 = 0.4 cm³/s。速率下降是因为盐酸在反应中被消耗，浓度降低。单位体积内酸粒子减少，镁与H⁺离子之间有效碰撞的频率下降，因此气体产生变慢。

---

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

Example: Balance the equation for the reaction of iron with oxygen to form iron(III) oxide: __Fe + __O₂ → __Fe₂O₃

例题：配平铁与氧气反应生成氧化铁的化学方程式：__Fe + __O₂ → __Fe₂O₃

Write the unbalanced equation and count atoms. On the right, there are 2 Fe atoms and 3 O atoms. On the left, oxygen comes as O₂ molecules (2 O atoms per molecule). To balance oxygen, find the lowest common multiple of 2 and 3, which is 6. Place a coefficient 3 in front of O₂ to give 6 O atoms, and a coefficient 2 in front of Fe₂O₃ to give 6 O atoms on the right. Now the right side has 2 × 2 = 4 Fe atoms, so place a coefficient 4 in front of Fe on the left. The balanced equation is:

写出未配平方程式并统计原子。右侧有2个Fe原子和3个O原子。左侧氧以O₂分子（每个含2个O原子）形式存在。要配平氧，找2和3的最小公倍数6。在O₂前放系数3得到6个O原子，在Fe₂O₃前放系数2使右侧也有6个O原子。此时右侧有2×2=4个Fe原子，因此左侧Fe前放系数4。配平后的方程式为：

4Fe + 3O₂ → 2Fe₂O₃

Always check your final atom count: 4 Fe on each side, 6 O on each side.

最后检查原子数：每侧4个Fe，6个O。

---

4. Applying Ohm’s Law | 应用欧姆定律

Example: A resistor in a circuit has a potential difference of 12 V across it and a current of 0.50 A flowing through it. Calculate its resistance. If the voltage is doubled to 24 V while the temperature remains constant, what is the new current?

例题：一个电阻两端的电位差为12 V，通过它的电流为0.50 A。计算其电阻。如果在温度不变的情况下电压加倍到24 V，新的电流是多少？

Ohm’s Law states V = I × R, so resistance R = V / I = 12 V / 0.50 A = 24 Ω. For a fixed resistor at constant temperature, resistance stays the same. When the voltage is increased to 24 V, the current I = V / R = 24 V / 24 Ω = 1.0 A. Thus doubling the voltage doubles the current.

欧姆定律表达式为V = I × R，因此电阻R = V / I = 12 V / 0.50 A = 24 Ω。对于恒定温度下的固定电阻器，电阻不变。当电压升高到24 V时，电流I = V / R = 24 V / 24 Ω = 1.0 A。因此电压加倍，电流也加倍。

---

5. Sankey Diagrams and Energy Efficiency | 桑基图与能量效率

Example: An electric motor lifts a load. The electrical energy supplied to the motor is 500 J. The useful work done in raising the load is 300 J. Draw a Sankey diagram description and calculate the efficiency.

例题：一台电动机提升重物。供给电动机的电能为500 J。提升重物所做的有用功为300 J。描述桑基图并计算效率。

A Sankey diagram represents energy transfers using arrows. The input arrow is drawn to scale and splits into useful output and wasted energy arrows. For this motor, the input arrow represents 500 J. It branches into a useful output arrow of 300 J pointing forward and a wasted energy arrow of 200 J branching downwards or sideways. The wasted energy is dissipated mainly as heat and sound. Efficiency = useful output energy / total input energy = 300 J / 500 J = 0.60 (or 60%). This means 40% of the input energy is wasted.

桑基图用箭头表示能量转移。输入箭头按比例画出，并分成有用输出和浪费能量箭头。对该电动机，输入箭头代表500 J。它分成一个向前的300 J有用输出箭头，以及一个向下或侧边分支的200 J浪费能量箭头。浪费的能量主要以热和声音的形式散失。效率 = 有用输出能量 / 总输入能量 = 300 J / 500 J = 0.60（即60%）。这意味着40%的输入能量被浪费了。

Efficiency = Useful output energy / Total input energy × 100%

---

6. The Reflex Arc | 反射弧

Example: Describe the pathway of a nerve impulse when a person accidentally touches a hot object and quickly withdraws the hand.

例题：描述当人不小心碰到热物体并迅速缩手时，神经冲动的传递路径。

Stimulus (heat) is detected by receptors in the skin. These receptors generate an impulse that travels along a sensory neurone to the spinal cord. In the spinal cord, the impulse passes across a synapse to a relay neurone. The relay neurone passes the impulse across another synapse to a motor neurone. The motor neurone carries the impulse to an effector – the biceps muscle in the arm. The muscle contracts, pulling the hand away from the hot object. This is a reflex action; it is rapid and involuntary because the decision is made in the spinal cord, not the brain, saving vital time.

刺激（热）被皮肤中的感受器察觉。感受器产生神经冲动，沿着感觉神经元传到脊髓。在脊髓中，冲动通过突触传递给中间神经元。中间神经元再将冲动经另一个突触传给运动神经元。运动神经元将冲动传至效应器——手臂的肱二头肌。肌肉收缩，将手拉离热物体。这是一个反射动作；它迅速且不自主，因为决策在脊髓而非大脑作出，节省了关键时间。

• Stimulus → Receptor → Sensory neurone → Relay neurone → Motor neurone → Effector → Response
• 刺激 → 感受器 → 感觉神经元 → 中间神经元 → 运动神经元 → 效应器 → 反应

---

7. Moles and Mass Calculations | 摩尔与质量计算

Example: Calculate the mass of carbon dioxide (CO₂) produced when 12 g of carbon is completely burned in excess oxygen. The equation for the reaction is C + O₂ → CO₂. Relative atomic masses: C = 12, O = 16.

例题：计算12 g碳在过量氧气中完全燃烧时产生的二氧化碳(CO₂)的质量。反应方程式为C + O₂ → CO₂。相对原子质量：C = 12，O = 16。

Step 1: Calculate the number of moles of carbon used. Number of moles = mass (g) / molar mass (g/mol) = 12 g / 12 g/mol = 1.0 mol. Step 2: Use the balanced equation to find the mole ratio. The equation shows that 1 mole of carbon produces 1 mole of CO₂. Therefore 1.0 mol of carbon produces 1.0 mol of CO₂. Step 3: Calculate the molar mass of CO₂. Mᵣ (CO₂) = 12 + (16 × 2) = 44 g/mol. Step 4: Convert moles of CO₂ to mass. Mass = moles × molar mass = 1.0 mol × 44 g/mol = 44 g. So 12 g of carbon yields 44 g of carbon dioxide.

第一步：计算所用碳的摩尔数。摩尔数 = 质量(g) / 摩尔质量(g/mol) = 12 g / 12 g/mol = 1.0 mol。第二步：用配平方程式确定物质的量之比。方程式显示1摩尔碳生成1摩尔CO₂。因此1.0 mol碳生成1.0 mol CO₂。第三步：计算CO₂的摩尔质量。Mᵣ (CO₂) = 12 + (16 × 2) = 44 g/mol。第四步：将CO₂的摩尔数换算为质量。质量 = 摩尔数 × 摩尔质量 = 1.0 mol × 44 g/mol = 44 g。因此12 g碳生成44 g二氧化碳。

---

8. Half-life from a Decay Graph | 根据衰变图求半衰期

Example: A radioactive sample is placed next to a Geiger‑Müller tube. The background count rate is 40 counts per minute (cpm). The recorded count rate is shown every 10 minutes:

例题：将一块放射性样品放在盖革-米勒计数管旁。背景计数率为每分钟40次(counts per minute, cpm)。每隔10分钟记录一次计数率：

Time (min)	Recorded count rate (cpm)
0	840
10	440
20	240
30	140

Determine the half‑life of the sample and predict the recorded count rate at 40 minutes.

求出该样品的半衰期，并预测40分钟时的记录计数率。

First, calculate the corrected count rate by subtracting the background (40 cpm) from each recorded value: 0 min → 800 cpm; 10 min → 400 cpm; 20 min → 200 cpm; 30 min → 100 cpm. The corrected count rate halves from 800 to 400 cpm in 10 minutes, and from 400 to 200 cpm in the next 10 minutes. Therefore the half‑life is 10 minutes. After each half‑life, the activity falls by half. After a further half‑life (total 40 min), the corrected count rate will halve again from 100 to 50 cpm. Adding back the background gives a predicted recorded count rate of 50 + 40 = 90 cpm.

首先，从每个记录值中减去背景计数率(40 cpm)，得到校正计数率：0 min → 800 cpm；10 min → 400 cpm；20 min → 200 cpm；30 min → 100 cpm。校正计数率在10分钟内从800减半至400 cpm，再10分钟又从400减半至200 cpm。所以半衰期为10分钟。每经过一个半衰期，活度减半。再经过一个半衰期（总共40分钟），校正计数率将从100再次减半至50 cpm。加上背景后，预测的记录计数率为50 + 40 = 90 cpm。

Published by TutorHao | Science Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）