📚 Year 8 Edexcel Science: Cross-Disciplinary Integrated Question Practice | 跨学科综合题型训练
In Year 8 Edexcel Science, you are often presented with questions that blend ideas from biology, chemistry and physics. These cross-disciplinary integrated questions push you to connect concepts, analyse data from multiple angles, and sharpen your scientific reasoning. The following training sections provide authentic practice, building your confidence for assessments and deepening your understanding of how science works across traditional boundaries.
在八年级爱德思科学中，你常常会遇到融合生物、化学和物理知识的题目。这些跨学科综合题型要求你会通概念、从多角度分析数据，并锤炼科学推理能力。下面的训练模块提供贴近真题的练习，帮助你在备考中建立信心，同时加深对科学跨界运作方式的理解。
1. Data Analysis: Temperature and Enzyme Activity | 图表解读：温度与酶活性
A student investigated the effect of temperature on catalase, an enzyme extracted from potato tissue. The rate of oxygen production was recorded using a gas syringe. The table below shows the results.
一位学生探究了温度对土豆过氧化氢酶的影响，用气体注射器记录氧气生成速率。下表为实验结果。
| Temperature (°C) | Rate of O₂ production (cm³ min⁻¹) |
|---|---|
| 10 | 1.2 |
| 20 | 2.9 |
| 30 | 5.6 |
| 40 | 8.1 |
| 50 | 4.3 |
| 60 | 0.3 |
Questions: (a) Plot these data as a line graph. (b) Describe the relationship between temperature and reaction rate between 10 °C and 40 °C. (c) Explain why the rate decreases sharply above 50 °C by referring to enzyme structure. (d) State the optimum temperature for this enzyme.
问题：(a) 将这些数据绘制成折线图。(b) 描述10 °C至40 °C之间温度与反应速率的关系。(c) 结合酶的结构解释为什么50 °C以上速率急剧下降。(d) 指出该酶的最适温度。
Between 10 °C and 40 °C, the rate increases because higher temperature gives molecules more kinetic energy, leading to more frequent and more energetic collisions between enzyme and substrate. This is a classic application of particle theory from physics. Beyond the optimum temperature (around 40 °C), the enzyme’s delicate three‑dimensional shape is disrupted – the active site denatures. Chemical bonds holding the protein together break, so the substrate no longer fits. Denaturation is permanent, hence the near‑zero rate at 60 °C.
10 °C到40 °C之间，反应速率升高，因为温度升高赋予分子更大的动能，酶与底物碰撞更频繁、更有力，这运用了物理学的粒子理论。超过最适温度（约40 °C）后，酶精细的三维结构被破坏——活性位点变性。维持蛋白质结构的化学键断裂，底物不再契合。变性是不可逆的，因此在60 °C时速率几乎为零。
2. Forces in the Human Body: Levers and Muscles | 人体中的力：杠杆与肌肉
The forearm and hand have a mass of 1.5 kg, and their centre of mass is 15 cm from the elbow pivot. The biceps tendon attaches 5 cm from the elbow. Calculate the upward force the biceps must exert to hold the arm stationary. Take gravitational field strength g = 10 N kg⁻¹.
二头肌通过肘关节提起前臂，这是一个生物杠杆系统。假设某人手臂水平，前臂和手质量1.5 kg，重心距肘支点15 cm。二头肌肌腱附着点距肘5 cm。计算二头肌需施加多大的向上力才能保持手臂静止。取重力场强度 g = 10 N kg⁻¹。
ced by the anticlockwise moment from the biceps force. Weight W = m × g = 1.5 × 10 = 15 N.
该问题将物理的力矩与解剖学结合起来。前臂重量产生的顺时针力矩必须由二头肌力的逆时针力矩平衡。重力W = m × g = 1.5 × 10 = 15 N。
Clockwise moment = 15 N × 0.15 m = 2.25 N m
For equilibrium, Fbiceps × 0.05 m = 2.25 N m, so Fbiceps vide a force three times the weight lifted, explaining why real limbs require strong muscles near the joint.
要达到平衡，F二头肌 × 0.05 m = 2.25 N m，因此F二头肌 = 2.25 ÷ 0.05 = 45 N。肌肉必须提供三倍于所提重物的力，这说明了为什么真实肢体靠近关节处需要强大的肌肉。
3. States of Matter and Energy: Heating Curves | 物质状态与能量：加热曲线
A beaker of ice at −10 °C is heated steadily until the water boils and vapour reaches 120 °C. The heating curve shows two flat regions at 0 °C and 100 °C. (a) Explain why the temperature does not rise during melting, using the concept of intermolecular forces. (b) Calculate the energy needed to melt 10 g of ice given a specific latent heat of fusion of 334 J g⁻¹.
从−10 °C的冰开始均匀加热，直到水沸腾、蒸气达到120 °C。加热曲线上出现0 °C和100 °C两个平台。(a) 利用分子间作用力的概念解释为什么融化过程中温度不升高。(b) 已知熔化比潜热为334 J g⁻¹，计算融化10 g冰所需的能量。
During a change of state, the energy supplied goes into overcoming the attractive forces between particles, not into raising their kinetic energy. Hence the temperature stays constant. This illustrates a bridge between chemical bonding and physical thermodynamics.
在状态变化期间，能量用于克服粒子间的吸引力，而不是增加动能，因此温度保持恒定。这体现了化学键与物理热力学的跨学科联系。
Q = m × L = 10 g × 334 J g⁻¹ = 3340 J
So 3340 joules are absorbed just to turn 10 g of ice at 0 °C into water at 0 °C without any temperature change.
因此，仅将10 g 0 °C的冰变为0 °C的水就吸收了3340焦耳，期间温度不变。
4. Food Chains and Bioaccumulation | 食物链与生物积累
The concentration of a persistent pesticide (DDT) was measured in organisms from a lake ecosystem. Data: phytoplankton 0.04 ppm, zooplankton 0.23 ppm, small fish 2.0 ppm, large fish 25 ppm, osprey 160 ppm. (a) Draw a bar chart representing these levels. (b) Explain the trend using the idea of biomass transfer and chemical properties of DDT.
测量了某湖生态系统中生物体内持久性杀虫剂(DDT)的浓度。数据：浮游植物0.04 ppm，浮游动物0.23 ppm，小鱼2.0 ppm，大鱼25 ppm，鱼鹰160 ppm。(a) 绘制表达这些水平的条形图。(b) 运用生物量传递和DDT的化学性质解释这一趋势。
DDT is fat‑soluble and chemically stable, so it is not easily broken down in living organisms. As each consumer eats many prey items, the chemical accumulates and becomes more concentrated at higher trophic levels – a process called biomagnification. This combines ecological pyramid principles with organic chemistry and environmental toxicity.
DDT是脂溶性的且化学性质稳定，不易在生物体内分解。当各级消费者取食大量猎物时，该化学物质不断累积并在更高营养级中浓度增大——这一过程称为生物放大。这融合了生态金字塔原理、有机化学与环境毒理学。
5. Investigating Resistance: Length of a Wire | 探究电阻：导线长度
A pupil investigates how the length of a nichrome wire affects its resistance. She records the potential difference (V) and current (I) for five lengths.
| Length (m) | V (V) | I (A) | R = V / I (Ω) |
|---|---|---|---|
| 0.20 | 1.0 | 2.5 | 0.40 |
| 0.40 | 1.6 | 2.0 | 0.80 |
| 0.60 | 1.8 | 1.5 | 1.20 |
| 0.80 | 1.6 | 1.0 | 1.60 |
| 1.
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