📚 Year 10 CIE Chemistry Interdisciplinary Question Training | 跨学科综合题型训练
Interdisciplinary questions in CIE IGCSE Chemistry test your ability to connect chemical principles with concepts from physics, biology, environmental science and mathematics. This article provides a series of worked examples that combine multiple subjects, helping you sharpen your analytical skills for the final exam. Each section presents a real‑world scenario, followed by a detailed solution.
跨学科综合题是 CIE IGCSE 化学考试中检验你能否将化学原理与物理、生物、环境科学和数学概念联系起来的重要题型。本文提供一系列融合多学科知识的例题与解析，帮助你提升分析能力，为考试做好充分准备。每个小节都给出真实情境问题并配有详细解答。
1. Chemistry & Physics: Energy Changes and Heat Transfer | 化学与物理：能量变化与热传递
ied using a simple calorimeter. The question often asks you to calculate the heat energy released using the formula q = m×c×ΔT, a concept from physics.
镁条与稀盐酸反应时溶液温度升高。这是一个放热反应，可用简易量热计进行研究。题目常要求使用物理中的公式 q = m×c×ΔT 计算释放的热能。
Example: 0.60 g of magnesium ribbon is added to 50.0 cm³ of 1.0 mol/dm³ HCl (excess). The temperature rises from 22.0 °C to 38.6 °C. Assume the solution has a density of 1.0 g/cm³ and specific heat capacity 4.2 J/g°C. Calculate the heat energy released and the molar enthalpy change for Mg.
例题：将 0.60 g 镁条加入 50.0 cm³ 1.0 mol/dm³ HCl（过量）中，温度从 22.0 °C 升至 38.6 °C。假设溶液密度为 1.0 g/cm³，比热容为 4.2 J/g°C。计算释放的热能及 Mg 的摩尔焓变。
Solution: mass of solution = 50.0 cm³ × 1.0 g/cm³ = 50.0 g. ΔT = 38.6 – °C = 3486 J ≈ 3.49 kJ. Moles of Mg = 0.60 g ÷ 24.3 g/mol = 0.0247 mol. Molar enthalpy ΔH = –q ÷ moles = –3.49 kJ ÷ 0.0247 mol = –141 kJ/mol (negative for exothermic).
解析：溶液质量 = 50.0 cm³ × 1.0 g/cm³ = 50.0 g。ΔT = 16.6 °C。释放热量 q = 50.0 × 4.2 × 16.6 = 3486 J ≈ 3.49 kJ。Mg 的物质的量 = 0.60 ÷ 24.3 = 0.0247 mol。摩尔焓变 ΔH = –3.49 kJ ÷ 0.0247 mol ≈ –141 kJ/mol（负号表示放热）。
2. Chemistry & Biology: Photosynthesis and Glucose Combustion | 化学与生物：光合作用与葡萄糖燃烧
Photosynthesis produces glucose and oxygen from carbon dioxide and water. In chemistry, glucose undergoes complete combustion, releasing energy. Questions often link these two processes to calculate the volume of CO₂ absorbed and O₂ released, and then relate to plant growth.
光合作用利用二氧化碳和水生成葡萄糖和氧气。化学中葡萄糖完全燃烧释放能量。题目常常将这两个过程结合起来，计算吸收的 CO₂ 和释放的 O₂ 体积，并联系植物生长。
r.t.p.) absorbed. Then, if this glucose is burned completely, what volume of O₂ is needed? (Molar volume at r.t.p. = 24 dm³)
例题：某植物通过光合作用生成 1.80 g 葡萄糖。计算在室温常压下吸收的 CO₂ 体积。若该葡萄糖被完全燃烧，需要多少体积的 O₂？（室温常压下气体摩尔体积 = 24 dm³）
ed = 6 × 0.0100 = 0.0600 mol. Volume CO₂ = 0.0600 × 24 dm³ = 1.44 dm³. Combustion: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. Moles of O₂ needed = 6 × 0.0100 = 0.0600 mol. Volume O₂ = 0.0600 × 24 = 1.44 dm³.
解析：葡萄糖摩尔质量 = 180 g/mol，物质的量 = 0.0100 mol。光合作用方程式中 CO₂ 与葡萄糖的计量比为 6:1，因此吸收的 CO₂ 为 0.0600 mol，体积 1.44 dm³。燃烧反应中 O₂ 与葡萄糖计量比也是 6:1，所需 O₂ 同样为 0.0600 mol，体积 1.44 dm³。
3. Chemistry & Environmental Science: Acid Rain and Limestone | 化学与环境科学：酸雨与石灰石
Acid rain contains sulfuric acid and nitric acid. It reacts with limestone (calcium carbonate) buildings and statues. These problems combine environmental knowledge with chemical equations and mass calculations.
酸雨含有硫酸和硝酸，会与石灰石（碳酸钙）建筑物和雕像反应。此类问题将环境知识与化学方程式及质量计算结合在一起。
Example: A marble statue weighing 200 kg contains pure CaCO₃. Acid rain containing 0.050 mol/dm³ H₂SO₄ drips onto it. What volume of this acid rain is required to react completely with 10.0% of the statue’s mass?
例题：一尊重 200 kg 的大理石雕像为纯 CaCO₃。含有 0.050 mol/dm³ H₂SO₄ 的酸雨滴落在雕像上。问需要多少体积的此酸雨才能与雕像质量的 10.0% 完全反应？
Solution: Mass of CaCO₃ reacting = 200 kg × 10.0% equired = 199.8 mol. Volume = moles ÷ concentration = 199.8 ÷ 0.050 = 3996 dm³ ≈ 4000 dm³.
解析：反应的 CaCO₃ 质量为 20.0 kg = 20000 g，物质的量约 199.8 mol。反应方程式中 CaCO₃ 与 H₂SO₄ 的计量比为 1:1，故需 H₂SO₄ 199.8 mol。所需酸雨体积 = 199.8 ÷ 0.050 ≈ 4000 dm³。
4. Chemistry & Mathematics: Stoichiometry and Yield Calculations | 化学与数学：化学计量与产率计算
Mathematics appears in nearly every chemistry question through mole calculations, percentage yield, and atom economy. These problems require careful arithmetic and clear logical steps.
数学几乎出现在每一道化学题中，包括摩尔计算、百分产率和原子经济。这些问题需要细致的算术和清晰的逻辑步骤。
o produce 7.50 g of magnesium oxide. Calculate the percentage yield. (Aᵣ: Mg=24.3, O=16.0)
例题：4.80 g 镁与过量氧气反应生成 7.50 g 氧化镁。计算百分产率。（相对原子质量：Mg=24.3，O=16.0）
retical mass of MgO = 0.1975 × 40.3 = 7.96 g. Percentage yield = (actual/theoretical)×100 = (7.50 ÷ 7.96)×100 ≈ 94.2%.
解析：2Mg + O₂ → 2MgO。镁的物质的量 = 4.80 ÷ 24.3 ≈ 0.1975 mol，根据方程式 MgO 的理论物质的量相同，质量 = 0.1975 × 40.3 ≈ 7.96 g。百分产率 = (7.50 ÷ 7.96) × 100% ≈ 94.2%。
5. Chemistry & Biology: Enzyme Activity and Optimal pH | 化学与生物：酶活性与最适 pH
Enzymes are biological catalysts that work within narrow pH ranges. Understanding acid‑base chemistry helps explain why enzyme activity drops in extreme pH conditions. Questions may ask you to interpret graphs and connect pH changes to chemical species.
酶是在较窄的 pH 范围内发挥作用的生物催化剂。理解酸碱化学有助于解释为何 pH 极端时酶活性下降。此类题目可能要求解释图表并将 pH 变化与化学物种联系起来。
Example: The enzyme pepsin works best at pH 2.0 in the stomach. If the stomach pH rises to 4.0, the hydrogen ion concentration changes by a factor of 100. Explain how this pH change is calculated and why it affects enzyme structure.
例题：胃蛋白酶在 pH 2.0 时活性最佳。若胃内 pH 升至 4.0，氢离子浓度改变为原来的百分之一。解释这一 pH 变化如何计算，以及它为何影响酶的结构。
Solution: pH = – , [H⁺] = 10⁻² mol/dm³. At pH 4.0, [H⁺] = 10⁻⁴ mol/dm³. The concentration has decreased 100‑ alters charges on amino acid side chains, breaking these bonds and denaturing the enzyme.
解析：pH = –log[H⁺]。pH 2.0 时 [H⁺] = 10⁻² mol/dm³，pH 4.0 时 [H⁺] = 10⁻⁴ mol/dm³，浓度降为原来的 1/100。酶是蛋白质，其三级结构由离子键和氢键维持。H⁺ 浓度降低会改变氨基酸侧链的电荷，破坏这些键，导致酶变性。
6. Chemistry & Physics: Electrolysis and Electric Circuits | 化学与物理：电解与电路
Electrolysis requires a direct current to drive non‑spontaneous reactions. Physics concepts such as current, charge and power are often combined with Faraday’s laws. CIE questions frequently ask students to calculate the mass of metal deposited given the current and time.
电解需用直流电驱动非自发的反应。物理中的电流、电荷和功率常与法拉第定律一同出现。CIE 试题经常要求根据电流和时间计算析出金属的质量。
30 minutes. Calculate the mass of aluminium formed. (1 mol e⁻ = 96500 C; Al³⁺ + 3e⁻ → Al)
例题：用 2.0 A 的电流电解熔融氧化铝 30 分钟。计算生成的铝的质量。（1 mol e⁻ = 96500 C；Al³⁺ + 3e⁻ → Al）
Moles of Al = 0.03731 ÷ 3 = 0.01244 mol. Mass of Al = 0.01244 mol × 27.0 g/mol = 0.336 g.
解析：电荷量 Q = 2.0 A × 1800 s = 3600 C。电子物质的量 = 3600 ÷ 96500 ≈ 0.03731 mol。生成 1 mol Al 需 3 mol 电子，故 Al 的物质的量 = 0.03731 ÷ 3 ≈ 0.01244 mol，质量 = 0.01244 × 27.0 ≈ 0.336 g。
7. Chemistry & Earth Science: The Limestone Cycle and Thermal Decomposition | 化学与地球科学：石灰石循环与热分解
The limestone cycle involves calcium carbonate, calcium oxide and calcium hydroxide. Geologically, these reactions occur in rock formation and the carbon cycle. Questions can interweave chemical equations with environmental implications, such as cement manufacturing.
石灰石循环涉及碳酸钙、氧化钙和氢氧化钙。地质学上，这些反应发生在岩石形成和碳循环过程中。题目常将化学方程式与水泥制造等环境影响交织在一起。
Example: A cement kiln decomposes 500 kg of CaCO₃ into CaO and CO₂. The CO₂ released is captured and bubbled through water to form carbonic acid. Calculate the volume of CO₂ (at r.t.p.) produced and the mass of carbonic acid (H₂CO₃) that could form if 80% of the CO₂ dissolves and reacts.
例题：一座水泥窑将 500 kg CaCO₃ 分解为 CaO 和 CO₂。释放的 CO₂ 被捕集并通入水中生成碳酸。计算产生的 CO₂ 体积（室温常压）以及若有 80% CO₂ 溶解并反应，能形成多少质量的碳酸 (H₂CO₃)。
es of CO₂ = 4995 mol. Volume CO₂ at r.t.p. = 4995 × 24 dm³ = 119880 dm³ ≈ 1.20×10⁵ dm³. 80% s H₂CO₃ = 3996 mol. Molar mass H₂CO₃ = 62.0 g/mol; mass = 3996 × 62.0 ≈ 247752 g ≈ 248 kg.
解析：CaCO₃ 物质的量 ≈ 4995 mol，生成 CO₂ 同量，体积约 1.20×10⁵ dm³。80% 的 CO₂ 物质的量为 3996 mol，全部转化为 H₂CO₃，质量 = 3996 × 62.0 ≈ 248 kg。
8. Chemistry & Everyday Life: Water Treatment and Health | 化学与生活：水处理与健康
Disinfection of drinking water often uses chlorine. While chlorine kills bacteria, it can react with organic matter to form trihalomethanes, which are health concerns. Understanding redox and substitution reactions is crucial here.
饮用水消毒常使用氯气。氯能杀灭细菌，但也可能与有机物反应生成三卤甲烷，带来健康风险。理解氧化还原和取代反应在这里至关重要。
late the mass of chlorine required. Then explain one potential chemical risk of using chlorine.
例题：一水处理厂向 5.0×10⁶ dm³ 水体内通入氯气，达到 2.5×10⁻⁴ mol/dm³ 的浓度。计算所需氯气的质量，并解释使用氯气的一个潜在化学风险。
Risk: Chlorine can react with organic compounds to form toxic chlorinated hydrocarbons such as trihalomethanes (THMs), which are suspected carcinogens.
解析：Cl₂ 总物质的量 = 5.0×10⁶ × 2.5×10⁻⁴ = 1250 mol。质量 = 1250 × 71.0 = 88.75 kg。风险：氯可能与有机物反应生成有毒的氯化烃如三卤甲烷 (THM)，它们被怀疑是致癌物。
9. Chemistry & Geography: Carbon Footprint and Global Warming | 化学与地理：碳足迹与全球变暖
The burning of fossil fuels releases CO₂, a greenhouse gas. Calculating carbon footprints involves converting masses of fuels into moles and then into CO₂ masses. These problems merge chemical equations with geographical data on climate change.
化石燃料燃烧释放温室气体 CO₂。计算碳足迹需将燃料质量转换为物质的量，再换算为 CO₂ 质量。此类题目将化学方程式与气候变化的地理数据融合。
³ of natural gas (assumed to be pure methane, CH₄) per day, measured at r.t.p. Determine the mass of CO₂ emitted per day. Then calculate the annual CO₂ emission in tonnes and comment on its contribution to the greenhouse effect.
例题：某城市每天消耗 2.00×10⁸ dm³ 天然气（假设为纯甲烷 CH₄，室温常压下计）。求每天排放的 CO₂ 质量。然后计算年排放量（吨），并评论其对温室效应的贡献。
mol; daily mass = 8.33×10⁶ × 44.0 = 3.67×10⁸ g = 367 tonnes. Annual mass = 367 × 365 ≈ 1.34×10⁵ tonnes. This CO₂ absorbs infrared radiation, trapping heat and contributing significantly to global warming.
解析：CH₄ 物质的量 = 8.33×10⁶ mol，完全燃烧产生等量 CO₂，日排放质量 = 367 吨，年排放约 1.34×10⁵ 吨。这些 CO₂ 吸收红外线，留存热量，对全球变暖有显著贡献。
10. Chemistry & Physics: Fuel Efficiency and Energy Density | 化学与物理：燃料效率与能量密度
Different fuels release different amounts of energy per gram when burned. Energy density is a physics concept that in chemistry links to bond energies and enthalpy of combustion. Questions compare fuels like hydrogen and methane.
不同燃料燃烧时每克释放的能量不同。能量密度是一个物理概念，在化学中与键能和燃烧焓相关。题目常比较氢和甲烷等燃料。
Example: The enthalpy of combustion of H₂ is –286 kJ/mol and that of CH₄ is – ased per gram for each fuel and suggest which is more energy-dense. (Mᵣ: H₂=2.0, CH₄=16.0)
例题：H₂ 的燃烧焓为 –286 kJ/mol，CH₄ 的燃烧焓为 –890 kJ/mol。计算每种燃料每克释放的能量，并判断哪种能量密度更高。
y per gram = 890 kJ ÷ 16.0 g = 55.6 kJ/g. Therefore, hydrogen is significantly more energy‑ is why it is attractive for clean energy applications, though storage remains a challenge.
解析：H₂ 每克释放能量 = 286 ÷ 2.0 = 143 kJ/g；CH₄ 每克释放 890 ÷ 16.0 ≈ 55.6 kJ/g。可见氢的能量密度远高于甲烷，这正是其作为清洁能源备受青睐的原因，尽管储存仍具挑战性。
11. Chemistry & Biology: Iron in Diet and Redox Chemistry | 化学与生物：膳食铁与氧化还原化学
Iron is an essential mineral for haemoglobin. The body absorbs Fe²⁺ ions, while Fe³⁺ is less available. Redox chemistry explains iron’s reactivity and why vitamin C (ascorbic acid) enhances absorption by reducing Fe³⁺ to Fe²⁺.
铁是血红蛋白的必需矿物质。人体吸收 Fe²⁺ 离子，而 Fe³⁺ 利用率较低。氧化还原化学解释了铁的反应性，以及维生素 C（抗坏血酸）为何能将 Fe³⁺ 还原为 Fe²⁺ 从而促进吸收。
⁺ + 2e⁻). Then calculate the mass of vitamin C needed to reduce all the iron in the tablet. (Aᵣ: Fe=55.8, O=16.0, C=12.0, H=1.0)
例题：某补铁片含 250 mg 氧化铁（Fe₂O₃）。写出维生素 C 将 Fe³⁺ 还原为 Fe²⁺ 的离子方程式（已知 C₆H₈O₆ → C₆H₆O₆ + 2H⁺ + 2e⁻），并计算还原该片剂中全部铁所需的维生素 C 质量。
3134 mol. Reduction: Fe³⁺ + e⁻ → Fe²⁺. Electrons needed = 0.003134 mol. Vitamin C provides 2e⁻ per molecule, so moles of vitamin C = 0.003134 ÷ 2 = 0.001567 mol. Molar mass C₆H₈O₆ = 176.1 g/mol; mass = 0.001567 × 176.1 ≈ 0.276 g = 276 mg.
解析：Fe₂O₃ 摩尔质量 159.6 g/mol，物质的量 0.001567 mol，所含 Fe³⁺ 物质的量为其两倍，即 0.003134 mol。每个 Fe³⁺ 需 1 个电子还原，电子总量 0.003134 mol。维生素 C 每个分子提供 2 个电子，故需维生素 C 0.001567 mol，质量 0.001567 × 176.1 ≈ 0.276 g（276 mg）。
12. Chemistry & Mathematics: Empirical Formula and Combustion Analysis | 化学与数学：经验式与燃烧分析
Combustion analysis is used to determine the empirical formula of organic compounds. It requires precise mass measurements and mole ratio calculations. This process is a pure exercise in mathematical rigor applied to chemistry.
燃烧分析用于确定有机化合物的经验式，需要精确的质量测量和摩尔比计算。这是一个将数学严谨性应用于化学的纯粹练习。
Example: 0.600 g of a hydrocarbon (CₓHᵧ) is completely burned. The products are 1.88 g of CO₂ and 0.770 g of H₂O. Determine the empirical formula of the hydrocarbon.
例题：某烃（CₓHᵧ）0.600 g 完全燃烧，生成 1.88 g CO₂ 和 0.770 g H₂O。求该烃的经验式。
s H = 2 × 0.04278 = 0.08556 mol. Ratio C:H = 0.04273 : 0.08556 ≈ 1 : 2.00. Empirical formula is CH₂.
解析：CO₂ 物质的量 = 1.88 ÷ 44.0 ≈ 0.04273 mol，C 原子的物质的量相同；H₂O 物质的量 = 0.770 ÷ 18.0 ≈ 0.04278 mol，H 原子物质的量为其两倍即 0.08556 mol。C:H 摩尔比 ≈ 1:2，经验式为 CH₂。
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