📚 IGCSE CIE Science: Calculation Questions Training | IGCSE CIE 科学：计算题专项训练
Calculation questions form a significant part of the IGCSE CIE Science papers, spanning Physics, Chemistry, and Biology. Success depends on mastering core formulae, understanding units, and practising step‑by‑step problem‑solving. This guide provides targeted training across all major calculation topics, with paired English–Chinese explanations so you can learn the concepts and language simultaneously.
计算题在 IGCSE CIE 科学考卷中占据很大比重，涵盖物理、化学和生物三科。成功的关键在于掌握核心公式、理解单位，并进行分步骤的解题训练。本指南提供所有主要计算专题的针对性训练，采用中英对照讲解，帮助你同时掌握概念和双语表达。
1. Speed, Velocity and Acceleration | 速度、速率与加速度计算
Speed is the distance travelled per unit time. The average speed is given by v = s ÷ t, where s is distance and t is time. Velocity includes direction, but the magnitude is calculated similarly. Acceleration is the rate of change of velocity: a = (v – u) ÷ t, where v is final velocity, u is initial velocity, and t is time taken.
速率是单位时间内移动的距离。平均速率公式为 v = s ÷ t，其中 s 表示距离，t 为时间。速度包含方向，但其大小计算方式相同。加速度是速度的变化率：a = (v – u) ÷ t，其中 v 为末速度，u 为初速度，t 为所用时间。
Example: A car travels 150 km in 2.5 hours. Calculate its average speed in km/h. Also convert this speed to m/s.
例题：一辆汽车用 2.5 小时行驶了 150 km。计算它的平均速率（km/h）。并将此速率转换为 m/s。
Solution: average speed = 150 km ÷ 2.5 h = 60 km/h. To convert, 60 km/h = 60 × (1000 m ÷ 3600 s) = 16.7 m/s.
解答：平均速率 = 150 km ÷ 2.5 h = 60 km/h。单位换算，60 km/h = 60 × (1000 m ÷ 3600 s) = 16.7 m/s。
Acceleration example: A cyclist accelerates from rest to 12 m/s in 6 seconds. a = (12 – 0) ÷ 6 = 2 m/s².
加速度例题：一位骑行者从静止加速到 12 m/s，用时 6 秒。a = (12 – 0) ÷ 6 = 2 m/s²。
2. Density, Mass and Volume | 密度、质量与体积
Density is mass per unit volume. Density = mass ÷ volume (ρ = m ÷ V). Units: kg/m³ or g/cm³. You must be able to rearrange for mass or volume and convert between units.
密度是单位体积的质量。密度 = 质量 ÷ 体积 (ρ = m ÷ V)。单位：kg/m³ 或 g/cm³。你需要能够变换公式求出质量或体积，并能进行单位换算。
Example: A block of aluminium has a mass of 540 g and a volume of 200 cm³. Calculate its density in g/cm³ and in kg/m³.
例题：一块铝的质量为 540 g，体积为 200 cm³。计算它的密度（g/cm³ 和 kg/m³）。
Solution: ρ = 540 ÷ 200 = 2.7 g/cm³. To convert, 2.7 g/cm³ = 2.7 × 1000 kg/m³ = 2700 kg/m³.
解答：ρ = 540 ÷ 200 = 2.7 g/cm³。换算，2.7 g/cm³ = 2.7 × 1000 kg/m³ = 2700 kg/m³。
3. Pressure and Force | 压强与压力
Pressure is force per unit area. p = F ÷ A. In solids, the force is usually weight (mg). For fluids, p = hρg applies (hydrostatic pressure). Units: Pascal (Pa), where 1 Pa = 1 N/m².
压强是单位面积上受到的压力。p = F ÷ A。在固体中，力通常是重力 (mg)。对于流体，适用 p = hρg（液体压强）。单位：帕斯卡 (Pa)，1 Pa = 1 N/m²。
Example: A box weighing 800 N rests on an area of 0.5 m². Calculate the pressure exerted on the floor.
例题：一个重 800 N 的箱子放在 0.5 m² 的面积上。计算它对地面的压强。
Solution: p = 800 ÷ 0.5 = 1600 Pa (or 1.6 kPa).
解答：p = 800 ÷ 0.5 = 1600 Pa（或 1.6 kPa）。
For a liquid column: p = hρg. If water (ρ = 1000 kg/m³) is 10 m deep, p = 10 × 1000 × 9.8 = 98 000 Pa.
液柱压强：p = hρg。若水（ρ = 1000 kg/m³）深 10 m，p = 10 × 1000 × 9.8 = 98 000 Pa。
4. Work, Energy and Power | 功、能量与功率
Work is done when a force moves an object. W = F × d (force × distance moved in direction of force). Gravitational potential energy (GPE) = mgh. Kinetic energy (KE) = ½mv². Power is the rate of doing work: P = W ÷ t or P = E ÷ t. The unit of power is the watt (W).
当力使物体移动时做了功。W = F × d（力 × 沿力方向移动的距离）。重力势能 (GPE) = mgh。动能 (KE) = ½mv²。功率是做功的速率：P = W ÷ t 或 P = E ÷ t。功率的单位是瓦特 (W)。
Example: A crane lifts a 200 kg mass through 15 m. Calculate the work done. Then find the power if this takes 30 seconds.
例题：一台起重机将 200 kg 的重物吊起 15 m。计算做功多少。接着求出若用时 30 秒，功率是多少。
Solution: F = weight = mg = 200 × 9.8 = 1960 N. W = 1960 × 15 = 29 400 J. Power = 29 400 ÷ 30 = 980 W.
解答：F = 重力 = mg = 200 × 9.8 = 1960 N。W = 1960 × 15 = 29 400 J。功率 = 29 400 ÷ 30 = 980 W。
Kinetic energy example: An 800 kg car moves at 20 m/s. KE = ½ × 800 × (20)² = 160 000 J.
动能例题：一辆 800 kg 的汽车以 20 m/s 行驶。KE = ½ × 800 × (20)² = 160 000 J。
5. Electrical Circuits: Ohm’s Law | 电路计算：欧姆定律
Ohm’s law relates voltage (V), current (I), and resistance (R): V = I × R. Units: volts (V), amperes (A), and ohms (Ω). For series circuits, resistances add up: R_total = R₁ + R₂ + … In parallel, 1/R_total = 1/R₁ + 1/R₂ + …
欧姆定律关联电压 (V)、电流 (I) 和电阻 (R)：V = I × R。单位：伏特 (V)、安培 (A) 和欧姆 (Ω)。串联电路中，电阻相加：R_total = R₁ + R₂ + … 并联电路中，1/R_total = 1/R₁ + 1/R₂ + …
Example: A resistor of 120 Ω has a current of 0.5 A flowing through it. Calculate the voltage across it.
例题：一个 120 Ω 的电阻中有 0.5 A 的电流通过。计算它两端的电压。
Solution: V = 0.5 × 120 = 60 V.
解答：V = 0.5 × 120 = 60 V。
For two parallel 6 Ω resistors: 1/R_total = 1/6 + 1/6 = 2/6, so R_total = 3 Ω.
两个 6 Ω 的电阻并联：1/R_total = 1/6 + 1/6 = 2/6，因此 R_total = 3 Ω。
6. Electrical Power and Energy Transfer | 电功率与电能转移
Power in an electrical circuit can be calculated using P = I × V. Also, P = I²R or P = V²/R if you know resistance. Energy transferred is E = P × t, often measured in joules (J) or kilowatt‑hours (kWh).
电路中的电功率可用 P = I × V 计算。若已知电阻，也可用 P = I²R 或 P = V²/R。转移的电能为 E = P × t，常用焦耳 (J) 或千瓦时 (kWh) 度量。
Example: A 240 V kettle draws a current of 8 A. Calculate its power and the energy used if it runs for 5 minutes.
例题：一个 240 V 的电热水壶工作电流为 8 A。计算其功率以及工作 5 分钟消耗的电能。
Solution: P = 8 × 240 = 1920 W. t = 5 min = 300 s. E = 1920 × 300 = 576 000 J (or 576 kJ).
解答：P = 8 × 240 = 1920 W。t = 5 min = 300 s。E = 1920 × 300 = 576 000 J（或 576 kJ）。
7. Mole Calculations (Chemistry) | 化学：摩尔计算
The mole is the unit for amount of substance. Number of moles = mass ÷ molar mass (n = m ÷ M). For gases at r.t.p. (room temperature and pressure), 1 mole occupies 24 dm³. For solutions, n = c × V (where c is concentration in mol/dm³ and V in dm³).
摩尔是物质的量的单位。摩尔数 = 质量 ÷ 摩尔质量 (n = m ÷ M)。对于在室温常压 (r.t.p.) 下的气体，1 摩尔占据 24 dm³。对于溶液，n = c × V（c 是浓度，单位 mol/dm³；V 是体积，单位 dm³）。
Example: Calculate the number of moles in 49 g of sulfuric acid, H₂SO₄. (Ar: H=1, S=32, O=16)
例题：计算 49 g 硫酸 (H₂SO₄) 中的摩尔数。（相对原子质量：H=1, S=32, O=16）
Solution: molar mass M = (2×1) + 32 + (4×16) = 98 g/mol. n = 49 ÷ 98 = 0.5 mol.
解答：摩尔质量 M = (2×1) + 32 + (4×16) = 98 g/mol。n = 49 ÷ 98 = 0.5 mol。
Gas volume example: What volume does 2 moles of CO₂ occupy at r.t.p.? V = 2 × 24 = 48 dm³.
气体体积例题：2 摩尔的 CO₂ 在室温常压下占据多少体积？V = 2 × 24 = 48 dm³。
8. Concentration of Solutions | 溶液浓度计算
Concentration is often expressed in mol/dm³ or g/dm³. Concentration = amount of solute ÷ volume of solution. For mol/dm³: c = n ÷ V, where V is in dm³. You can also calculate concentration in g/dm³ directly: conc. = mass of solute (g) ÷ volume of solution (dm³).
浓度通常用 mol/dm³ 或 g/dm³ 表示。浓度 = 溶质的量 ÷ 溶液的体积。对于 mol/dm³：c = n ÷ V，其中 V 的单位为 dm³。也可以直接计算 g/dm³ 浓度：浓度 = 溶质质量 (g) ÷ 溶液体积 (dm³)。
Example: 5.85 g of NaCl is dissolved in water to make 250 cm³ of solution. Find the concentration in mol/dm³. (M of NaCl = 58.5 g/mol)
例题：将 5.85 g 氯化钠溶于水，配成 250 cm³ 的溶液。计算其浓度（mol/dm³）。(NaCl 摩尔质量 = 58.5 g/mol)
Solution: n = 5.85 ÷ 58.5 = 0.1 mol. Volume = 250 cm³ = 0.25 dm³. c = 0.1 ÷ 0.25 = 0.4 mol/dm³.
解答：n = 5.85 ÷ 58.5 = 0.1 mol。体积 = 250 cm³ = 0.25 dm³。c = 0.1 ÷ 0.25 = 0.4 mol/dm³。
9. Magnification in Biology | 生物：放大率计算
Magnification (M) relates image size and actual size: M = I ÷ A (Image size ÷ Actual size). Both must be in the same unit. If a scale bar is given, use it to find the conversion. A typical question supplies a diagram and asks for actual size or magnification.
放大率 (M) 关联图像尺寸与实物尺寸：M = I ÷ A（图像尺寸 ÷ 实物尺寸）。两者必须使用相同单位。如果给出比例尺，可使用它找到换算关系。典型题目提供图示，要求计算实物尺寸或放大率。
Example: A cell image measures 60 mm on paper, and the actual cell length is 0.15 mm. Calculate the magnification.
例题：一个细胞在纸面上的图像长度为 60 mm，而实际细胞长度为 0.15 mm。计算放大率。
Solution: M = 60 ÷ 0.15 = 400 ×.
解答：M = 60 ÷ 0.15 = 400 倍。
To find actual size: If an image is 20 mm and magnification is 500×, A = 20 ÷ 500 = 0.04 mm = 40 µm.
求实物尺寸：若图像长度 20 mm，放大率为 500 倍，A = 20 ÷ 500 = 0.04 mm = 40 µm。
10. Energy in Food (Calorimetry) | 食物能量计算（量热法）
The energy content of food can be estimated by burning a sample and heating water. Energy = mass of water × specific heat capacity × temperature rise. For water, c = 4.2 J/g°C. Often the energy per gram is found and expressed in J/g or kJ/g.
食物中的能量可以通过燃烧样品并加热水来估算。能量 = 水的质量 × 比热容 × 温升。对于水，c = 4.2 J/g°C。通常求出每克能量，以 J/g 或 kJ/g 表示。
Example: 0.5 g of a cheese puff is burnt; it heats 20 g of water from 22°C to 48°C. Calculate the energy released per gram.
例题：0.5 g 的芝士条燃烧，使 20 g 的水从 22°C 升至 48°C。计算每克释放的能量。
Solution: Q = 20 × 4.2 × (48–22) = 20 × 4.2 × 26 = 2184 J. Per gram = 2184 ÷ 0.5 = 4368 J/g ≈ 4.37 kJ/g.
解答：Q = 20 × 4.2 × (48–22) = 20 × 4.2 × 26 = 2184 J。每克能量 = 2184 ÷ 0.5 = 4368 J/g ≈ 4.37 kJ/g。
11. Moments and Equilibrium | 力矩与平衡
A moment is the turning effect of a force: Moment = force × perpendicular distance from pivot. For equilibrium, total clockwise moments = total anticlockwise moments. This principle is used to find unknown forces or distances in levers and beams.
力矩是力的转动效果：力矩 = 力 × 到支点的垂直距离。平衡时，总顺时针力矩 = 总逆时针力矩。该原理用于求杠杆和横梁中未知的力或距离。
Example: A see‑saw has a boy weighing 400 N sitting 2 m from the pivot. How far from the pivot on the other side must a 500 N girl sit to balance it?
例题：跷跷板上，一个重 400 N 的男孩坐在距支点 2 m 处。重 500 N 的女孩需坐在另一侧距支点多远才能平衡？
Solution: clockwise moment = 400 × 2 = 800 Nm. For balance, 500 × d = 800 → d = 1.6 m.
解答：顺时针力矩 = 400 × 2 = 800 Nm。为平衡，500 × d = 800 → d = 1.6 m。
12. Wave Speed Equation | 波速方程
For all waves, v = f × λ, where v is wave speed (m/s), f is frequency (Hz), and λ is wavelength (m). This equation applies to sound waves, water waves, and electromagnetic waves. Pay attention to units: frequency may be given in kHz or MHz; convert to Hz.
对于所有波，v = f × λ，其中 v 为波速 (m/s)，f 为频率 (Hz)，λ 为波长 (m)。此方程适用于声波、水波和电磁波。注意单位：频率可能以 kHz 或 MHz 给出，需转换为 Hz。
Example: A radio wave has a frequency of 100 MHz and a wavelength of 3 m. Calculate its speed.
例题：一列无线电波频率为 100 MHz，波长为 3 m。计算其波速。
Solution: f = 100 × 10⁶ Hz = 1 × 10⁸ Hz. v = (1 × 10⁸) × 3 = 3 × 10⁸ m/s.
解答：f = 100 × 10⁶ Hz = 1 × 10⁸ Hz。v = (1 × 10⁸) × 3 = 3 × 10⁸ m/s。
If a sound wave travels at 340 m/s with frequency 170 Hz, λ = v ÷ f = 340 ÷ 170 = 2 m.
若声波波速 340 m/s，频率 170 Hz，λ = v ÷ f = 340 ÷ 170 = 2 m。
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