📚 GCSE WJEC Science: Mastering Calculation Questions | GCSE WJEC 科学：计算题专项训练
Calculation questions form a significant part of WJEC GCSE Science exams, testing your ability to apply mathematical skills in physics, chemistry and biology. From unit conversions to rearranging formulas, these problems require a systematic approach. This guide provides step-by-step practice across all major topics, helping you build confidence and accuracy.
计算题是 WJEC GCSE 科学考试的重要组成部分，考查你在物理、化学和生物中运用数学技能的能力。从单位换算到公式变形，这些问题需要系统的方法。本指南涵盖所有主要主题的分步练习，帮助你建立信心和准确性。
1. Unit Conversions and Symbols | 单位与符号
Mastering units and standard symbols is the foundation of all calculations. Always convert quantities to SI units before substituting into formulas. Common prefixes: kilo (k = 10³), centi (c = 10⁻²), milli (m = 10⁻³), micro (µ = 10⁻⁶).
掌握单位和标准符号是所有计算的基础。在代入公式之前，务必将量转换为国际单位。常见前缀：千 (k = 10³)，厘 (c = 10⁻²)，毫 (m = 10⁻³)，微 (µ = 10⁻⁶)。
Below is a quick reference for the most common conversions you will encounter in WJEC Science questions.
下表是 WJEC 科学题目中最常见换算的快速参考。
| Quantity (English) | Conversion | 量 (中文) | 换算 |
|---|---|---|---|
| Length | 1 km = 1000 m, 1 m = 100 cm, 1 cm = 10 mm | 长度 | 1 千米 = 1000 米, 1 米 = 100 厘米, 1 厘米 = 10 毫米 |
| Time | 1 hour = 60 min = 3600 s | 时间 | 1 小时 = 60 分 = 3600 秒 |
| Mass | 1 tonne = 1000 kg, 1 kg = 1000 g, 1 g = 1000 mg | 质量 | 1 吨 = 1000 千克, 1 千克 = 1000 克, 1 克 = 1000 毫克 |
| Volume | 1 m³ = 1000 dm³, 1 dm³ = 1000 cm³, 1 litre = 1 dm³ = 1000 cm³ | 体积 | 1 立方米 = 1000 立方分米, 1 立方分米 = 1000 立方厘米, 1 升 = 1 立方分米 = 1000 立方厘米 |
| Density | 1 g/cm³ = 1000 kg/m³ | 密度 | 1 克/立方厘米 = 1000 千克/立方米 |
2. Density Calculations | 密度计算
Density (ρ) is mass per unit volume. The formula is crucial for both physics and chemistry problems. Always check whether the question expects g/cm³ or kg/m³.
密度 (ρ) 是单位体积的质量。该公式对物理和化学问题都很关键。始终检查题目期望的单位是 g/cm³ 还是 kg/m³。
ρ = m ÷ V
Rearrange to find mass or volume: m = ρ × V and V = m ÷ ρ.
变形后求质量或体积：m = ρ × V 和 V = m ÷ ρ。
Example: A block of metal has a mass of 540 g and a volume of 200 cm³. Calculate its density in g/cm³ and kg/m³.
示例：一块金属的质量为 540 g，体积为 200 cm³。计算其密度，分别以 g/cm³ 和 kg/m³ 表示。
ρ = 540 g ÷ 200 cm³ = 2.7 g/cm³. To convert to kg/m³, multiply by 1000: 2.7 × 1000 = 2700 kg/m³.
ρ = 540 g ÷ 200 cm³ = 2.7 g/cm³。转换为 kg/m³，乘以 1000：2.7 × 1000 = 2700 kg/m³。
3. Speed and Acceleration | 速度与加速度
These motion equations appear frequently in both physics and biology (e.g. nerve impulses). Always use SI units: metres (m) and seconds (s).
这些运动方程经常出现在物理和生物题目中（如神经冲动）。始终使用国际单位：米 (m) 和秒 (s)。
v = d ÷ t
Where v = speed (m/s), d = distance (m), t = time (s).
其中 v = 速度 (m/s)，d = 距离 (m)，t = 时间 (s)。
a = (v – u) ÷ t
v = final velocity, u = initial velocity, t = time taken.
v = 末速度，u = 初速度，t = 所用时间。
Example: A car accelerates from 5 m/s to 25 m/s in 4 s. Find its acceleration.
示例：一辆汽车在 4 秒内从 5 m/s 加速到 25 m/s。求其加速度。
a = (25 – 5) ÷ 4 = 20 ÷ 4 = 5 m/s².
a = (25 – 5) ÷ 4 = 20 ÷ 4 = 5 m/s²。
4. Newton’s Second Law | 牛顿第二定律
Force calculations link mass and acceleration. Remember weight is a type of force, so use W = m × g where g = 9.8 m/s² (often rounded to 10 m/s² in WJEC questions).
力的计算将质量和加速度联系起来。记住重力是一种力，因此使用 W = m × g，g = 9.8 m/s²（WJEC 题目中常取整为 10 m/s²）。
F = m × a
W = m × g
Example: A force of 120 N acts on a 40 kg trolley. Calculate the acceleration.
示例：一个 120 N 的力作用在质量为 40 kg 的小车上。计算加速度。
a = F ÷ m = 120 N ÷ 40 kg = 3 m/s².
a = F ÷ m = 120 N ÷ 40 kg = 3 m/s²。
Example: A book has a mass of 0.8 kg. What is its weight on Earth? (g = 10 m/s²)
示例：一本书的质量为 0.8 kg。它在地球上的重量是多少？(g = 10 m/s²)
W = 0.8 kg × 10 m/s² = 8 N.
W = 0.8 kg × 10 m/s² = 8 N。
5. Work, Energy and Power | 功、能与功率
These concepts allow you to analyse energy transfers. The key formulas are linked and often combined in multi-step problems.
这些概念使你可以分析能量转移。关键公式相互关联，经常在多步问题中组合使用。
W = F × d
Eₚ = m × g × h
Eₖ = ½ × m × v²
P = W ÷ t = E ÷ t
Work done (W) is measured in joules (J), power (P) in watts (W).
做功 (W) 以焦耳 (J) 为单位，功率 (P) 以瓦特 (W) 为单位。
Example: A crane lifts a 200 kg load vertically through 15 m. g = 10 m/s². Calculate the work done and the power if the lift takes 30 s.
示例：一台起重机将 200 kg 的重物垂直提升 15 m。g = 10 m/s²。计算做功的大小，如果提升用时 30 s，求功率。
Weight = mg = 200 × 10 = 2000 N. Work done W = F × d = 2000 N × 15 m = 30 000 J. Power P = 30 000 J ÷ 30 s = 1000 W.
重力 = mg = 200 × 10 = 2000 N。做功 W = 2000 N × 15 m = 30 000 J。功率 P = 30 000 J ÷ 30 s = 1000 W。
6. Ohm’s Law and Circuit Calculations | 欧姆定律与电路计算
Ohm’s law is the most fundamental equation in electricity. It is vital to know how to rearrange it for current or resistance.
欧姆定律是电学中最基本的方程。必须知道如何将其变形以求解电流或电阻。
V = I × R
V = potential difference (volts), I = current (amperes), R = resistance (ohms, Ω).
V = 电势差（伏特），I = 电流（安培），R = 电阻（欧姆，Ω）。
Example: A lamp connected to a 6 V supply draws a current of 0.4 A. Find its resistance.
示例：一盏灯连接在 6 V 电源上，流过的电流为 0.4 A。求其电阻。
R = V ÷ I = 6 V ÷ 0.4 A = 15 Ω.
R = 6 V ÷ 0.4 A = 15 Ω。
7. Electrical Power and Energy | 电功率与电能
These formulas calculate how quickly electrical energy is transferred and how much is used over time. Energy companies use kilowatt-hours (kWh), but for most WJEC calculations you will work in joules or watts.
这些公式计算电能转移的快慢以及随时间消耗的总能量。能源公司使用千瓦时 (kWh)，但大多数 WJEC 计算中你将使用焦耳或瓦特。
P = I × V
P = I² × R
E = P × t
Example: An electric heater has a power rating of 2200 W. It is used for 5 minutes. Calculate the energy transferred in joules.
示例：一个电暖器的额定功率为 2200 W，使用了 5 分钟。计算以焦耳为单位的转移能量。
Time in seconds = 5 × 60 = 300 s. E = 2200 W × 300 s = 660 000 J (or 660 kJ).
时间以秒计 = 5 × 60 = 300 s。E = 2200 W × 300 s = 660 000 J（或 660 kJ）。
8. Wave Speed Equation | 波速公式
The wave equation links frequency, wavelength and speed. It is used for both mechanical waves (e.g. sound, water) and electromagnetic waves. Make sure frequency is in hertz (Hz) and wavelength in metres (m).
波速方程将频率、波长和速度联系起来。它适用于机械波（如声波、水波）和电磁波。确保频率以赫兹 (Hz)、波长以米 (m) 为单位。
v = f × λ
Example: A water wave has a frequency of 2 Hz and a wavelength of 0.8 m. Calculate its speed.
示例：一个水波的频率为 2 Hz，波长为 0.8 m。计算其速度。
v = 2 Hz × 0.8 m = 1.6 m/s.
v = 2 Hz × 0.8 m = 1.6 m/s。
9. Reacting Masses and Moles | 化学反应质量与摩尔
Chemical calculations require careful use of molar mass (Mᵣ) and the mole concept. The relative formula mass is the sum of relative atomic masses (Aᵣ) from the periodic table.
化学计算需要仔细使用摩尔质量 (Mᵣ) 和摩尔概念。相对式量是周期表中相对原子质量 (Aᵣ) 的总和。
n = m ÷ M
n = number of moles (mol), m = mass (g), M = molar mass (g/mol).
n = 摩尔数 (mol)，m = 质量 (g)，M = 摩尔质量 (g/mol)。
Example: Calculate the number of moles in 22 g of carbon dioxide (CO₂). Mᵣ of CO₂ = 12 + (16 × 2) = 44.
示例：计算 22 g 二氧化碳 (CO₂) 中的摩尔数。CO₂ 的 Mᵣ = 12 + (16 × 2) = 44。
n = 22 g ÷ 44 g/mol = 0.5 mol.
n = 22 g ÷ 44 g/mol = 0.5 mol。
For reacting masses, use molar ratios from the balanced equation: e.g. 2H₂ + O₂ → 2H₂O means 2 moles of H₂ react with 1 mole of O₂.
对于反应质量，使用配平方程式中的摩尔比：例如 2H₂ + O₂ → 2H₂O，意味着 2 mol H₂ 与 1 mol O₂ 反应。
10. Concentration of Solutions | 溶液浓度
Concentration calculations appear in titrations and chemical analysis. The formula relates moles of solute to volume of solution. Volume must be in dm³; if given in cm³, divide by 1000.
浓度计算出现在滴定和化学分析中。该公式将溶质的摩尔数与溶液体积联系起来。体积必须用 dm³；如果给出的是 cm³，则除以 1000。
c = n ÷ V
c = concentration (mol/dm³), n = moles of solute, V = volume (dm³).
c = 浓度 (mol/dm³)，n = 溶质摩尔数，V = 体积 (dm³)。
Example: 0.05 moles of sodium hydroxide (NaOH) are dissolved in 250 cm³ of water. Find the concentration.
示例：0.05 mol 氢氧化钠 (NaOH) 溶解在 250 cm³ 水中。求浓度。
V = 250 cm³ ÷ 1000 = 0.25 dm³. c = 0.05 mol ÷ 0.25 dm³ = 0.2 mol/dm³.
V = 250 cm³ ÷ 1000 = 0.25 dm³。c = 0.05 mol ÷ 0.25 dm³ = 0.2 mol/dm³。
11. Mixed Calculations and Common Pitfalls | 综合计算与常见误区
WJEC exams often combine multiple steps, for example finding kinetic energy then power, or calculating moles then concentration. A structured approach is essential: 1) List known quantities, 2) Convert all units to SI, 3) Write the relevant formula, 4) Substitute numbers, 5) Check the answer’s units and significant figures.
WJEC 考试经常组合多个步骤，例如先求动能再求功率，或先计算摩尔数再算浓度。结构化的解题方法至关重要：1) 列出已知量，2) 将所有单位转换为国际单位，3) 写出相关公式，4) 代入数值，5) 检查答案的单位和有效数字。
Common mistakes include forgetting to convert grams to kilograms, using cm³ instead of m³ in density, confusing mass and weight, misplacing decimal points in mole calculations, and mixing up speed and velocity directions. Always double-check that your derived unit matches the quantity you are solving for.
常见错误包括忘记将克转换为千克，密度中使用 cm³ 而非 m³，混淆质量和重量，摩尔计算中小数点位置错误，以及混淆速度和速度方向。始终反复检查你推导出的单位是否与你要求解的量一致。
Example of a mixed
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