GCSE Science: Calculation Questions Intensive Training | GCSE 科学：计算题专项训练

📚 GCSE Science: Calculation Questions Intensive Training | GCSE 科学：计算题专项训练

Calculation questions form the backbone of GCSE Science, appearing across Physics, Chemistry and Biology papers. Mastering them requires not just memorising formulas but understanding how to select the correct equation, rearrange it, convert units and present answers to the right number of significant figures. This intensive training article walks you through the most common types of calculation problems, with worked examples and bilingual explanations to build both confidence and accuracy.

计算题是 GCSE 科学的支柱，出现在物理、化学和生物试卷中。要掌握它们，不仅需要记住公式，还需要知道如何选择正确的方程、变形、转换单位，并以正确的有效数字表示答案。这篇文章带你练习最常见的计算题型，通过例题和双语讲解，帮助你建立信心并提高准确率。

1. Understanding Units and Formula Triangles | 理解单位和公式三角形

Before tackling any calculation, you must be comfortable with SI units and conversions. Common units include metres (m), seconds (s), kilograms (kg), joules (J), newtons (N), amperes (A), volts (V), moles (mol) and pascals (Pa). Prefixes such as kilo (10³), centi (10⁻²) and milli (10⁻³) frequently appear. Always convert given quantities into standard SI units before substituting into equations unless the question specifies otherwise.

在处理任何计算之前，你必须熟悉国际单位制和单位换算。常见单位有米(m)、秒(s)、千克(kg)、焦耳(J)、牛顿(N)、安培(A)、伏特(V)、摩尔(mol)和帕斯卡(Pa)。前缀如千(10³)、厘(10⁻²)和毫(10⁻³)经常出现。除非题目另有说明，否则在代入方程前，务必将已知量转换为标准国际单位。

A formula triangle is a handy memory aid for three-variable equations. Cover the quantity you need; the remaining two symbols show the operation. For instance, the speed triangle has s (distance) at the top, v (speed) and t (time) in the bottom corners. Cover v and you see s ÷ t. Cover s and you see v × t. Cover t and you see s ÷ v. Always write the triangle in your working to help rearrange.

公式三角形是记忆三变量方程的好帮手。遮住你要求的量；其余两个符号就展示了运算方式。例如，速度三角形的顶部是 s (距离)，底部是 v (速度) 和 t (时间)。遮住 v 就会看到 s ÷ t。遮住 s 就会看到 v × t。遮住 t 就会看到 s ÷ v。解题时画出三角形有助于变形。

2. Speed, Distance, Time and Acceleration | 速度、距离、时间和加速度

Average speed is defined as total distance travelled divided by the time taken. The equation is often written as:

平均速度定义为总路程除以所用时间。方程通常写作：

v = s ÷ t

where v is speed (m/s), s is distance (m) and t is time (s). Acceleration is the rate of change of velocity. Its formula is:

其中 v 是速度(m/s)，s 是距离(m)，t 是时间(s)。加速度是速度的变化率。其公式为：

a = (v – u) ÷ t

with u as initial velocity (m/s), v as final velocity (m/s) and a as acceleration (m/s²).

其中 u 是初速度(m/s)，v 是末速度(m/s)，a 是加速度(m/s²)。

Worked example: A car accelerates from 5 m/s to 25 m/s in 4 seconds. Calculate its acceleration.

例题：一辆汽车在 4 秒内从 5 m/s 加速到 25 m/s。计算其加速度。

Step 1: Identify u = 5 m/s, v = 25 m/s, t = 4 s.

步骤一：确定 u = 5 m/s，v = 25 m/s，t = 4 s。

Step 2: Substitute into a = (v – u) ÷ t = (25 – 5) ÷ 4 = 20 ÷ 4 = 5 m/s².

步骤二：代入公式 a = (v – u) ÷ t = (25 – 5) ÷ 4 = 20 ÷ 4 = 5 m/s²。

Always check the unit: m/s² means metres per second per second, which correctly represents a change in velocity over time.

务必检查单位：m/s² 表示米每二次方秒，正确地代表速度随时间的变化。

3. Forces and Newton’s Second Law | 力与牛顿第二定律

Newton’s Second Law links resultant force, mass and acceleration:

牛顿第二定律将合力、质量和加速度联系起来：

F = m a

where F is resultant force in newtons (N), m is mass in kilograms (kg) and a is acceleration in m/s². This is a fundamental equation that appears in many contexts, from braking distances to rocket launches.

其中 F 是合力，单位牛顿(N)，m 是质量，单位千克(kg)，a 是加速度，单位为 m/s²。这是一个基础方程，出现在从刹车距离到火箭发射等许多情境中。

Worked example: A 1200 kg car experiences a resultant force of 3600 N. Find its acceleration.

例题：一辆 1200 千克的汽车受到 3600 牛顿的合力。求其加速度。

Rearrange: a = F ÷ m = 3600 N ÷ 1200 kg = 3 m/s².

变形：a = F ÷ m = 3600 N ÷ 1200 kg = 3 m/s²。

For problems involving weight, use W = m g where g is gravitational field strength (9.8 N/kg on Earth, often rounded to 10 N/kg in GCSE). Never confuse weight (N) with mass (kg).

对于涉及重量的问题，用 W = m g，其中 g 是引力场强度（地球取 9.8 N/kg，GCSE 常近似为 10 N/kg）。切勿混淆重量(N)和质量(kg)。

4. Work Done, Energy Transferred and Power | 做功、能量转移与功率

Work done is equal to the energy transferred when a force moves an object over a distance:

做功等于力使物体移动一段距离时所转移的能量：

W = F d

with W in joules (J), F in newtons (N) and d in metres (m). Power is the rate of doing work or transferring energy:

其中 W 的单位是焦耳(J)，F 是牛顿(N)，d 是米(m)。功率是做功或能量转移的速率：

P = E ÷ t or P = W ÷ t

where P is power in watts (W), E is energy in joules (J) and t is time in seconds (s).

其中 P 是功率，单位瓦特(W)，E 是能量，单位焦耳(J)，t 是时间，单位秒(s)。

Worked example: A crane lifts a 200 kg load through a vertical height of 15 m in 10 seconds. Calculate the power output. (Take g = 10 N/kg)

例题：一辆起重机在 10 秒内将 200 千克的重物垂直提升 15 米。计算输出功率。（取 g = 10 N/kg）

First, find the weight force: F = m g = 200 × 10 = 2000 N.

首先求重量力：F = m g = 200 × 10 = 2000 N。

Work done = F × d = 2000 N × 15 m = 30 000 J.

做功 = F × d = 2000 N × 15 m = 30 000 J。

Power = W ÷ t = 30 000 J ÷ 10 s = 3000 W (or 3 kW).

功率 = W ÷ t = 30 000 J ÷ 10 s = 3000 W (或 3 kW)。

5. Density and Specific Heat Capacity | 密度和比热容

Density is mass per unit volume. The equation is:

密度是单位体积的质量。方程为：

ρ = m ÷ V

where ρ is density (kg/m³ or g/cm³), m is mass (kg or g) and V is volume (m³ or cm³). Note that 1 g/cm³ = 1000 kg/m³. Common GCSE problems involve irregular solids and the displacement method.

其中 ρ 是密度(kg/m³ 或 g/cm³)，m 是质量(kg 或 g)，V 是体积(m³ 或 cm³)。注意 1 g/cm³ = 1000 kg/m³。GCSE 常见问题涉及不规则固体和排水法。

Specific heat capacity is the energy required to raise the temperature of 1 kg of a substance by 1°C. The formula is:

比热容是将 1 千克物质升高 1°C 所需的能量。公式为：

ΔE = m c Δθ

ΔE is change in thermal energy (J), m is mass (kg), c is specific heat capacity (J/kg°C) and Δθ is temperature change (°C).

ΔE 是热能变化(J)，m 是质量(kg)，c 是比热容(J/kg°C)，Δθ 是温度变化(°C)。

Worked example: An aluminium block of mass 0.5 kg is heated. Its temperature rises from 22°C to 42°C. The specific heat capacity of aluminium is 900 J/kg°C. Find the energy supplied.

例题：一块质量为 0.5 kg 的铝块被加热，温度从 22°C 升至 42°C。铝的比热容为 900 J/kg°C。求提供的能量。

Δθ = 42 – 22 = 20°C. ΔE = 0.5 × 900 × 20 = 9000 J.

Δθ = 42 – 22 = 20°C。ΔE = 0.5 × 900 × 20 = 9000 J。

6. Ohm’s Law and Electrical Power | 欧姆定律和电功率

Ohm’s Law states that the current through a resistor is directly proportional to the potential difference across it, provided temperature remains constant:

欧姆定律指出，在温度不变的情况下，通过电阻器的电流与两端的电势差成正比：

V = I R

where V is voltage (V), I is current (A) and R is resistance (Ω). Two other key electrical equations are charge and power:

其中 V 是电压(V)，I 是电流(A)，R 是电阻(Ω)。另外两个关键电学方程是电荷和功率：

Q = I t

P = I V and E = I V t

Q is charge (coulombs, C), t is time (s), P is power (W) and E is energy (J).

Q 是电荷(库仑, C)，t 是时间(s)，P 是功率(W)，E 是能量(J)。

Worked example: A 12 V lamp draws a current of 2.5 A. Calculate its power and the energy transferred in 5 minutes.

例题：一个 12 V 的灯泡通过的电流为 2.5 A。计算其功率和 5 分钟内转移的能量。

P = I V = 2.5 × 12 = 30 W. t = 5 × 60 = 300 s. E = I V t = 2.5 × 12 × 300 = 9000 J.

P = I V = 2.5 × 12 = 30 W。t = 5 × 60 = 300 s。E = I V t = 2.5 × 12 × 300 = 9000 J。

Always convert minutes to seconds in energy calculations, as the watt is a joule per second.

在能量计算中，总要将分钟转换为秒，因为瓦特是焦耳每秒。

7. Wave Speed, Frequency and Wavelength | 波速、频率和波长

All waves obey the wave equation:

所有波都遵循波动方程：

v = f λ

where v is wave speed (m/s), f is frequency (hertz, Hz) and λ is wavelength (metres, m). This applies to sound waves, water waves and electromagnetic waves.

其中 v 是波速(m/s)，f 是频率(赫兹, Hz)，λ 是波长(米, m)。这适用于声波、水波和电磁波。

Worked example: A sound wave has a frequency of 440 Hz and a wavelength of 0.75 m. Determine the speed of sound in air.

例题：某声波频率为 440 Hz，波长为 0.75 m。求空气中的声速。

v = f λ = 440 × 0.75 = 330 m/s.

v = f λ = 440 × 0.75 = 330 m/s。

If you are given the period T (time for one complete wave), remember that f = 1 ÷ T. Use the correct units: 1 kHz = 1000 Hz and 1 cm = 0.01 m.

若给出周期 T (一个完整波的时间)，记住 f = 1 ÷ T。使用正确单位：1 kHz = 1000 Hz，1 cm = 0.01 m。

8. Mole Calculations: Mass and Molar Mass | 摩尔计算：质量与摩尔质量

The mole is the unit for amount of substance. The linking equation is:

摩尔是物质的量的单位。联系方程为：

n = m ÷ M

where n is number of moles (mol), m is mass (g) and M is molar mass (g/mol). Molar mass equals the relative formula mass (Mᵣ) in grams. Always use the periodic table to find atomic masses.

其中 n 是摩尔数(mol)，m 是质量(g)，M 是摩尔质量(g/mol)。摩尔质量等于相对式量(Mᵣ)的克数。务必使用元素周期表查找原子量。

Worked example: How many moles are present in 8.0 g of sulfur dioxide, SO₂? (S = 32, O = 16)

例题：8.0 g 二氧化硫(SO₂)中有多少摩尔？(S = 32, O = 16)

Mᵣ of SO₂ = 32 + (2 × 16) = 64. M = 64 g/mol. n = 8.0 ÷ 64 = 0.125 mol.

SO₂ 的 Mᵣ = 32 + (2 × 16) = 64。M = 64 g/mol。n = 8.0 ÷ 64 = 0.125 mol。

For reacting mass calculations, use the mole ratio from the balanced equation: e.g. 2Mg + O₂ → 2MgO. If 48 g of Mg (M = 24 g/mol, so 2 mol) react, they produce 2 mol of MgO, so mass of MgO = 2 × (24+16) = 80 g.

在反应质量计算中，使用平衡方程式中的摩尔比：例如 2Mg + O₂ → 2MgO。如果 48 g Mg (M = 24 g/mol，即 2 mol) 反应，生成 2 mol MgO，则 MgO 的质量 = 2 × (24+16) = 80 g。

9. Concentration of Solutions | 溶液浓度

Concentration can be expressed in g/dm³ or mol/dm³. The key equations are:

浓度可以用 g/dm³ 或 mol/dm³ 表示。关键方程有：

Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)

Concentration (mol/dm³) = number of moles (mol) ÷ volume (dm³)

Remember that 1 dm³ = 1000 cm³, so divide cm³ by 1000 to convert to dm³.

记住 1 dm³ = 1000 cm³，因此 cm³ 除以 1000 可转化为 dm³。

Worked example: 5.85 g of sodium chloride (NaCl) is dissolved in water to make 250 cm³ of solution. Find the concentration in mol/dm³. (Na = 23, Cl = 35.5)

例题：将 5.85 g 氯化钠(NaCl)溶于水，配成 250 cm³ 溶液。求以 mol/dm³ 为单位的浓度。(Na = 23, Cl = 35.5)

Mᵣ of NaCl = 23 + 35.5 = 58.5, so M = 58.5 g/mol. Moles n = 5.85 ÷ 58.5 = 0.100 mol. Volume = 250 ÷ 1000 = 0.250 dm³. Concentration = 0.100 mol ÷ 0.250 dm³ = 0.400 mol/dm³.

NaCl 的 Mᵣ = 23 + 35.5 = 58.5，故 M = 58.5 g/mol。摩尔 n = 5.85 ÷ 58.5 = 0.100 mol。体积 = 250 ÷ 1000 = 0.250 dm³。浓度 = 0.100 mol ÷ 0.250 dm³ = 0.400 mol/dm³。

Titration calculations combine this with the mole ratio; practice using n = c × V (with V in dm³) for both solutions, then apply the stoichiometric ratio.

滴定计算会结合摩尔比；练习对两种溶液都使用 n = c × V (V 用 dm³)，然后应用化学计量比。

10. Magnification in Biology | 生物中的放大率

In microscopy, magnification tells you how many times larger the image is compared to the actual specimen:

在显微镜中，放大倍率表示图像比实际标本大多少倍：

Magnification = image size ÷ actual size

Both sizes must be in the same unit. Always convert mm, µm and nm carefully: 1 mm = 1000 µm, 1 µm = 1000 nm. Use standard form for very small numbers.

两个尺寸必须使用相同单位。务必仔细转换 mm、µm 和 nm：1 mm = 1000 µm，1 µm = 1000 nm。对于极小的数字使用标准形式。

Worked example: An image of a cell measures 15 mm across. The actual cell diameter is 5 µm. Calculate the magnification.

例题：一个细胞的图像直径为 15 mm。实际细胞直径为 5 µm。计算放大倍率。

Convert image size to µm: 15 mm = 15 × 1000 = 15 000 µm. Magnification = 15 000 ÷ 5 = ×3000.

将图像尺寸转换为 µm：15 mm = 15 × 1000 = 15 000 µm。放大倍率 = 15 000 ÷ 5 = ×3000。

You can also be asked to rearrange: actual size = image size ÷ magnification. Always include the multiplication sign (×) when stating magnification.

你也可能需要变形：实际尺寸 = 图像尺寸 ÷ 放大倍率。表述放大倍率时，始终要加上乘号(×)。

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GCSE Science: Calculation Questions Intensive Training | GCSE 科学：计算题专项训练