📚 AQA Year 10 Physics Unit Test: Particle Model of Matter Mock Exam Analysis | AQA 十年级物理物质粒子模型单元测试模拟卷解析
This article walks you through a full mock unit test for the AQA Year 10 Physics topic 'Particle Model of Matter'. Each question is broken down with step-by-step explanations, common mistakes, and essential keyword tips to help you maximise your marks in the real assessment.
本文带你完整拆解一份AQA十年级“物质粒子模型”单元测试模拟卷。每一道题都配有逐步解析、常见错误提点以及核心关键词提示，帮助你在真实考试中拿到高分。
1. Density Calculation | 密度计算
Question 1 gives the mass of a metal cube as 50 g and each side as 2.0 cm. The first part asks for its density in g/cm³. Density is mass per unit volume, so we use ρ = m / V 0 cm³ = 6.25 g/cm³. Many students forget to cube the side length or use cm instead of cm³.
第1题给出一个金属立方体的质量为50克，边长为2.0厘米。第一部分要求以g/cm³为单位计算密度。密度是单位体积的质量，使用ρ = m / V。立方体的体积是边长的三次方，(2.0 cm)³ = 8.0 cm³。因此密度 = 50 g ÷ 8.0 cm³ = 6.25 g/cm³。很多学生忘记将边长立方化，或混淆单位。
ρ = m / V = 50 g / (2.0 cm)³ = 6.25 g/cm³
ρ = m / V = 50 g / (2.0 cm)³ = 6.25 g/cm³
it separately: 50 g = 0.050 kg, 8.0 cm³ = 8.0 × 10⁻⁶ m³, giving 0.050 / 8.0×10⁻⁶ = 6250 kg/m³. Always show the conversion factor clearly.
第二部分通常要求以kg/m³为单位表示密度。换算时乘以1000：6.25 g/cm³ × 1000 = 6250 kg/m³。也可以逐个单位换算：50 g = 0.050 kg，8.0 cm³ = 8.0 × 10⁻⁶ m³，计算得6250 kg/m³。务必清晰写出换算因子。
2. Measuring Density by Displacement | 排水法测量密度
Question 2 describes an irregular stone and asks for the experimental steps to find its density. First, measure the mass of the stone using a balance. Record the value in grams. Second, partly fill a measuring cylinder with water and note the initial volume. Third, gently lower the stone into the water, ensuring it is fully submerged, and record the new volume. The volume of the stone is the difference between the final and initial volumes. Finally, calculate density using ρ = mass / volume difference. A common error is reading the water level incorrectly; always read the bottom of the meniscus at eye level.
第2题给了一块不规则石块，要求写出测量其密度的实验步骤。首先，用天平测量石块的质量并记录克数。其次，在量筒中装入适量水，记下初始体积。第三，将石块轻轻放入水中，确保完全浸没，记录新的体积。石块的体积为最终体积与初始体积之差。最后，用ρ = 质量 / 体积差算出密度。常见错误是读数不准——务必视线与凹液面最低处齐平。
Volume of stone = V₂ − V₁
石块体积 = V₂ − V₁
If the stone floats, you must use a sinker or push it under with a pin (mentioning this gains extra marks for technique). The calculated density should be written with the correct unit, usually g/cm³ or kg/m³, and to an appropriate number of significant figures based on the precision of the instruments.
如果石块漂浮，需要使用沉块或用细针将其压入水中（提及这一点能获得操作技巧方面的额外分数）。计算出的密度应写上正确单位，通常为g/cm³或kg/m³，并根据仪器的精度保留合适的有效数字位数。
3. Specific Heat Capacity | 比热容计算
Question 3 provides an electric kettle heating 0.5 kg of water from 20°C to 40°C. The specific heat capacity of water is 4200 J/kg°C. The energy required is calculated using E = m c Δθ kJ). Always include the formula, show the substitution clearly, and state the final unit.
第3题给出一个电热水壶将0.5千克水从20°C加热至40°C，水的比热容为4200 J/kg°C。所需能量用公式E = m c Δθ计算。Δθ = 40 − 20 = 20°C。因此E = 0.5 × 4200 × 20 = 42 000 J（或42 kJ）。务必写出公式，清晰代入数值，并注明最终单位。
E = m c Δθ = 0.5 × 4200 × 20 = 42 000 J
E = m c Δθ = 0.5 × 4200 × 20 = 42 000 J
A common mistake is forgetting that Δθ is the change in temperature, not the final temperature. Some students also confuse specific heat capacity with specific latent heat. The mark scheme often awards marks for stating the assumption of no heat loss to the surroundings, so mention this in an evaluate question.
常见错误是忘记Δθ是温度变化而非最终温度。也有学生把比热容与比潜热混淆。评分方案中通常在评价性题目里给分，条件是要说明假设没有热量散失到环境中，所以遇到此类题目要记得提及这一点。
4. Specific Latent Heat of Fusion | 熔化比潜热
Question 4 asks for the energy required to melt 0.10 kg of ice at 0°C, given that the specific latent heat of fusion of water is 334 kJ/kg. Since melting happens at constant temperature, we use E = m L xpects the answer in kJ as the data is given in kJ/kg, but converting to joules (33 400 J) may be required for consistency.
第4题要求计算熔化0.10千克0°C冰所需的能量，已知水的熔化比潜热为334 kJ/kg。由于熔化在恒定温度下进行，使用E = m L。于是E = 0.10 × 334 = 33.4 kJ。因数据以kJ/kg给出，考试中可能期望答案用kJ，但为了单位一致有时也需转换为焦耳（33 400 J）。
E = m L = 0.10 × 334 kJ = 33.4 kJ
E = m L = 0.10 × 334 kJ = 33.4 kJ
Make sure to underline the fact that the temperature of the substance does not change during the state change; all energy supplied goes to breaking intermolecular bonds, increasing internal energy but not kinetic energy. Many marks are lost by saying the temperature continues to rise.
一定要强调在状态变化期间物质的温度保持不变；所有能量都用于打破分子间作用力，增加内能但不改变动能。很多学生因为说温度继续上升而丢分。
5. Internal Energy & Particle Arrangement | 内能与粒子排列
Question 5 is a short-answer question comparing the internal energy and particle arrangement of water in solid and liquid states. In the solid (ice), particles are held in a fixed, regular lattice by strong intermolecular forces; they vibrate about fixed positions and have low kinetic energy. In the liquid, particles are still close but can slide past each other; they have higher kinetic energy and the arrangement is irregular. The internal energy of the liquid is higher because additional energy has been supplied to break the bonds during melting, even though the temperature is the same (0°C). Always link internal energy to both kinetic energy and potential energy of particles.
第5题是一道简答题，比较固态和液态水中内能与粒子排列。在固态（冰）中，粒子被强分子间作用力固定在规则的晶格中，只能在固定位置附近振动，动能较低。在液态中，粒子仍然紧密但可以相互滑动，动能较高，排列不规则。液体的内能更高，因为在熔化时额外能量被用来打破键合，即使温度相同（0°C）。务必将内能与粒子的动能和势能都联系起来。
A classic mistake is to define internal energy only as the kinetic energy of particles. The AQA specification states that internal energy is the total kinetic and potential energy of all the particles in a system. Emphasise the potential energy contribution when bonds are broken or formed.
一个典型的错误是把内能仅仅定义为粒子的动能。AQA课程大纲明确指出，内能是系统中所有粒子的总动能与总势能之和。当键被打破或形成时，务必强调势能这部分贡献。
6. Gas Pressure & Temperature | 气体压强与温度
Question 6 asks to explain, using the particle model, why the pressure of a gas in a sealed container increases when its temperature is raised. At a higher temperature, gas particles have a higher average kinetic energy, so they move faster. They collide with the container walls more frequently and with greater force per collision. Since pressure is the total force per unit area exerted by particles on the walls, both the increased frequency and force lead to a greater pressure. The container is sealed, so volume and number of particles remain constant.
第6题要求用粒子模型解释，为什么密闭容器中气体升高温度时压强会增大。温度较高时，气体粒子的平均动能更大，运动更快。它们与容器壁的碰撞更加频繁，每次碰撞的力也更大。由于压强是粒子施加在壁上的单位面积总力，频率和力同时增加导致压强更大。容器密封，因此体积和粒子数目不变。
A successful answer must include the words 'kinetic energy', 'more frequent collisions', and 'greater force per collision'. Do not just say 'particles move faster'. For top marks, relate the increased momentum change to the force using Newton's laws.
一份满分的答案必须包含“动能”、“更频繁的碰撞”和“每次碰撞的力更大”这几个关键词。不要只说“粒子运动变快”。想拿最高分，要用牛顿定律将动量变化增加与力联系起来。
7. Boyle's Law: Pressure & Volume | 玻意耳定律：压强与体积
Question 7 presents a gas at constant temperature with initial pressure 100 kPa and volume 2.0 m³. It is compressed to a volume of 1.0 m³; find the new pressure. For a fixed mass of gas at constant temperature, pV = constant, so p₁V₁ = p₂V₂ (100 kPa × 2.0 m³) / 1.0 m³ = 200 kPa. The pressure doubles because the volume is halved.
第7题给出恒温下气体初始压强100 kPa、体积2.0 m³，压缩至体积1.0 m³，求新的压强。对于恒定温度下一定质量的气体，pV = 常数，所以p₁V₁ = p₂V₂。变形得：p₂ = (p₁V₁) / V₂ = (100 kPa × 2.0 m³) / 1.0 m³ = 200 kPa。体积减半，压强加倍。
p₁V₁ = p₂V₂ → p₂ = 200 kPa
p₁V₁ = p₂V₂ → p₂ = 200 kPa
Watch out for unit conversions: if volumes are given in cm³, you can leave them in cm³ as they cancel. But if pressure is in Pa and volume in m³, ensure consistency. Also, explain why temperature must be constant—so that the average kinetic energy of particles does not change, and changes in pressure are purely due to volume change affecting collision frequency.
注意单位换算：如果体积以cm³给出，可以保留cm³因为会约掉。但如果压强用Pa而体积用m³，要确保一致。此外，还要解释为什么温度必须恒定——这样粒子的平均动能才不会改变，压强的改变完全由体积变化导致碰撞频率改变引起。
8. Combining Latent Heat & Specific Heat Capacity | 潜热与比热容综合
Question 8 is a multi-step problem: 0.20 kg of ice at 0°C is melted and then the resulting water is heated to 10°C. The specific latent heat of fusion is 334 kJ/kg, and the specific heat capacity of water is 4200 J/kg°C. Calculate the total energy required. Step 1: Energy to melt ice, E₁ = m L = 0.20 × 334 = 66.8 kJ. Step 2: Energy to heat water from 0°C to 10°C, E₂ = m c Δθ k scheme gives credit for clearly separating the two stages and checking unit consistency.
第8题为多步计算：0.20千克0°C的冰熔化成水，再将水加热至10°C。熔化比潜热为334 kJ/kg，水的比热容为4200 J/kg°C。计算所需总能量。步骤一：熔化冰的能量 E₁ = m L = 0.20 × 334 = 66.8 kJ。步骤二：将水从0°C加热至10°C的能量 E₂ = m c Δθ = 0.20 × 4200 × 10 = 8400 J = 8.4 kJ。总能量 = 66.8 kJ + 8.4 kJ = 75.2 kJ。也可以统一用焦耳表示：66 800 J + 8 400 J = 75 200 J。评分方案认可清晰分步计算并确保单位一致。
Etotal = m L + m c Δθ = 75.2 kJ
总能量 = m L + m c Δθ = 75.2 kJ
A typical error is mixing kJ and J without conversion, leading to answers that are wrong by a factor of 1000. Some students also forget that the ice is already at 0°C, so no energy is needed to first warm the ice. Always highlight the temperature at which the state change occurs.
典型错误之一是未换算单位就混合使用kJ和J，导致答案相差1000倍。有些学生忘记冰已经是0°C，所以不需要先给冰加热。务必强调状态变化发生的温度点。
Published by TutorHao | Physics Revision Series | aleveler.com
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