GCSE物理必做实验1到10全解析 密度 比热容
AQA GCSE Physics 大纲把实验能力列为独立考查维度，Paper 1 与 Paper 2 各有必做实验（Required Practicals），合计约占 15% 的分值。考试不直接让你背步骤，而是考你如何设计方法、控制变量、处理数据、评估误差。按 AQA 官方编号，1 到 5 号实验多出现在 Paper 1（能量、电学、物质），6 到 10 号与 Paper 2（波、力、辐射）联系更紧，但任何编号都可能在两份卷子中被追问。这篇指南按 1 到 10 号逐一拆解，每个实验讲清装置、步骤、结论、一处常见误差与一道高频追问。
The AQA GCSE Physics course assesses practical skills as a separate strand. Ten required practicals appear across Paper 1 and Paper 2 and together account for roughly fifteen percent of the total marks. In the official numbering, practicals 1 to 5 sit closest to Paper 1 topics such as energy, electricity and matter, while practicals 6 to 10 link to the waves, forces and radiation content of Paper 2, although any practical can be examined in either paper. Examiners rarely ask you to recite a method; they want to see that you can plan a fair test, control variables, process measurements and evaluate errors. This guide works through practicals 1 to 10 one by one, covering apparatus, procedure, results, one typical error and one common follow-up question for each.
1. 测定比热容 Specific Heat Capacity
用绝缘材料包裹已知质量的金属块，把电加热器插入块内，用焦耳计记录输入能量，同时用温度计每隔一段时间读取温度。根据 E = mcΔθ 变形得 c = E / (mΔθ) 计算比热容。关键控制：加热器完全插入、金属块包隔热棉、顶部加绝缘盖减少热散失。能量损失会使算出的比热容偏大，这是本实验最常被问的误差来源。
Wrap an insulated metal block of known mass, insert an electric heater and record the energy supplied with a joulemeter while a thermometer reads the temperature rise. Rearrange E = mcΔθ to find c = E / (mΔθ). To cut heat loss, keep the heater fully inserted, lag the block with insulating cotton and cover the top. Energy lost to the surroundings makes the measured value of c too high, and this is the error examiners ask about most.
计算示例：0.5 kg 的铝块温度升高 8.0 摄氏度需要 3600 J 能量，c = 3600 / (0.5 × 8.0) = 900 J/(kg·摄氏度)，与铝的参考值吻合。作答时务必写出代入过程与单位，2 分的计算题单位漏写会扣 1 分。
Worked example: heating a 0.5 kg aluminium block by 8.0 degrees Celsius needs 3600 J, so c = 3600 / (0.5 × 8.0) = 900 J per kilogram per degree Celsius, matching the accepted value for aluminium. Always show the substitution and the units, because a two-mark calculation loses a mark if the unit is missing.
2. 保温效果 Thermal Insulation
向几个相同的烧杯倒入等量开水，分别用不同材料包裹杯壁（或对比有无盖子），记录初始温度后每两三分钟测一次，持续约二十分钟，画出冷却曲线比较保温强弱。必须控制的变量包括水量、初始温度、室温以及温度计插入的深度，否则对比不公平。水温越高散热越快，所以冷却曲线前段陡后段平，属于正常现象，不要误判为测量错误。
Pour equal volumes of just-boiled water into identical beakers, wrap each in a different insulating material (and compare with and without a lid), then record the temperature every two to three minutes for about twenty minutes. Plot cooling curves to compare which material keeps the water warmest. Control the water volume, starting temperature, room temperature and thermometer depth so the comparison is fair. The curve falls steeply at first and flattens later because a hotter drink loses energy faster; that shape is expected, not a sign of a faulty reading.
3. 电阻与导线长度 Resistance and Wire Length
把电阻丝接入电路，电流表串联、电压表并联，用鳄鱼夹改变接入长度，读取多组电压和电流，R = V/I 计算电阻。保持导线材料与粗细不变，结论是电阻与长度成正比。多次测量取平均、每次测量间隔让导线冷却，可减小随机误差；直尺估读误差属于系统性的读数误差。
Connect a length of wire in series with an ammeter and place a voltmeter across the wire. Move a crocodile clip to change the length in the circuit, record V and I at each length and calculate R = V/I. With material and thickness fixed, resistance is proportional to length. Repeating readings at each length and letting the wire cool between readings reduces random error, while parallax when judging the length on the ruler is a systematic reading error.
高频追问：为什么实验中要避免导线过热？因为温度升高使金属电阻增大，会破坏单一变量，让电阻与长度不再呈干净的正比关系。升温后先等导线冷却再读下一组数据是标准做法。
Common question: why must the wire not get too hot? Heating raises the resistance of the metal and introduces a second variable, so the graph of resistance against length is no longer a clean straight line. Waiting for the wire to cool between readings is the standard procedure.
4. I-V 特性曲线 I-V Characteristics
用滑动变阻器分压，分别测量电阻器、小灯泡和二极管的电流与电压，把结果画成 I-V 图。电阻器的图线是过原点的直线，符合欧姆定律；小灯泡的图线变弯，因为灯丝温度升高使电阻增大；二极管只在正向导通，反向电流几乎为零。三个元件要在同一张图或三张图中分别标注，注意电流表内接外接不会在 GCSE 层面考查，不要写无关内容。
Use a variable resistor as a potential divider to sweep the voltage across a fixed resistor, a filament lamp and a diode, recording I at each V and plotting I-V graphs. The fixed resistor gives a straight line through the origin, which is ohmic behaviour. The lamp graph curves because the filament heats up and its resistance rises. The diode conducts only in the forward direction, with almost zero current in reverse. Label the three graphs clearly; ammeter placement debates are beyond GCSE, so do not pad answers with irrelevant detail.
图线题拿分法：问某电压下的电阻，就在该点读对应的 I，用 R = V/I 计算；问灯泡电阻如何变化，就说明图线斜率变小代表电阻增大，因为温度升高。不要只写一句图线是曲线，必须联系电阻变化的原因。
Scoring the graph question: to find the resistance at a chosen voltage, read the current at that point and use R = V/I. If asked how the lamp resistance changes, state that the falling gradient of the curve means rising resistance caused by the higher temperature of the filament. A bare comment that the line is curved earns no mark unless you link it to the reason.
5. 密度测量 Density
规则固体用天平称质量，用直尺量出长宽高计算体积，ρ = m/V 求密度；不规则固体用排水法，或放入盛满水的溢水杯（Eureka can）收集溢出的水测体积；液体则用量筒读取体积，用烧杯加液体与空烧杯的质量差求质量。读数时视线与液面最低处齐平，避免视差。溢出杯外壁沾水会使收集的水偏少，应擦干后称量。
For a regular solid, measure mass on a balance and volume from length, width and height with a ruler, then use ρ = m/V. For an irregular solid, use the displacement method or drop it into an overflowing Eureka can and collect the displaced water. For a liquid, read the volume in a measuring cylinder and find the mass by difference using a balance. Read the meniscus at eye level to avoid parallax, and dry the outside of the can because spilled water clinging to it reduces the collected volume.
数字示例：一块边长 2 cm、3 cm、4 cm 的规则固体体积为 24 立方厘米，天平称得 64.8 g，密度 = 64.8 / 24 = 2.7 g/cm³，可以判断材料接近铝。换算到 kg/m³ 时要乘以 1000，即 2700 kg/m³，这是常见的 1 分换算点。
Example: a regular solid measuring 2 cm by 3 cm by 4 cm has a volume of 24 cubic centimetres; if the balance reads 64.8 g then the density is 64.8 / 24 = 2.7 g per cubic centimetre, close to aluminium. Converting to kilograms per cubic metre means multiplying by 1000 to give 2700, a frequent one-mark trap.
6. 光的反射与折射 Reflection and Refraction
用光源加单缝产生一束细光，射向平面镜，用量角器测量入射角与反射角，验证反射定律：反射角等于入射角。再让光束斜射过半圆形玻璃块，观察光线的弯折方向：光从空气进入玻璃时向法线方向弯折，从玻璃进入空气时向远离法线方向弯折，入射角增大到临界角后发生全反射。作图时光线要画箭头，法线用虚线，角度以法线为基准量取。
Produce a narrow beam with a ray box and single slit, direct it at a plane mirror and measure the angles of incidence and reflection with a protractor to confirm that the angle of reflection equals the angle of incidence. Then pass the beam through a semi-circular glass block and observe the bending: entering the glass the ray bends towards the normal, leaving it bends away, and beyond the critical angle the ray is totally internally reflected. In diagrams draw the rays with arrows, mark the normal as a dashed line and measure all angles from the normal, not from the surface.
7. 力与弹簧伸长 Force and Extension
弹簧上端固定，下端挂一个钩码后读取指针位置，逐次增加钩码并记录每次的伸长量，画出力与伸长量关系图。在弹性限度内图线是过原点的直线，满足胡克定律 F = ke，斜率就是劲度系数；超过弹性限度后图线弯曲，弹簧发生永久形变。测量前先挂一个砝码让弹簧拉直，指针读数要视线平齐刻度，减小视差。
Clamp a spring vertically, hang masses on the lower end and read the pointer position after each addition, then plot force against extension. Within the elastic limit the graph is a straight line through the origin, obeying Hooke's law F = ke, and its gradient equals the spring constant. Past the elastic limit the line curves and the spring is permanently deformed. Hang one mass first to straighten the spring, and read the pointer at eye level to avoid parallax.
计算示例：弹簧挂 0.5 N 伸长 2.0 cm，挂 1.0 N 伸长 4.0 cm，则 k = 1.0 / 0.04 = 25 N/m。若题中给出的是伸长总量而非原始长度，要先把弹簧原长减去；很多学生把总长当伸长量，图线不过原点还找不出原因。
Example: a spring extends 2.0 cm under 0.5 N and 4.0 cm under 1.0 N, so k = 1.0 / 0.04 = 25 newtons per metre. If the question gives the total length rather than the extension, subtract the original length first; confusing total length with extension is why many students draw a graph that fails to pass through the origin.
8. 加速度与牛顿第二定律 Acceleration
让小车在轨道上被砝码通过滑轮拉动，用两个光电门或运动传感器记录通过时间，算出速度与加速度。保持小车质量不变、逐次增大拉力，验证加速度与合力成正比；再保持合力不变、加大小车质量，验证加速度与质量成反比。若用打点计时器，v-t 图的斜率就是加速度。每次实验重复三次取平均，释放小车前确保轨道水平可减小摩擦力干扰。
Pull a trolley along a runway with masses hanging over a pulley and use two light gates or a motion sensor to measure the times from which velocity and acceleration are calculated. Keeping the trolley mass fixed while increasing the driving force shows that acceleration is proportional to the resultant force; keeping the force fixed while adding mass shows that acceleration is inversely proportional to mass. With a ticker timer, the gradient of a velocity-time graph gives the acceleration. Repeat each run three times and average, and level the runway so friction does not distort the results.
分析题套路：题目问为什么图线不过原点，通常答存在摩擦力或轨道未调平；问为什么小质量砝码方案更准，答砝码重力近似等于绳上合力。加速度实验是 6 分大题的高频载体，方法和结论各占一半分数，控制变量必须逐条写明。
Analysis patterns: if the graph does not pass through the origin, answer that friction was present or the runway was not level. If asked why small hanging masses are better, answer that their weight approximates the resultant force on the trolley more closely. This experiment is a favourite home for the six-mark question, with method and conclusion each earning half the marks, so list the controlled variables explicitly.
9. 水波与波动 Waves
在波纹槽中产生水波，用频闪灯照明使波看起来静止，用标尺测量多个相邻波峰之间的距离取平均得到波长，从频闪频率读出波的频率，再用 v = fλ 计算波速。量多个波长取平均可减小读数误差；也可以用振动绳演示横波，观察波速随绳张力和线密度的变化。
Generate water waves in a ripple tank and light them with a strobe lamp so the pattern appears frozen; measure the distance between several adjacent wave crests with a ruler and divide to find the average wavelength, read the frequency from the strobe setting, and calculate wave speed with v = fλ. Averaging over several wavelengths reduces reading error. A vibrating string demonstrates transverse waves and shows how speed depends on the tension and the linear density of the string.
数值示例：频闪频率 5 Hz，测得相邻波峰平均间距 3.0 cm 即 0.03 m，则 v = 5 × 0.03 = 0.15 m/s。把厘米换算成米是本题最高频的失分点，单位不统一直接导致答案差 100 倍。
Example: with a strobe frequency of 5 Hz and an average crest spacing of 3.0 cm, which is 0.03 m, the speed is v = 5 × 0.03 = 0.15 metres per second. Converting centimetres to metres is the most common error here, since forgetting it changes the answer by a factor of one hundred.
10. 红外辐射的吸收与发射 Infrared Radiation
用表面处理不同的金属容器（黑色粗糙、白色、光亮）装入等量热水，或在相同距离放置温度计与红外传感器，每隔固定时间记录温度，比较降温速度。结论：黑色粗糙表面是红外辐射最强的吸收体和发射体，光亮表面最弱。用莱斯利立方体演示时，四个侧面温度差异来自表面性质而非材料本身，因为立方体是同一块金属，这是排除变量的巧妙设计。
Fill identical containers with different surface finishes (matt black, white and shiny) with equal volumes of hot water, or place a thermometer or infrared sensor at the same distance from each, and record the temperature at fixed intervals to compare cooling rates. Matt black surfaces are the best absorbers and emitters of infrared radiation while shiny surfaces are the worst. With a Leslie cube the four faces are one solid piece of metal, so any temperature difference comes from the surface finish alone, a neat way of removing material as a variable.
生活联系题常考：黑色散热器散热更快、银白色保温瓶内胆反射辐射、深色衣服夏天更热，本质上都是同一规律。答这类题先点明吸收或发射的强弱顺序，再结合具体场景说一句，即可拿满 2 分。
Everyday links are popular: dark radiators shed heat faster, shiny vacuum-flask walls reflect radiation, and dark clothing feels hotter in summer, all for the same reason. Name the order of absorbing or emitting strength first, then connect it to the scenario in one sentence, and the two marks are secure.
实验题拿分要点 Exam Technique
实验题失分多集中在三处：一是不写明控制变量或只写无关变量，二是结论没提数据关系或正反比方向，三是误差分析只写随机误差不提系统误差。答题模板：方法题写清装置连接与测量工具，结论题引用图线形状与具体数据，改进题写重复测量、减小间隔、增加保温、使用更精密仪器或更大样本。不确定度表述也有固定套路：测量值与真值的差来自仪器精度和估读，重复实验取平均只能减小随机误差，改变实验设计才能消除系统误差。
Marks are most often lost in three places: naming no controlled variables or only irrelevant ones, stating a conclusion without referring to the data trend, and discussing random error while ignoring systematic error. A reliable answering pattern: for method questions describe the circuit or apparatus and the measuring instruments; for conclusion questions quote the shape of the graph and the actual readings; for improvement questions mention repeating readings, using smaller intervals, adding insulation, choosing more sensitive equipment or taking a larger sample. Uncertainty has its own fixed logic: instrument precision and estimating produce the gap between a reading and the true value, averaging repeated readings only shrinks random error, and changing the design is what removes systematic error.
考前一周把每个实验的装置图重画一遍、把十条结论用一句话写出来、把每个实验的一处误差来源背熟，配合真题限时训练，实验分值就能稳定拿到。记住：考官要的是可重复、可对比、有依据的回答，而不是华丽的辞藻。
In the final week, redraw the apparatus for each practical from memory, compress the ten conclusions into single sentences, memorise one error source per experiment, and practise past-paper questions under timed conditions. Examiners reward answers that are repeatable, comparative and evidence-based, not flowery language.
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