Unit B1 · The Particulate Nature of MatterUnit B1 · 物质的粒子本质
Thermal Energy Transfers热能传递
IB-Style Practice QuestionsIB 风格练习题
MEDIUMHARDPaper 1Paper 1BPaper 2
Syllabus B1.1 to B1.6考纲 B1.1 至 B1.6PHYSICS HL
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PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 21 marks短结构题 · 可用计算器 · 21 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. Convert every temperature to kelvin before using the radiation laws, but use the temperature difference directly in $Q = mc\Delta T$. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。使用辐射定律前把温度换算为开尔文，但在 $Q = mc\Delta T$ 中直接用温度差。数值答案保留适当的有效数字。
A sealed sample of helium gas is at a temperature of $27\ ^{\circ}\mathrm{C}$.一份密封的氦气样品温度为 $27\ ^{\circ}\mathrm{C}$。
(a)Convert this temperature to the Kelvin scale, and state what happens to the average random kinetic energy of the molecules if the absolute temperature is doubled.把该温度换算为开尔文温标，并说明若绝对温度加倍，分子的平均随机动能会如何变化。[2]
(b)Define the internal energy of the gas, and state in molecular terms why warming the gas (with no change of phase) increases its internal energy.定义该气体的内能，并从分子角度说明为何加热气体（无相变）会增大其内能。[2]
An electric heater of constant power $50\ \mathrm{W}$ is embedded in a $0.40\ \mathrm{kg}$ metal block. After heating for $90\ \mathrm{s}$ the temperature of the block rises by $9.0\ \mathrm{K}$. Assume all the electrical energy is transferred to the block.一只恒定功率 $50\ \mathrm{W}$ 的电加热器嵌入一块 $0.40\ \mathrm{kg}$ 的金属块中。加热 $90\ \mathrm{s}$ 后金属块温度升高 $9.0\ \mathrm{K}$。设全部电能都传给金属块。
(a)Calculate the energy supplied to the block, and hence determine the specific heat capacity of the metal.计算供给金属块的能量，并由此求该金属的比热容。[3]
(b)In a real experiment some energy is lost to the surroundings. State whether the experimental value of the specific heat capacity would be an overestimate or an underestimate, and justify your answer.在真实实验中有部分能量损失到环境。说明所测比热容是偏大还是偏小，并给出理由。[2]
A vacuum flask keeps a hot drink warm. It has a double wall with a vacuum gap between the inner and outer layers, and the wall surfaces facing the gap are silvered (shiny).一只真空保温瓶用于保持饮料温热。它有双层壁，内外层之间是真空间隙，朝向间隙的壁面镀银（光亮）。
(a)State, with a reason, why neither conduction nor convection can transfer significant energy across the vacuum gap.说明为何传导与对流都无法越过真空间隙传递明显的能量，并给出理由。[3]
(b)Explain how the silvered surfaces reduce energy loss, naming the heat-transfer mechanism involved.解释镀银表面如何减少能量损失，并指明所涉及的热传递机制。[2]
(c)State which one of the three mechanisms can transfer energy through a vacuum.写出三种机制中哪一种能在真空中传递能量。[1]
Q4HARDPaper 1Stefan-Boltzmann + Wien scaling斯特藩-玻尔兹曼与维恩缩放[6 marks]
A small object is modelled as a black body. Its absolute temperature is increased so that it becomes exactly twice as hot.一个小物体被建模为黑体。其绝对温度升高，恰好变为原来的两倍。
(a)Using the Stefan-Boltzmann law $L = \sigma A T^{4}$, determine the factor by which the total power radiated increases.利用斯特藩-玻尔兹曼定律 $L = \sigma A T^{4}$，求辐射总功率增大的倍数。[2]
(b)Using Wien's displacement law, state the factor by which the peak wavelength of the emitted spectrum changes, and whether the peak moves to longer or shorter wavelengths.利用维恩位移定律，写出发射光谱峰值波长变化的倍数，以及峰值移向更长还是更短波长。[2]
(c)A surface at room temperature has an absolute temperature of about $300\ \mathrm{K}$. Show that its spectrum peaks in the infrared, well beyond the visible range ($400\ \mathrm{nm}$ to $700\ \mathrm{nm}$).室温表面绝对温度约 $300\ \mathrm{K}$。证明其光谱峰值落在红外，远超可见光范围（$400\ \mathrm{nm}$ 至 $700\ \mathrm{nm}$）。[2]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 20 marks图像 · 数据 · 不确定度 · 20 分
Graph and Data Questions图像与数据题
These items reward careful reading of heating curves and gradients, and correct handling of uncertainties. Quote uncertainties to one significant figure and round the value to match.这些题考查对加热曲线与斜率的细致读取以及对不确定度的正确处理。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
Q5HARDPaper 1Bheating curve: plateau and slope加热曲线：平台与斜段[10 marks]
A $0.10\ \mathrm{kg}$ sample of a pure solid is heated by a heater of constant power $40\ \mathrm{W}$. The temperature is recorded against time, giving the heating curve below. The solid first warms, then melts, then the liquid warms.用一只恒定功率 $40\ \mathrm{W}$ 的加热器加热一份 $0.10\ \mathrm{kg}$ 的纯固体样品。记录温度随时间的变化，得到下面的加热曲线。固体先升温，随后熔化，然后液体升温。
$t\ /\ \mathrm{s}$
$0$
$50$
$100$
$150$
$200$
$\theta\ /\ ^{\circ}\mathrm{C}$
$20$
$60$
$60$
$60$
$100$
(a)Identify the time interval over which the sample is melting, and state the melting point of the solid. Explain why the temperature stays constant during this interval in terms of molecular potential energy.指出样品正在熔化的时间区间，并写出该固体的熔点。用分子势能解释为何这段区间内温度保持不变。[3]
(b)Using the first sloped section ($0$ to $50\ \mathrm{s}$), calculate the specific heat capacity of the solid.利用第一段斜段（$0$ 至 $50\ \mathrm{s}$），计算该固体的比热容。[3]
(c)Using the flat plateau, calculate the specific latent heat of fusion of the substance.利用平台段，计算该物质的比熔化潜热。[2]
(d)The power of the heater is quoted as $40\ \mathrm{W} \pm 2\ \mathrm{W}$. State the percentage uncertainty in the power, and explain in one sentence how it propagates to the latent heat found in (c).加热器功率标为 $40\ \mathrm{W} \pm 2\ \mathrm{W}$。写出功率的百分比不确定度，并用一句话说明它如何传递到 (c) 中求出的潜热。[2]
A student determines the specific heat capacity of water by heating $0.50\ \mathrm{kg}$ of water in an insulated beaker with an electric heater of constant power $90\ \mathrm{W}$, and plotting the temperature against time. The graph is a straight line of measured gradient $0.040\ \mathrm{K\,s^{-1}}$.某学生用恒定功率 $90\ \mathrm{W}$ 的电加热器在隔热烧杯中加热 $0.50\ \mathrm{kg}$ 的水来测定水的比热容，并作温度对时间的图。图为一条直线，测得斜率为 $0.040\ \mathrm{K\,s^{-1}}$。
(a)Starting from $Q = mc\Delta T$ and $Q = Pt$, show that the gradient of the temperature-time graph equals $\dfrac{P}{mc}$.从 $Q = mc\Delta T$ 与 $Q = Pt$ 出发，证明温度-时间图的斜率等于 $\dfrac{P}{mc}$。[3]
(b)Use the gradient to calculate the experimental specific heat capacity of the water.用斜率计算所测水的比热容。[2]
(c)The gradient has an uncertainty of $\pm 0.002\ \mathrm{K\,s^{-1}}$ and the power is known precisely. Calculate the percentage uncertainty in the experimental specific heat capacity, and hence state its absolute uncertainty.斜率的不确定度为 $\pm 0.002\ \mathrm{K\,s^{-1}}$，功率已精确知道。计算所测比热容的百分比不确定度，并由此写出其绝对不确定度。[3]
(d)The accepted value of the specific heat capacity of water is $4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$. State one experimental reason why the measured value tends to come out slightly too high if the beaker is not perfectly insulated.水的比热容公认值为 $4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$。说明若烧杯隔热不完美，所测值为何往往偏高的一个实验原因。[2]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 39 marks长结构题 · 可用计算器 · 39 分
Extended Structured Problems长结构问题
Set up each problem with a clear energy-conservation statement. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer.每题先写清楚能量守恒的表述。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。
Q7HARDPaper 2method of mixtures混合法[10 marks]
A $0.20\ \mathrm{kg}$ block of copper at $95\ ^{\circ}\mathrm{C}$ is dropped into $0.25\ \mathrm{kg}$ of water at $18\ ^{\circ}\mathrm{C}$ in an insulated cup. Take $c_{\text{Cu}} = 385\ \mathrm{J\,kg^{-1}\,K^{-1}}$ and $c_{\text{water}} = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$. Neglect the heat capacity of the cup. No phase change occurs.在隔热杯中把一块 $0.20\ \mathrm{kg}$、$95\ ^{\circ}\mathrm{C}$ 的铜块投入 $0.25\ \mathrm{kg}$、$18\ ^{\circ}\mathrm{C}$ 的水中。取 $c_{\text{Cu}} = 385\ \mathrm{J\,kg^{-1}\,K^{-1}}$、$c_{\text{water}} = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$。忽略杯的热容。无相变发生。
(a)State the principle of conservation of energy as it applies to this insulated system, and write the equation relating the energy lost by the copper to the energy gained by the water.陈述适用于该隔热系统的能量守恒原理，并写出铜放出能量与水吸收能量相联系的方程。[2]
(b)Calculate the final equilibrium temperature of the mixture.计算混合物的最终平衡温度。[4]
(c)Calculate the quantity of thermal energy transferred from the copper to the water.计算从铜传给水的热能。[2]
(d)Explain, using the masses and specific heat capacities, why the final temperature is much closer to the water's initial temperature than to the copper's.结合质量与比热容，解释为何末温更接近水的初温而非铜的初温。[2]
$0.030\ \mathrm{kg}$ of ice at $0\ ^{\circ}\mathrm{C}$ is added to $0.300\ \mathrm{kg}$ of water at $25\ ^{\circ}\mathrm{C}$ in a well-insulated container. Take $L_{f} = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$ and $c_{\text{water}} = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$.在隔热良好的容器中，把 $0.030\ \mathrm{kg}$、$0\ ^{\circ}\mathrm{C}$ 的冰加入 $0.300\ \mathrm{kg}$、$25\ ^{\circ}\mathrm{C}$ 的水中。取 $L_{f} = 3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$、$c_{\text{water}} = 4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$。
(a)Calculate the energy required to melt all the ice at $0\ ^{\circ}\mathrm{C}$.计算在 $0\ ^{\circ}\mathrm{C}$ 下熔化全部冰所需的能量。[2]
(b)Calculate the maximum energy the warm water could release if it cooled to $0\ ^{\circ}\mathrm{C}$, and hence show that all of the ice melts and the final temperature is above $0\ ^{\circ}\mathrm{C}$.计算温水冷却到 $0\ ^{\circ}\mathrm{C}$ 时可释放的最大能量，并由此证明全部冰都会熔化且末温高于 $0\ ^{\circ}\mathrm{C}$。[3]
(c)By writing a full energy-conservation equation (melting the ice, then warming the melt-water from $0\ ^{\circ}\mathrm{C}$, balanced against the cooling of the original water), calculate the final equilibrium temperature.写出完整的能量守恒方程（熔冰，再把化出的水从 $0\ ^{\circ}\mathrm{C}$ 升温，与原有水的降温相平衡），计算最终平衡温度。[5]
(d)State and explain how your final temperature would change if the container were not perfectly insulated and allowed heat in from a warm room.说明并解释若容器隔热不完美、允许温暖房间的热量进入，末温会如何变化。[2]
Q9HARDPaper 2star: Wien then Stefan-Boltzmann恒星：维恩接斯特藩-玻尔兹曼[11 marks]
A star behaves as a black body. Its spectrum peaks at a wavelength of $420\ \mathrm{nm}$, and its radius is $6.5\times10^{8}\ \mathrm{m}$. Take $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ and the Wien constant $2.9\times10^{-3}\ \mathrm{m\,K}$.某恒星表现为黑体。其光谱在波长 $420\ \mathrm{nm}$ 处达峰，半径为 $6.5\times10^{8}\ \mathrm{m}$。取 $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$，维恩常数 $2.9\times10^{-3}\ \mathrm{m\,K}$。
(a)Use Wien's displacement law to calculate the surface temperature of the star.用维恩位移定律计算恒星的表面温度。[2]
(b)Calculate the surface area of the star, treating it as a sphere.把恒星视为球体，计算其表面积。[2]
(c)Use the Stefan-Boltzmann law to calculate the luminosity (total power radiated) of the star.用斯特藩-玻尔兹曼定律计算恒星的光度（辐射总功率）。[3]
(d)The star is at a distance of $2.0\times10^{17}\ \mathrm{m}$ from Earth. Calculate the apparent brightness (power per unit area received) at Earth, stating the assumption you make about how the radiation spreads out.该恒星距地球 $2.0\times10^{17}\ \mathrm{m}$。计算地球处接收到的视亮度（单位面积接收功率），并写出你对辐射如何扩散所作的假设。[4]
Q10HARDPaper 2internal energy + radiated power内能与辐射功率[6 marks]
A lamp filament behaves as a black body of surface area $3.0\times10^{-5}\ \mathrm{m^{2}}$ and operates at a steady temperature of $2500\ \mathrm{K}$. Take $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$.某灯丝表现为表面积 $3.0\times10^{-5}\ \mathrm{m^{2}}$ 的黑体，在 $2500\ \mathrm{K}$ 的稳定温度下工作。取 $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$。
(a)Calculate the power radiated by the filament.计算灯丝辐射的功率。[2]
(b)The filament operates at a constant temperature, so its internal energy does not change with time. Explain how this is consistent with energy continuously entering the filament electrically and leaving it as radiation.灯丝在恒温下工作，故其内能不随时间变化。解释这与能量持续以电能进入、以辐射离开如何相容。[2]
(c)A second identical filament is run at $5000\ \mathrm{K}$. State the factor by which its radiated power exceeds that of the first filament, and justify your answer.第二根完全相同的灯丝在 $5000\ \mathrm{K}$ 下工作。写出其辐射功率超过第一根的倍数，并给出理由。[2]