PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 28 marks短结构题 · 可用计算器 · 28 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. Quote data-booklet relations explicitly and keep the geometry straight: a planet intercepts over its disc $\pi r^{2}$ but radiates from its whole surface $4 \pi r^{2}$. 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。明确引用数据手册中的关系式，并理清几何：行星以圆盘 $\pi r^{2}$ 截获、却从整个表面 $4 \pi r^{2}$ 辐射。数值答案保留适当的有效数字。
The Sun radiates with total power (luminosity) $P = 3.85\times 10^{26}\ \mathrm{W}$. Jupiter orbits at a mean distance $d = 7.78\times 10^{11}\ \mathrm{m}$ from the Sun and has radius $r = 6.99\times 10^{7}\ \mathrm{m}$.太阳以总功率（光度）$P = 3.85\times 10^{26}\ \mathrm{W}$ 辐射。木星绕太阳的平均距离为 $d = 7.78\times 10^{11}\ \mathrm{m}$，半径 $r = 6.99\times 10^{7}\ \mathrm{m}$。
(a)State the data-booklet relation for the intensity of radiation from a point source, and calculate the intensity of sunlight at Jupiter's orbit.写出数据手册中点源辐射强度的关系式，并计算木星轨道处的阳光强度。[3]
(b)Calculate the total power that Jupiter intercepts from the Sun, explaining which area you use.计算木星从太阳截获的总功率，并说明你用的是哪个面积。[3]
Q2MEDIUMPaper 1albedo: reflected vs absorbed反照率：反射与吸收[4 marks]
A planet intercepts solar power $P_{\text{in}} = 2.0\times 10^{17}\ \mathrm{W}$ over its cross-section. Its albedo is $\alpha = 0.35$.一颗行星以其横截面截获太阳功率 $P_{\text{in}} = 2.0\times 10^{17}\ \mathrm{W}$。其反照率为 $\alpha = 0.35$。
(a)Define albedo, and calculate the power reflected back to space.定义反照率，并计算反射回太空的功率。[2]
(b)Calculate the power absorbed by the planet, and state which of the two powers enters the planetary energy balance.计算行星吸收的功率，并说明这两个功率中哪一个进入行星能量平衡。[2]
A spherical planet of radius $r = 3.4\times 10^{6}\ \mathrm{m}$ has a uniform surface temperature $T = 210\ \mathrm{K}$ and behaves as a grey body of emissivity $e = 0.90$.一颗半径 $r = 3.4\times 10^{6}\ \mathrm{m}$ 的球形行星，表面温度均匀为 $T = 210\ \mathrm{K}$，表现为发射率 $e = 0.90$ 的灰体。
(a)Define emissivity, and state the value of emissivity for a perfect black body.定义发射率，并写出理想黑体的发射率值。[2]
(b)Calculate the total power radiated by the planet, using the surface area appropriate to emission.用适合发射的表面积，计算行星辐射的总功率。[3]
(c)State by what factor the radiated power would change if the temperature alone were doubled.写出若仅将温度加倍，辐射功率将变为原来的几倍。[1]
Q4HARDPaper 1equilibrium temperature平衡温度[6 marks]
An airless planet receives solar radiation of intensity $S = 1.0\times 10^{3}\ \mathrm{W\,m^{-2}}$ at the top of its (negligible) atmosphere and has albedo $\alpha = 0.25$.一颗无大气行星，在其（可忽略的）大气层顶接收强度 $S = 1.0\times 10^{3}\ \mathrm{W\,m^{-2}}$ 的太阳辐射，反照率 $\alpha = 0.25$。
(a)By equating absorbed power to radiated power, show that the equilibrium temperature is $T = \left[ (1 - \alpha) S / (4 e \sigma) \right]^{1/4}$, explaining where the factor of $4$ comes from.通过令吸收功率等于辐射功率，证明平衡温度为 $T = \left[ (1 - \alpha) S / (4 e \sigma) \right]^{1/4}$，并解释因子 $4$ 的来源。[3]
(b)Treating the planet as a black body ($e = 1$), calculate its equilibrium temperature.把行星当作黑体（$e = 1$），计算其平衡温度。[2]
(c)State and explain how the equilibrium temperature would change if the albedo were increased.说明并解释若反照率增大，平衡温度将如何变化。[1]
A vibrational mode of a carbon dioxide molecule absorbs infrared radiation of wavelength $\lambda = 14\ \mathrm{\mu m}$. Take $c = 3.0\times 10^{8}\ \mathrm{m\,s^{-1}}$ and $h = 6.63\times 10^{-34}\ \mathrm{J\,s}$.二氧化碳分子的某振动模式吸收波长 $\lambda = 14\ \mathrm{\mu m}$ 的红外辐射。取 $c = 3.0\times 10^{8}\ \mathrm{m\,s^{-1}}$、$h = 6.63\times 10^{-34}\ \mathrm{J\,s}$。
(a)Calculate the frequency and the energy of a photon of this radiation.计算该辐射光子的频率与能量。[3]
(b)Explain, in terms of resonance, why this molecule absorbs strongly at this particular wavelength.从共振的角度解释，为何该分子在这一特定波长强烈吸收。[2]
(c)State, with a reason, why nitrogen $\mathrm{N_2}$ is not a greenhouse gas.写出氮气 $\mathrm{N_2}$ 不是温室气体的原因。[1]
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 linearisation, reading of gradients, and correct handling of uncertainties. When a power law is given, choose axes that straighten the data. Quote uncertainties to one significant figure and round the value to match.这些题考查细致的线性化、对斜率的读取以及对不确定度的正确处理。给出幂律时，选取能把数据拉直的坐标轴。不确定度保留 1 位有效数字，并使数值的末位与之对齐。
Q6HARDPaper 1Bintensity vs distance: linearised graph + uncertainty强度与距离：线性化图像与不确定度[10 marks]
A space probe measures the intensity $I$ of sunlight at four distances $d$ from the Sun. The data are tabulated below, with $d$ in units of $10^{11}\ \mathrm{m}$.一艘空间探测器在距太阳四个距离 $d$ 处测量阳光强度 $I$。数据列于下表，$d$ 以 $10^{11}\ \mathrm{m}$ 为单位。
$d\ /\ 10^{11}\ \mathrm{m}$
$1.0$
$1.5$
$2.0$
$3.0$
$I\ /\ \mathrm{W\,m^{-2}}$
$3060$
$1360$
$766$
$340$
(a)Starting from $I = P / (4 \pi d^{2})$, state what should be plotted on each axis so that the data give a straight line through the origin, and state what the gradient represents.从 $I = P / (4 \pi d^{2})$ 出发，写出两坐标轴上各应作什么量，使数据给出过原点的直线，并说明斜率代表什么。[3]
(b)Using the points at $d = 1.0\times 10^{11}\ \mathrm{m}$ and $d = 3.0\times 10^{11}\ \mathrm{m}$, calculate the gradient of your linearised graph.用 $d = 1.0\times 10^{11}\ \mathrm{m}$ 与 $d = 3.0\times 10^{11}\ \mathrm{m}$ 两点，计算你线性化图像的斜率。[3]
(c)Hence determine the luminosity $P$ of the Sun from the gradient.由斜率求太阳的光度 $P$。[2]
(d)The intensity reading at $d = 1.0\times 10^{11}\ \mathrm{m}$ has an absolute uncertainty of $\pm 60\ \mathrm{W\,m^{-2}}$. Calculate its percentage uncertainty.$d = 1.0\times 10^{11}\ \mathrm{m}$ 处的强度读数有绝对不确定度 $\pm 60\ \mathrm{W\,m^{-2}}$。计算其百分比不确定度。[2]
Q7HARDPaper 1Balbedo from data · energy balance · uncertainty propagation由数据求反照率与能量平衡及不确定度传递[10 marks]
A satellite above a planet measures the incoming solar intensity as $S = 1.50\times 10^{3}\ \mathrm{W\,m^{-2}}$ and the reflected (scattered) intensity as $0.45\times 10^{3}\ \mathrm{W\,m^{-2}}$. The planet is modelled as a black body in radiative equilibrium.一颗行星上空的卫星测得入射太阳强度 $S = 1.50\times 10^{3}\ \mathrm{W\,m^{-2}}$，反射（散射）强度 $0.45\times 10^{3}\ \mathrm{W\,m^{-2}}$。该行星建模为处于辐射平衡的黑体。
(a)Calculate the albedo of the planet from these measurements.由这些测量计算行星的反照率。[2]
(b)Calculate the equilibrium temperature of the planet ($e = 1$).计算行星的平衡温度（$e = 1$）。[3]
(c)The solar intensity $S$ has a percentage uncertainty of $4\%$. Using $T \propto S^{1/4}$, determine the percentage uncertainty in the equilibrium temperature.太阳强度 $S$ 的百分比不确定度为 $4\%$。用 $T \propto S^{1/4}$，求平衡温度的百分比不确定度。[3]
(d)State one reason why the actual surface temperature of a real planet would differ from this calculated value.写出真实行星的实际地表温度与该计算值不同的一个原因。[2]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 32 marks长结构题 · 可用计算器 · 32 分
Extended Structured Problems长结构问题
Set up each problem with a clear statement of the energy-balance model. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer. Take $\sigma = 5.67\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ throughout.每题先清楚陈述能量平衡模型。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。全程取 $\sigma = 5.67\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$。
Q8HARDPaper 2full Earth energy balance with albedo含反照率的完整地球能量平衡[12 marks]
For the Earth take the solar constant $S = 1.36\times 10^{3}\ \mathrm{W\,m^{-2}}$, albedo $\alpha = 0.30$, and radius $r = 6.37\times 10^{6}\ \mathrm{m}$.对地球取太阳常数 $S = 1.36\times 10^{3}\ \mathrm{W\,m^{-2}}$、反照率 $\alpha = 0.30$、半径 $r = 6.37\times 10^{6}\ \mathrm{m}$。
(a)Calculate the total solar power intercepted by the Earth, stating the area used.计算地球截获的太阳总功率，并写出所用的面积。[2]
(b)Hence calculate the power absorbed by the Earth.由此计算地球吸收的功率。[2]
(c)Treating the Earth as a black body, calculate its equilibrium temperature.把地球当作黑体，计算其平衡温度。[3]
(d)The observed mean surface temperature is $288\ \mathrm{K}$. State the name of the effect that accounts for the difference, and calculate the emissivity a grey-body model would require to reproduce $288\ \mathrm{K}$.实测平均地表温度为 $288\ \mathrm{K}$。写出造成该差异的效应名称，并计算灰体模型重现 $288\ \mathrm{K}$ 所需的发射率。[4]
(e)State why the cross-sectional area $\pi r^{2}$ is used for the incoming power but the full surface area $4 \pi r^{2}$ is used for the outgoing power.说明为何入射功率用横截面积 $\pi r^{2}$、而外逸功率用整个表面积 $4 \pi r^{2}$。[1]
A vibrational mode of a methane molecule $\mathrm{CH_4}$ absorbs infrared radiation of wavelength $\lambda = 7.7\ \mathrm{\mu m}$. Take $c = 3.0\times 10^{8}\ \mathrm{m\,s^{-1}}$ and $h = 6.63\times 10^{-34}\ \mathrm{J\,s}$.甲烷分子 $\mathrm{CH_4}$ 的某振动模式吸收波长 $\lambda = 7.7\ \mathrm{\mu m}$ 的红外辐射。取 $c = 3.0\times 10^{8}\ \mathrm{m\,s^{-1}}$、$h = 6.63\times 10^{-34}\ \mathrm{J\,s}$。
(a)Calculate the energy of a photon at this wavelength.计算该波长光子的能量。[2]
(b)Explain, by reference to the Sun's spectrum and the Earth's spectrum, why greenhouse gases warm the surface even though they let sunlight through.结合太阳光谱与地球光谱，解释温室气体为何在让阳光透过的同时仍使地表变暖。[4]
(c)Distinguish between the natural greenhouse effect and the enhanced greenhouse effect.区分自然温室效应与增强温室效应。[2]
(d)A rise in greenhouse-gas concentration lowers the Earth's effective emissivity to space from $e_{1} = 0.61$ to $e_{2} = 0.58$. The grey-body equilibrium temperature was $288\ \mathrm{K}$. Using $T \propto e^{-1/4}$, calculate the new equilibrium temperature and the resulting rise.温室气体浓度上升使地球向太空的有效发射率从 $e_{1} = 0.61$ 降到 $e_{2} = 0.58$。原灰体平衡温度为 $288\ \mathrm{K}$。用 $T \propto e^{-1/4}$，计算新的平衡温度及由此产生的升温。[4]
Q10HARDPaper 2enhanced effect · energy imbalance · consequences增强效应、能量失衡与后果[8 marks]
A planet sits in radiative equilibrium with absorbed power $P_{\text{in}}$ equal to emitted power $P_{\text{out}}$. Its atmosphere then gains greenhouse gas, lowering its effective emissivity from $e_{1}$ to a smaller value $e_{2}$, while the temperature is momentarily unchanged at $T_{1}$.一颗行星处于辐射平衡，吸收功率 $P_{\text{in}}$ 等于发射功率 $P_{\text{out}}$。随后其大气增加温室气体，使有效发射率从 $e_{1}$ 降到更小的 $e_{2}$，而温度瞬间仍为 $T_{1}$。
(a)State, in terms of $P_{\text{in}}$ and $P_{\text{out}}$, the energy-balance situation in the instant just after the emissivity falls, and deduce whether the planet warms or cools.用 $P_{\text{in}}$ 与 $P_{\text{out}}$ 写出发射率下降后那一瞬间的能量平衡状况，并推断行星是升温还是降温。[3]
(b)Explain, using the fact that $P_{\text{out}} \propto T^{4}$, how a new equilibrium is reached, and state whether the new temperature is higher or lower than $T_{1}$.利用 $P_{\text{out}} \propto T^{4}$ 这一事实，解释新的平衡如何达到，并说明新温度比 $T_{1}$ 更高还是更低。[3]
(c)State two observable consequences for the Earth of the enhanced greenhouse effect.写出增强温室效应对地球的两个可观测后果。[2]