Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus B1.1 to B1.6考纲 B1.1 至 B1.6PHYSICS HL
Helium gas at $27\ ^{\circ}\mathrm{C}$. (a) convert to kelvin and state the effect of doubling absolute temperature on average molecular KE; (b) define internal energy and explain why warming raises it.氦气处于 $27\ ^{\circ}\mathrm{C}$。(a) 换算为开尔文，并说明绝对温度加倍对平均分子动能的影响；(b) 定义内能，并解释加热为何使其增大。
Use $T(\mathrm{K}) = \theta(^{\circ}\mathrm{C}) + 273$: $T = 27 + 273 = 300\ \mathrm{K}$. (A1)
The average random kinetic energy of the molecules is directly proportional to the absolute temperature, so doubling $T$ (to $600\ \mathrm{K}$) doubles the average molecular kinetic energy. (A1)
The internal energy $U$ is the total of the random kinetic energies of all the molecules plus the intermolecular potential energies: $U = \mathrm{KE}_{\text{molecular}} + \mathrm{PE}_{\text{molecular}}$. (A1)
Warming the gas with no phase change leaves the molecular PE unchanged but raises the molecular KE (the molecules move faster), so the internal energy increases through its kinetic part. (A1)
用 $T(\mathrm{K}) = \theta(^{\circ}\mathrm{C}) + 273$：$T = 27 + 273 = 300\ \mathrm{K}$。(A1)
分子的平均随机动能与绝对温度成正比，故 $T$ 加倍（至 $600\ \mathrm{K}$）使平均分子动能加倍。(A1)
内能 $U$ 是所有分子随机动能与分子间势能的总和：$U = \mathrm{KE}_{\text{molecular}} + \mathrm{PE}_{\text{molecular}}$。(A1)
无相变地加热气体时分子势能不变，但分子动能增大（分子运动更快），故内能通过其动能部分增大。(A1)
$50\ \mathrm{W}$ heater in a $0.40\ \mathrm{kg}$ block; $90\ \mathrm{s}$ gives $\Delta T = 9.0\ \mathrm{K}$. (a) energy supplied and specific heat capacity; (b) over- or under-estimate when heat is lost, with justification.$50\ \mathrm{W}$ 加热器嵌入 $0.40\ \mathrm{kg}$ 金属块；$90\ \mathrm{s}$ 使 $\Delta T = 9.0\ \mathrm{K}$。(a) 供能与比热容；(b) 有热损失时偏大还是偏小，并说明理由。
The energy supplied is $Q = Pt = (50)(90) = 4500\ \mathrm{J}$. (M1)
Apply $Q = mc\Delta T$ and solve for $c$: (M1)
$$ c = \frac{Q}{m\,\Delta T} = \frac{4500}{(0.40)(9.0)} = 1250\ \mathrm{J\,kg^{-1}\,K^{-1}}. $$(A1)
If energy leaks to the surroundings, less than $4500\ \mathrm{J}$ actually goes into the block, yet the calculation assumes all of it does. (R1)
Dividing the full input energy by the measured $\Delta T$ therefore gives a value of $c$ that is too large, so the experimental value is an overestimate. (A1)
供给的能量为 $Q = Pt = (50)(90) = 4500\ \mathrm{J}$。(M1)
用 $Q = mc\Delta T$ 解出 $c$：(M1)
$$ c = \frac{Q}{m\,\Delta T} = \frac{4500}{(0.40)(9.0)} = 1250\ \mathrm{J\,kg^{-1}\,K^{-1}}. $$(A1)
若能量泄漏到环境，实际进入金属块的能量不足 $4500\ \mathrm{J}$，而计算却假定全部进入。(R1)
用全部输入能量除以测得的 $\Delta T$，故得到的 $c$ 偏大，即实验值偏高。(A1)
Vacuum flask: double wall with a vacuum gap, silvered surfaces. (a) why conduction and convection fail across the gap; (b) how silvering helps and which mechanism; (c) which mechanism crosses a vacuum.真空保温瓶：双层壁带真空间隙，表面镀银。(a) 为何传导与对流无法越过间隙；(b) 镀银如何帮助及涉及哪种机制；(c) 哪种机制能穿越真空。
Conduction transfers energy by particle-to-particle interaction, and convection transfers energy by the bulk movement of a fluid; both require matter. (M1)
A vacuum contains almost no particles, so there is no chain of colliding particles to conduct energy across the gap. (A1)
With no fluid in the gap, no convection current of rising warm fluid and sinking cool fluid can form either. (A1)
The relevant mechanism is thermal radiation, which crosses the vacuum as electromagnetic (infrared) waves. (A1)
A shiny silvered surface has a low emissivity, so it is a poor emitter and a good reflector of infrared: it radiates little energy outward and reflects radiation back toward the drink, reducing radiative loss. (A1)
Radiation is the only mechanism that needs no medium and so can transfer energy through a vacuum. (B1)
热传导靠粒子间相互作用传能，对流靠流体整体移动传能；两者都需要物质。(M1)
真空中几乎没有粒子，故没有可碰撞的粒子链来跨越间隙传导能量。(A1)
间隙中没有流体，也无法形成受热流体上升、较冷流体下沉的对流循环。(A1)
所涉机制是热辐射，它以电磁（红外）波形式穿越真空。(A1)
光亮的镀银表面发射率低，故是不良发射体、良好反射体：向外辐射很少能量，并把辐射反射回饮料一侧，从而减少辐射损失。(A1)
辐射是唯一无需介质、因而能在真空中传能的机制。(B1)
Black body whose absolute temperature is doubled. (a) factor by which radiated power rises; (b) factor and direction of the peak-wavelength shift; (c) show a $300\ \mathrm{K}$ surface peaks in the infrared.黑体的绝对温度加倍。(a) 辐射功率增大的倍数；(b) 峰值波长变化的倍数与方向；(c) 证明 $300\ \mathrm{K}$ 表面在红外达峰。
Power scales as the fourth power of absolute temperature, with $A$ fixed: $\dfrac{L_2}{L_1} = \left(\dfrac{2T}{T}\right)^{4}$. (M1)
$$ \frac{L_2}{L_1} = 2^{4} = 16. $$The radiated power increases by a factor of $16$. (A1)
By $\lambda_{\max} T = $ constant, $\lambda_{\max} \propto 1/T$, so doubling $T$ multiplies $\lambda_{\max}$ by $\tfrac{1}{2}$. (M1)
The peak wavelength halves, so the peak moves to shorter wavelengths (toward the blue / ultraviolet end). (A1)
Apply Wien's law at $T = 300\ \mathrm{K}$: (M1)
$$ \lambda_{\max} = \frac{2.9\times10^{-3}}{300} \approx 9.7\times10^{-6}\ \mathrm{m} = 9.7\ \mathrm{\mu m}. $$This is far longer than the visible band ($400$ to $700\ \mathrm{nm}$), so the peak lies well into the infrared. (A1)
$A$ 固定时功率按绝对温度的四次方变化：$\dfrac{L_2}{L_1} = \left(\dfrac{2T}{T}\right)^{4}$。(M1)
$$ \frac{L_2}{L_1} = 2^{4} = 16. $$辐射功率增大为原来的 $16$ 倍。(A1)
由 $\lambda_{\max} T = $ 常数，$\lambda_{\max} \propto 1/T$，故 $T$ 加倍使 $\lambda_{\max}$ 乘以 $\tfrac{1}{2}$。(M1)
峰值波长减半，故峰值移向更短波长（朝蓝 / 紫外一端）。(A1)
在 $T = 300\ \mathrm{K}$ 应用维恩定律：(M1)
$$ \lambda_{\max} = \frac{2.9\times10^{-3}}{300} \approx 9.7\times10^{-6}\ \mathrm{m} = 9.7\ \mathrm{\mu m}. $$这远长于可见光波段（$400$ 至 $700\ \mathrm{nm}$），故峰值深入红外。(A1)
$0.10\ \mathrm{kg}$ solid heated at $40\ \mathrm{W}$; temperatures at $t = 0, 50, 100, 150, 200\ \mathrm{s}$ are $20, 60, 60, 60, 100\ ^{\circ}\mathrm{C}$. (a) melting interval, melting point, why $T$ is constant; (b) $c$ of the solid from the first slope; (c) latent heat of fusion from the plateau; (d) percentage uncertainty in power and its propagation.$0.10\ \mathrm{kg}$ 固体以 $40\ \mathrm{W}$ 加热；$t = 0, 50, 100, 150, 200\ \mathrm{s}$ 处温度为 $20, 60, 60, 60, 100\ ^{\circ}\mathrm{C}$。(a) 熔化区间、熔点、温度恒定的原因；(b) 由首段斜率求固体比热容；(c) 由平台求熔化潜热；(d) 功率的百分比不确定度及其传递。
The temperature is flat at $60\ ^{\circ}\mathrm{C}$ from $t = 50\ \mathrm{s}$ to $t = 150\ \mathrm{s}$, so that plateau is the phase change: the substance melts over this interval. (A1)
The melting point is the plateau temperature, $60\ ^{\circ}\mathrm{C}$. (A1)
During melting the supplied energy goes entirely into breaking intermolecular bonds, raising the molecular potential energy, not the molecular kinetic energy. Since temperature tracks only the kinetic energy, it stays constant. (R1)
Over the first $50\ \mathrm{s}$ the energy supplied is $Q = Pt = (40)(50) = 2000\ \mathrm{J}$, and the temperature rises by $\Delta T = 60 - 20 = 40\ \mathrm{K}$. (M1)
Apply $Q = mc\Delta T$: (M1)
$$ c = \frac{Q}{m\,\Delta T} = \frac{2000}{(0.10)(40)} = 500\ \mathrm{J\,kg^{-1}\,K^{-1}}. $$(A1)
The plateau lasts from $50\ \mathrm{s}$ to $150\ \mathrm{s}$, a duration of $100\ \mathrm{s}$, so $Q = Pt = (40)(100) = 4000\ \mathrm{J}$ all goes into melting. Apply $Q = mL_{f}$: (M1)
$$ L_{f} = \frac{Q}{m} = \frac{4000}{0.10} = 4.0\times10^{4}\ \mathrm{J\,kg^{-1}}. $$(A1)
Percentage uncertainty in power: $\dfrac{2}{40}\times 100\% = 5\%$. (A1)
Since $L_{f} = Pt/m$ is directly proportional to $P$, the $5\%$ uncertainty in $P$ carries straight through, giving a $5\%$ uncertainty in $L_{f}$ (with $t$ and $m$ assumed precise). (R1)
温度从 $t = 50\ \mathrm{s}$ 到 $t = 150\ \mathrm{s}$ 在 $60\ ^{\circ}\mathrm{C}$ 处保持水平，故该平台即相变：物质在此区间熔化。(A1)
熔点即平台温度 $60\ ^{\circ}\mathrm{C}$。(A1)
熔化期间供给的能量全部用于断裂分子间键、提升分子势能，而非分子动能。由于温度只反映动能，故保持不变。(R1)
前 $50\ \mathrm{s}$ 内供能 $Q = Pt = (40)(50) = 2000\ \mathrm{J}$，温升 $\Delta T = 60 - 20 = 40\ \mathrm{K}$。(M1)
用 $Q = mc\Delta T$：(M1)
$$ c = \frac{Q}{m\,\Delta T} = \frac{2000}{(0.10)(40)} = 500\ \mathrm{J\,kg^{-1}\,K^{-1}}. $$(A1)
平台从 $50\ \mathrm{s}$ 持续到 $150\ \mathrm{s}$，历时 $100\ \mathrm{s}$，故 $Q = Pt = (40)(100) = 4000\ \mathrm{J}$ 全部用于熔化。用 $Q = mL_{f}$：(M1)
$$ L_{f} = \frac{Q}{m} = \frac{4000}{0.10} = 4.0\times10^{4}\ \mathrm{J\,kg^{-1}}. $$(A1)
功率的百分比不确定度：$\dfrac{2}{40}\times 100\% = 5\%$。(A1)
由于 $L_{f} = Pt/m$ 与 $P$ 成正比，$P$ 的 $5\%$ 不确定度直接传过去，使 $L_{f}$ 也有 $5\%$ 不确定度（设 $t$、$m$ 精确）。(R1)
$0.50\ \mathrm{kg}$ water, $90\ \mathrm{W}$ heater; temperature-time graph straight with gradient $0.040\ \mathrm{K\,s^{-1}}$. (a) show gradient $= P/(mc)$; (b) experimental $c$; (c) percentage and absolute uncertainty from gradient $\pm 0.002$; (d) why an imperfect beaker gives a value too high.$0.50\ \mathrm{kg}$ 水，$90\ \mathrm{W}$ 加热器；温度-时间图为直线，斜率 $0.040\ \mathrm{K\,s^{-1}}$。(a) 证明斜率 $= P/(mc)$；(b) 实测 $c$；(c) 由斜率 $\pm 0.002$ 求百分比与绝对不确定度；(d) 隔热不完美为何使值偏高。
The heater delivers energy at a steady rate, so in time $t$ it supplies $Q = Pt$, and this energy heats the water: $Pt = mc\,\Delta T$. (M1)
Rearrange for the temperature change as a function of time: $\Delta T = \dfrac{P}{mc}\,t$. (M1)
This has the straight-line form $\Delta T = (\text{gradient})\times t$, so the gradient of the temperature-time graph is $\dfrac{P}{mc}$. (A1)
Rearrange gradient $= \dfrac{P}{mc}$ for $c$: (M1)
$$ c = \frac{P}{m\times\text{gradient}} = \frac{90}{(0.50)(0.040)} = 4500\ \mathrm{J\,kg^{-1}\,K^{-1}}. $$(A1)
With $P$ and $m$ precise, $c$ depends only on the gradient, and $c \propto 1/\text{gradient}$, so the fractional uncertainty in $c$ equals that in the gradient: (M1)
$$ \frac{\Delta c}{c} = \frac{0.002}{0.040} = 0.05 = 5\%. $$(A1)
Absolute uncertainty: $\Delta c = 0.05\times 4500 = 200\ \mathrm{J\,kg^{-1}\,K^{-1}}$, so $c = (4500 \pm 200)\ \mathrm{J\,kg^{-1}\,K^{-1}}$. (A1)
If the beaker is not perfectly insulated, some energy leaks to the surroundings, so the water heats up more slowly and the measured gradient is smaller than the ideal $P/(mc)$. (B1)
Because $c \propto 1/\text{gradient}$, a smaller gradient gives a larger computed $c$, so the value comes out too high. (R1)
加热器以恒定速率供能，故时间 $t$ 内供给 $Q = Pt$，此能量加热水：$Pt = mc\,\Delta T$。(M1)
解出温升随时间的关系：$\Delta T = \dfrac{P}{mc}\,t$。(M1)
此式形如 $\Delta T = (\text{斜率})\times t$，故温度-时间图的斜率为 $\dfrac{P}{mc}$。(A1)
由斜率 $= \dfrac{P}{mc}$ 解出 $c$：(M1)
$$ c = \frac{P}{m\times\text{斜率}} = \frac{90}{(0.50)(0.040)} = 4500\ \mathrm{J\,kg^{-1}\,K^{-1}}. $$(A1)
$P$、$m$ 精确时 $c$ 只依赖斜率，且 $c \propto 1/\text{斜率}$，故 $c$ 的分数不确定度等于斜率的：(M1)
$$ \frac{\Delta c}{c} = \frac{0.002}{0.040} = 0.05 = 5\%. $$(A1)
绝对不确定度：$\Delta c = 0.05\times 4500 = 200\ \mathrm{J\,kg^{-1}\,K^{-1}}$，故 $c = (4500 \pm 200)\ \mathrm{J\,kg^{-1}\,K^{-1}}$。(A1)
若烧杯隔热不完美，部分能量泄漏到环境，故水升温更慢，测得斜率小于理想的 $P/(mc)$。(B1)
因 $c \propto 1/\text{斜率}$，斜率偏小使算得的 $c$ 偏大，故测值偏高。(R1)
$0.20\ \mathrm{kg}$ copper at $95\ ^{\circ}\mathrm{C}$ into $0.25\ \mathrm{kg}$ water at $18\ ^{\circ}\mathrm{C}$, insulated cup; $c_{\text{Cu}} = 385$, $c_{\text{water}} = 4180$. (a) state conservation of energy and the balance equation; (b) final temperature; (c) energy transferred; (d) why $T_f$ is near the water's temperature.$0.20\ \mathrm{kg}$、$95\ ^{\circ}\mathrm{C}$ 的铜投入 $0.25\ \mathrm{kg}$、$18\ ^{\circ}\mathrm{C}$ 的水，隔热杯；$c_{\text{Cu}} = 385$、$c_{\text{water}} = 4180$。(a) 陈述能量守恒与平衡方程；(b) 末温；(c) 传递的能量；(d) 为何末温接近水温。
In an insulated cup no energy leaves the system, so by conservation of energy the thermal energy lost by the cooling copper equals the thermal energy gained by the warming water. (R1)
Writing $T_{f}$ for the final temperature: $m_{\text{Cu}} c_{\text{Cu}} (95 - T_{f}) = m_{w} c_{w} (T_{f} - 18)$. (A1)
Compute each heat-capacity product: $m_{\text{Cu}} c_{\text{Cu}} = (0.20)(385) = 77.0\ \mathrm{J\,K^{-1}}$ and $m_{w} c_{w} = (0.25)(4180) = 1045\ \mathrm{J\,K^{-1}}$. (M1)
Substitute: $77.0(95 - T_{f}) = 1045(T_{f} - 18)$. (M1)
$$ 7315 - 77.0\,T_{f} = 1045\,T_{f} - 18810. $$Collect terms: $26125 = 1122\,T_{f}$. (M1)
$$ T_{f} = \frac{26125}{1122} \approx 23.3\ ^{\circ}\mathrm{C}. $$(A1)
Use the copper's loss (equivalently the water's gain): $Q = m_{\text{Cu}} c_{\text{Cu}} (95 - T_{f}) = 77.0\,(95 - 23.3)$. (M1)
$$ Q = 77.0\times 71.7 \approx 5.5\times10^{3}\ \mathrm{J}. $$(A1)
The final temperature lands at the heat-capacity-weighted average of the two starting temperatures, weighted by each body's $mc$. (R1)
The water's $mc = 1045\ \mathrm{J\,K^{-1}}$ is about $14$ times the copper's $77.0\ \mathrm{J\,K^{-1}}$, because the water has both more mass and a far larger specific heat capacity, so the mixture settles close to the water's $18\ ^{\circ}\mathrm{C}$. (A1)
隔热杯中没有能量离开系统，故由能量守恒，降温的铜放出的热能等于升温的水吸收的热能。(R1)
设末温为 $T_{f}$：$m_{\text{Cu}} c_{\text{Cu}} (95 - T_{f}) = m_{w} c_{w} (T_{f} - 18)$。(A1)
先算各热容乘积：$m_{\text{Cu}} c_{\text{Cu}} = (0.20)(385) = 77.0\ \mathrm{J\,K^{-1}}$，$m_{w} c_{w} = (0.25)(4180) = 1045\ \mathrm{J\,K^{-1}}$。(M1)
代入：$77.0(95 - T_{f}) = 1045(T_{f} - 18)$。(M1)
$$ 7315 - 77.0\,T_{f} = 1045\,T_{f} - 18810. $$合并同类项：$26125 = 1122\,T_{f}$。(M1)
$$ T_{f} = \frac{26125}{1122} \approx 23.3\ ^{\circ}\mathrm{C}. $$(A1)
用铜的放热（等于水的吸热）：$Q = m_{\text{Cu}} c_{\text{Cu}} (95 - T_{f}) = 77.0\,(95 - 23.3)$。(M1)
$$ Q = 77.0\times 71.7 \approx 5.5\times10^{3}\ \mathrm{J}. $$(A1)
末温落在两个初温按各自 $mc$ 加权的平均处。(R1)
水的 $mc = 1045\ \mathrm{J\,K^{-1}}$ 约为铜的 $77.0\ \mathrm{J\,K^{-1}}$ 的 $14$ 倍，因为水质量更大、比热容也大得多，故混合结果靠近水的 $18\ ^{\circ}\mathrm{C}$。(A1)
$0.030\ \mathrm{kg}$ ice at $0\ ^{\circ}\mathrm{C}$ added to $0.300\ \mathrm{kg}$ water at $25\ ^{\circ}\mathrm{C}$, well insulated; $L_{f} = 3.34\times10^{5}$, $c_{w} = 4180$. (a) energy to melt the ice; (b) max energy the water can release and show all ice melts; (c) final temperature; (d) effect of heat leaking in from a warm room.$0.030\ \mathrm{kg}$、$0\ ^{\circ}\mathrm{C}$ 的冰加入 $0.300\ \mathrm{kg}$、$25\ ^{\circ}\mathrm{C}$ 的水，隔热良好；$L_{f} = 3.34\times10^{5}$、$c_{w} = 4180$。(a) 熔冰所需能量；(b) 水可释放的最大能量并证明冰全熔；(c) 末温；(d) 温暖房间漏入热量的影响。
Melting at constant $0\ ^{\circ}\mathrm{C}$ uses $Q = m_{i} L_{f}$: (M1)
$$ Q = (0.030)(3.34\times10^{5}) = 1.0\times10^{4}\ \mathrm{J}. $$(A1)
The most the warm water can give up is by cooling all the way to $0\ ^{\circ}\mathrm{C}$: $Q_{\max} = m_{w} c_{w}\,\Delta T = (0.300)(4180)(25)$. (M1)
$$ Q_{\max} = 3.135\times10^{4}\ \mathrm{J} \approx 3.1\times10^{4}\ \mathrm{J}. $$(A1)
Since $3.1\times10^{4}\ \mathrm{J}$ available exceeds the $1.0\times10^{4}\ \mathrm{J}$ needed to melt the ice, all the ice melts and energy is still left over, so the final temperature lies above $0\ ^{\circ}\mathrm{C}$. (R1)
Energy conservation: the warm water cools from $25\ ^{\circ}\mathrm{C}$ to $T_{f}$, supplying both the melting of the ice and the warming of the melt-water from $0\ ^{\circ}\mathrm{C}$ to $T_{f}$. (M1)
$$ m_{w} c_{w}(25 - T_{f}) = m_{i} L_{f} + m_{i} c_{w}(T_{f} - 0). $$(M1 for the complete balance)
Substitute numbers: $1254(25 - T_{f}) = 1.002\times10^{4} + 125.4\,T_{f}$. (M1)
$$ 31350 - 1254\,T_{f} = 10020 + 125.4\,T_{f}. $$ $$ 21330 = 1379.4\,T_{f}. $$(A1)
$$ T_{f} = \frac{21330}{1379.4} \approx 15.5\ ^{\circ}\mathrm{C}. $$(A1)
If heat leaks in from a warmer room, the system gains extra energy beyond what the original water supplied. (R1)
That extra energy goes into the mixture after the ice has melted, raising the final equilibrium temperature above the ideal $15.5\ ^{\circ}\mathrm{C}$. (A1)
在恒定 $0\ ^{\circ}\mathrm{C}$ 熔化用 $Q = m_{i} L_{f}$：(M1)
$$ Q = (0.030)(3.34\times10^{5}) = 1.0\times10^{4}\ \mathrm{J}. $$(A1)
温水最多能放出的能量是冷却到 $0\ ^{\circ}\mathrm{C}$：$Q_{\max} = m_{w} c_{w}\,\Delta T = (0.300)(4180)(25)$。(M1)
$$ Q_{\max} = 3.135\times10^{4}\ \mathrm{J} \approx 3.1\times10^{4}\ \mathrm{J}. $$(A1)
由于可供的 $3.1\times10^{4}\ \mathrm{J}$ 超过熔冰所需的 $1.0\times10^{4}\ \mathrm{J}$，冰全部熔化且尚有能量剩余，故末温高于 $0\ ^{\circ}\mathrm{C}$。(R1)
能量守恒：温水从 $25\ ^{\circ}\mathrm{C}$ 冷却到 $T_{f}$，既供熔冰又把化出的水从 $0\ ^{\circ}\mathrm{C}$ 升温到 $T_{f}$。(M1)
$$ m_{w} c_{w}(25 - T_{f}) = m_{i} L_{f} + m_{i} c_{w}(T_{f} - 0). $$(完整平衡式得 M1)
代入数值：$1254(25 - T_{f}) = 1.002\times10^{4} + 125.4\,T_{f}$。(M1)
$$ 31350 - 1254\,T_{f} = 10020 + 125.4\,T_{f}. $$ $$ 21330 = 1379.4\,T_{f}. $$(A1)
$$ T_{f} = \frac{21330}{1379.4} \approx 15.5\ ^{\circ}\mathrm{C}. $$(A1)
若热量从较暖房间漏入，系统获得超出原有水所供的额外能量。(R1)
这部分额外能量在冰熔化后进入混合物，使末温高于理想的 $15.5\ ^{\circ}\mathrm{C}$。(A1)
Black-body star: spectral peak $420\ \mathrm{nm}$, radius $6.5\times10^{8}\ \mathrm{m}$. (a) surface temperature from Wien; (b) surface area; (c) luminosity from Stefan-Boltzmann; (d) apparent brightness at $d = 2.0\times10^{17}\ \mathrm{m}$ with the spreading assumption stated.黑体恒星：光谱峰值 $420\ \mathrm{nm}$，半径 $6.5\times10^{8}\ \mathrm{m}$。(a) 由维恩求表面温度；(b) 表面积；(c) 由斯特藩-玻尔兹曼求光度；(d) 距离 $d = 2.0\times10^{17}\ \mathrm{m}$ 处的视亮度并写出扩散假设。
Convert the peak to metres, $\lambda_{\max} = 420\ \mathrm{nm} = 4.20\times10^{-7}\ \mathrm{m}$, then use $\lambda_{\max} T = 2.9\times10^{-3}\ \mathrm{m\,K}$: (M1)
$$ T = \frac{2.9\times10^{-3}}{4.20\times10^{-7}} \approx 6900\ \mathrm{K}. $$(A1)
Treat the star as a sphere of radius $r = 6.5\times10^{8}\ \mathrm{m}$, $A = 4\pi r^{2}$: (M1)
$$ A = 4\pi (6.5\times10^{8})^{2} \approx 5.3\times10^{18}\ \mathrm{m^{2}}. $$(A1)
Use $L = \sigma A T^{4}$ with $T$ in kelvin: (M1)
$$ L = (5.67\times10^{-8})(5.3\times10^{18})(6900)^{4}. $$(M1 for substitution; $(6900)^{4} \approx 2.27\times10^{15}$)
$$ L \approx 6.8\times10^{26}\ \mathrm{W}. $$(A1)
Assume the radiation spreads uniformly over a sphere of radius $d$, so by the time it reaches Earth it is diluted over an area $4\pi d^{2}$. (M1)
Apparent brightness is power per unit area received: $b = \dfrac{L}{4\pi d^{2}}$. (M1)
$$ b = \frac{6.8\times10^{26}}{4\pi (2.0\times10^{17})^{2}} = \frac{6.8\times10^{26}}{5.03\times10^{35}}. $$(A1)
$$ b \approx 1.4\times10^{-9}\ \mathrm{W\,m^{-2}}. $$(A1)
把峰值换算成米，$\lambda_{\max} = 420\ \mathrm{nm} = 4.20\times10^{-7}\ \mathrm{m}$，再用 $\lambda_{\max} T = 2.9\times10^{-3}\ \mathrm{m\,K}$：(M1)
$$ T = \frac{2.9\times10^{-3}}{4.20\times10^{-7}} \approx 6900\ \mathrm{K}. $$(A1)
把恒星视为半径 $r = 6.5\times10^{8}\ \mathrm{m}$ 的球，$A = 4\pi r^{2}$：(M1)
$$ A = 4\pi (6.5\times10^{8})^{2} \approx 5.3\times10^{18}\ \mathrm{m^{2}}. $$(A1)
用 $L = \sigma A T^{4}$，$T$ 取开尔文：(M1)
$$ L = (5.67\times10^{-8})(5.3\times10^{18})(6900)^{4}. $$(代入得 M1；$(6900)^{4} \approx 2.27\times10^{15}$)
$$ L \approx 6.8\times10^{26}\ \mathrm{W}. $$(A1)
假设辐射均匀扩散到半径 $d$ 的球面上，故到达地球时已稀释到面积 $4\pi d^{2}$ 上。(M1)
视亮度即单位面积接收功率：$b = \dfrac{L}{4\pi d^{2}}$。(M1)
$$ b = \frac{6.8\times10^{26}}{4\pi (2.0\times10^{17})^{2}} = \frac{6.8\times10^{26}}{5.03\times10^{35}}. $$(A1)
$$ b \approx 1.4\times10^{-9}\ \mathrm{W\,m^{-2}}. $$(A1)
Black-body filament, area $3.0\times10^{-5}\ \mathrm{m^{2}}$, steady $2500\ \mathrm{K}$. (a) power radiated; (b) why constant internal energy is consistent with energy flowing in and out; (c) factor by which an identical filament at $5000\ \mathrm{K}$ radiates more, justified.黑体灯丝，面积 $3.0\times10^{-5}\ \mathrm{m^{2}}$，稳定 $2500\ \mathrm{K}$。(a) 辐射功率；(b) 为何恒定内能与能量进出相容；(c) 同样灯丝在 $5000\ \mathrm{K}$ 辐射更多的倍数及理由。
Use $L = \sigma A T^{4}$ with $T = 2500\ \mathrm{K}$: (M1)
$$ P = (5.67\times10^{-8})(3.0\times10^{-5})(2500)^{4} \approx 66\ \mathrm{W}. $$(A1)
The internal energy of the filament depends on its temperature, which is steady, so $U$ does not change with time. (A1)
This is consistent with energy flowing through the filament because the electrical power delivered to it equals the power it radiates away: the inflow and outflow are balanced, so there is no net change in internal energy. (R1)
The temperature ratio is $\dfrac{5000}{2500} = 2$, and power scales as $T^{4}$: (M1)
$$ \frac{P_2}{P_1} = 2^{4} = 16. $$The hotter filament radiates $16$ times the power of the first. (A1)
用 $L = \sigma A T^{4}$，$T = 2500\ \mathrm{K}$：(M1)
$$ P = (5.67\times10^{-8})(3.0\times10^{-5})(2500)^{4} \approx 66\ \mathrm{W}. $$(A1)
灯丝的内能取决于其温度，而温度稳定，故 $U$ 不随时间变化。(A1)
这与能量流经灯丝相容，因为供给它的电功率等于它辐射出去的功率：进出平衡，故内能无净变化。(R1)
温度比为 $\dfrac{5000}{2500} = 2$，功率按 $T^{4}$ 变化：(M1)
$$ \frac{P_2}{P_1} = 2^{4} = 16. $$更热的灯丝辐射功率为第一根的 $16$ 倍。(A1)