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Solutions详解

Thermodynamics and Heat · Solutions热力学与热 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 BC卑 AP-style MCQAP 风格选择题 AP-feeder FRQAP 衔接简答题 ON Provincial-style安大略省考风格 BC Provincial-style卑诗省考风格 Honors荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解AP MCQ + ON/BC short answer · 18 marksAP 选择题 + 安/卑省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Temperature vs thermal energy温度与热能 · HS-PS3-1 [3 marks][3 分]

Cup: $0.20$ kg water at $100\,^\circ\text{C}$. Bathtub: $80$ kg water at $40\,^\circ\text{C}$. Which statement is correct?杯子：$0.20$ kg 水，$100\,^\circ\text{C}$。浴缸：$80$ kg 水，$40\,^\circ\text{C}$。哪句话正确？

Answer:答案：  (A)

(a) Separate temperature from total thermal energy区分温度与总热能 M1·A1·A1

Temperature measures the average random kinetic energy per particle, so the cup ($100\,^\circ\text{C}$) is hotter. Thermal energy is the total internal energy of all particles, which scales with mass. Compare the heat-capacity products $mc$:温度衡量每个粒子的平均随机动能，故杯子（$100\,^\circ\text{C}$）更热。热能是所有粒子的总内能，与质量成正比。比较热容乘积 $mc$： $$ m_{\text{cup}} = 0.20\,\text{kg} \quad\text{vs}\quad m_{\text{tub}} = 80\,\text{kg} \;\; (400\times \text{ more mass}). $$ The bathtub holds $400$ times the mass at a temperature still well above absolute zero, so its total thermal energy is far greater. Option (A): cup has the higher temperature, bathtub has the greater total thermal energy.浴缸的质量是 $400$ 倍，且温度仍远高于绝对零度，故其总热能远大于杯子。选 (A)：杯子温度更高，浴缸总热能更多。
Why the distractors fail.干扰项分析。
(B): conflates "hotter" with "more energy"; the cup's tiny mass makes its total energy small.把"更热"误等于"能量更多"；杯子质量极小，其总能量很小。
(C): temperature and thermal energy are distinct quantities, not the same.温度与热能是不同的量，并非相同。
(D): more water does not mean higher temperature; the bathtub is measured at $40\,^\circ\text{C}$.水多不代表温度高；浴缸实测为 $40\,^\circ\text{C}$。
Temperature is intensive (per particle); thermal energy is extensive (scales with amount).温度是强度量（按粒子计）；热能是广延量（随物质量增加）。 A single spark at $1000\,^\circ\text{C}$ has a very high temperature but almost no thermal energy because its mass is minute; a swimming pool at $25\,^\circ\text{C}$ has enormous thermal energy because of its mass. Temperature tells you the direction heat will flow (always hot to cold); thermal energy tells you how much energy is available to transfer. Heat $Q$ is the energy actually in transit between the two. Keeping these three ideas distinct is the foundation of the whole unit and is exactly what NGSS HS-PS3-1 energy bookkeeping requires.$1000\,^\circ\text{C}$ 的单个火星温度极高，但因质量极小几乎没有热能；$25\,^\circ\text{C}$ 的游泳池因质量巨大而拥有庞大的热能。温度决定热量流动的方向（永远从热到冷）；热能决定可供传递的能量有多少。热量 $Q$ 是两者之间实际传递中的能量。把这三个概念分清，是整个单元的基础，也正是 NGSS HS-PS3-1 能量核算所要求的。
Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §2 Heat transfer modes热传递模式 · HS-PS3-2 [3 marks][3 分]

Solar energy crosses $\approx 150$ million km of near-vacuum to reach Earth. Which heat-transfer mode?太阳能穿越约 $1.5$ 亿 km 近真空到达地球。哪种热传递模式？

Answer:答案：  (C)  Radiation辐射

(a) Match the mechanism to the medium将机制与介质对应 M1·A1·A1

Conduction needs direct particle contact; convection needs a moving fluid. Space between the Sun and Earth is near-vacuum, so neither can operate. Only radiation (electromagnetic waves: infrared and visible light) travels through a vacuum. Option (C).传导需要粒子直接接触；对流需要流动的流体。日地之间是近真空，故两者都无法进行。只有辐射（电磁波：红外线和可见光）能穿越真空。选 (C)。
Why the distractors fail.干扰项分析。
(A): conduction requires touching particles, absent in a vacuum.传导需要粒子接触，真空中没有。
(B): convection requires a fluid to circulate, absent in a vacuum.对流需要流体循环，真空中没有。
(D): combines two mechanisms that both fail in a vacuum.把两种在真空中都无效的机制组合在一起。
Only radiation needs no medium; conduction and convection both require matter.只有辐射不需要介质；传导与对流都需要物质。 The quick test: "Is there matter between source and receiver?" If the path is empty space, the answer must be radiation. Every object above $0$ K emits thermal radiation, and hotter objects emit more (the Stefan-Boltzmann law $P = \sigma A T^4$ makes radiated power grow with the fourth power of absolute temperature). This is why a thermos flask uses a silvered vacuum gap: the vacuum blocks conduction and convection, and the reflective coating minimises radiation, so the contents stay hot for hours.快速判断："源与接收者之间有物质吗？"若路径是空旷的空间，答案必为辐射。每个温度高于 $0$ K 的物体都发射热辐射，温度越高发射越多（斯特藩-玻尔兹曼定律 $P = \sigma A T^4$ 使辐射功率随绝对温度的四次方增长）。这就是保温瓶采用镀银真空夹层的原因：真空阻断传导与对流，反射涂层最大限度减少辐射，故内容物能保温数小时。
Q3EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §1 Kelvin scale开尔文量程 · SPH3U D2 [4 marks][4 分]

Gas cooled from $27\,^\circ\text{C}$ to $-73\,^\circ\text{C}$. (a) Convert to kelvin. (b) Temperature change in K and $^\circ$C. (c) Why kelvin for the gas law?气体从 $27\,^\circ\text{C}$ 冷却至 $-73\,^\circ\text{C}$。(a) 换算为开尔文。(b) 温度变化（K 与 $^\circ$C）。(c) 气体定律为何用开尔文？

Answer:答案：  (a) $300\,\text{K}$ and和 $200\,\text{K}$  ·  (b) $100\,\text{K} = 100\,^\circ\text{C}$ drop降幅  ·  (c) the law uses absolute temperature该定律使用绝对温度

(a) Convert each temperature to kelvin将各温度换算为开尔文 M1·A1

$$ T_1 = 27 + 273 = 300\,\text{K}; \qquad T_2 = -73 + 273 = 200\,\text{K}. $$

(b) Size of the temperature change温度变化的大小 A1

$\Delta T = 300 - 200 = 100\,\text{K}$. Because a one-degree step is identical on both scales, the change is also $100\,^\circ\text{C}$.$\Delta T = 300 - 200 = 100\,\text{K}$。由于一度的间隔在两个量程上完全相同，该变化也等于 $100\,^\circ\text{C}$。

(c) Why kelvin is required为何必须用开尔文 A1

The ideal gas law $PV = nRT$ is derived from kinetic theory with $T$ as the absolute temperature, proportional to average kinetic energy. Celsius has an arbitrary, non-physical zero, so substituting a negative Celsius value would predict negative pressure or volume, which is impossible.理想气体定律 $PV = nRT$ 由动理论推导，其中 $T$ 是绝对温度，与平均动能成正比。摄氏的零点是任意且非物理的，因此代入负摄氏值会预测出负压力或负体积，这是不可能的。
Convert temperatures to kelvin for gas-law states, but a temperature difference is the same number in K or $^\circ$C.气体定律中的状态温度须换算为开尔文，但温度差在 K 与 $^\circ$C 中数值相同。 Two distinct facts trip students up. First, any absolute-temperature calculation ($PV = nRT$, Charles's law, Gay-Lussac's law, Carnot efficiency) must use kelvin because the physics depends on the ratio of absolute temperatures. Second, in $Q = mc\Delta T$ only a temperature difference appears, and since the kelvin and Celsius degrees are the same size, $\Delta T$ has the same numerical value in both, so no conversion is needed there. Knowing when conversion matters and when it does not is a frequently tested distinction.两个不同的事实常让学生出错。第一，任何涉及绝对温度的计算（$PV = nRT$、查理定律、盖-吕萨克定律、卡诺效率）都必须用开尔文，因为物理依赖绝对温度之比。第二，在 $Q = mc\Delta T$ 中只出现温度差，由于开尔文与摄氏的度数大小相同，$\Delta T$ 在两者中数值相同，故此处无需换算。分清何时需要换算、何时不需要，是常考的要点。
Q4MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §3 Specific heat比热容 · HS-PS3-1 [3 marks][3 分]

Heat to raise $0.50$ kg aluminium ($c = 900\ \text{J kg}^{-1}\text{K}^{-1}$) by $60\,^\circ\text{C}$?将 $0.50$ kg 铝（$c = 900\ \text{J kg}^{-1}\text{K}^{-1}$）升温 $60\,^\circ\text{C}$ 需要多少热量？

Answer:答案：  (A)  $27\,000\ \text{J}$

(a) Apply $Q = mc\Delta T$套用 $Q = mc\Delta T$ M1·A1·A1

$$ Q \;=\; mc\Delta T \;=\; 0.50 \times 900 \times 60 \;=\; 27\,000\ \text{J} \;=\; 27\ \text{kJ}. $$ Option (A). Note $\Delta T = 60\,^\circ\text{C} = 60\,\text{K}$, so no kelvin conversion of the difference is needed.选 (A)。注意 $\Delta T = 60\,^\circ\text{C} = 60\,\text{K}$，故温度差无需换算。
Why the distractors fail.干扰项分析。
(B) $54\,000\ \text{J}$: forgets the $0.50$ kg mass (uses $m = 1.0$ kg).忘记 $0.50$ kg 质量（误用 $m = 1.0$ kg）。
(C) $1080\ \text{J}$: uses $c = 900$ but $\Delta T$ wrongly (e.g. divides instead of multiplies).用 $c = 900$ 但 $\Delta T$ 处理错误（如误用除法）。
(D) $13\,500\ \text{J}$: halves the correct answer twice or uses $\Delta T = 30$.将正确答案再减半，或误用 $\Delta T = 30$。
$Q = mc\Delta T$ is the workhorse equation: substitute all three factors with correct units.$Q = mc\Delta T$ 是核心公式：代入三个因子并保证单位正确。 The most common error is dropping the mass or using grams instead of kilograms (since $c$ is given per kilogram). Always check units: $\text{kg} \times \text{J kg}^{-1}\text{K}^{-1} \times \text{K} = \text{J}$. Aluminium's specific heat ($900$) is far below water's ($4186$), so the same heat raises aluminium's temperature about $4.6$ times more than water's for equal mass. This is why metal pans heat up fast while water in them takes much longer.最常见的错误是漏掉质量，或用克而非千克（因为 $c$ 按千克给出）。务必核对单位：$\text{kg} \times \text{J kg}^{-1}\text{K}^{-1} \times \text{K} = \text{J}$。铝的比热容（$900$）远低于水（$4186$），故等质量下同样的热量使铝升温约为水的 $4.6$ 倍。这就是金属锅升温快、而锅中的水升温慢得多的原因。
Q5MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Boyle's law玻意耳定律 · Physics 11 [5 marks][5 分]

Constant $T$: $P_1 = 1.0 \times 10^5$ Pa, $V_1 = 2.0$ L, compressed to $V_2 = 0.50$ L. (a) Which law? (b) Find $P_2$. (c) Microscopic reason.恒温：$P_1 = 1.0 \times 10^5$ Pa，$V_1 = 2.0$ L，压缩至 $V_2 = 0.50$ L。(a) 何种定律？(b) 求 $P_2$。(c) 微观原因。

Answer:答案：  (a) Boyle's law玻意耳定律  ·  (b) $P_2 = 4.0 \times 10^5\ \text{Pa}$  ·  (c) collisions per second rise每秒碰撞次数增加

(a) Identify the gas law确定气体定律 A1

Temperature and amount of gas are constant, only $P$ and $V$ change, so Boyle's law applies: $P_1 V_1 = P_2 V_2$.温度与气体量不变，只有 $P$ 和 $V$ 变化，故适用玻意耳定律：$P_1 V_1 = P_2 V_2$。

(b) Solve for the new pressure求新压力 M1·A1·A1

$$ P_2 \;=\; P_1 \frac{V_1}{V_2} \;=\; (1.0 \times 10^5) \times \frac{2.0}{0.50} \;=\; (1.0 \times 10^5)\times 4.0 \;=\; 4.0 \times 10^5\ \text{Pa}. $$ The volume is reduced to one-quarter, so the pressure rises four-fold.体积减小到四分之一，故压力上升至四倍。

(c) Microscopic explanation微观解释 A1

At constant temperature the molecules keep the same average speed, but in a smaller volume they strike the walls more frequently per second, raising the pressure.恒温下分子的平均速率不变，但在更小的体积内每秒撞击器壁的次数更多，从而提高压力。
Boyle's law is an inverse proportion: $P \propto 1/V$ at fixed $T$ and $n$.玻意耳定律是反比关系：在 $T$、$n$ 固定时 $P \propto 1/V$。 Because $PV$ is constant, you can read off the answer by ratio: dividing $V$ by $4$ multiplies $P$ by $4$. Note that units of $V$ need not be converted to SI here because they cancel in the ratio $V_1/V_2$ (both in litres). This is a useful shortcut, but only when the same unit appears top and bottom. Temperature does not need conversion either, because it never enters Boyle's law. The microscopic picture (same speed, more frequent collisions) connects the macroscopic gas law back to kinetic theory, which BC Physics 11 emphasises.由于 $PV$ 恒定，可用比值直接读出答案：$V$ 除以 $4$，$P$ 就乘以 $4$。注意此处 $V$ 的单位无需换算为国际单位，因为在比值 $V_1/V_2$ 中相互约去（两者均为升）。这是一个有用的捷径，但仅当上下出现同一单位时才成立。温度也无需换算，因为它根本不进入玻意耳定律。微观图像（速率不变、碰撞更频繁）将宏观气体定律与动理论相联，这是卑诗物理 11 所强调的。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  简答题 · 详解AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 Calorimetry (mixing)量热法（混合） · HS-PS3-1 [8 marks][8 分]

Insulated mix: $0.20$ kg water at $80\,^\circ\text{C}$ with $0.30$ kg water at $20\,^\circ\text{C}$. $c = 4186$. (a) Principle. (b) Final $T$. (c) Heat transferred. (d) Why closer to $20\,^\circ\text{C}$.绝热混合：$0.20$ kg、$80\,^\circ\text{C}$ 水与 $0.30$ kg、$20\,^\circ\text{C}$ 水。$c = 4186$。(a) 原理。(b) 最终 $T$。(c) 传递的热量。(d) 为何更接近 $20\,^\circ\text{C}$。

Answer:答案：  (a) conservation of energy能量守恒  ·  (b) $T_f = 44\,^\circ\text{C}$  ·  (c) $Q \approx 3.0 \times 10^4\ \text{J}$  ·  (d) more cold-water mass冷水质量更大

(a) State the principle写出原理 A1

Conservation of energy in an isolated system: heat lost by the hot water equals heat gained by the cold water, $|Q_{\text{lost}}| = |Q_{\text{gained}}|$.孤立系统中的能量守恒：热水失去的热量等于冷水获得的热量，$|Q_{\text{失}}| = |Q_{\text{得}}|$。

(b) Find the equilibrium temperature求平衡温度 M1·M1·A1·A1

Both are water, so $c$ cancels. Set heat lost $=$ heat gained:两者均为水，$c$ 约去。令失热 $=$ 得热： $$ m_{\text{hot}}(80 - T_f) \;=\; m_{\text{cold}}(T_f - 20). $$ $$ 0.20(80 - T_f) \;=\; 0.30(T_f - 20). $$ $$ 16 - 0.20\,T_f \;=\; 0.30\,T_f - 6 \;\Longrightarrow\; 22 \;=\; 0.50\,T_f \;\Longrightarrow\; T_f \;=\; 44\,^\circ\text{C}. $$

(c) Heat transferred from hot to cold从热水向冷水传递的热量 M1·A1

$$ Q \;=\; m_{\text{hot}}\,c\,(80 - T_f) \;=\; 0.20 \times 4186 \times (80 - 44) \;=\; 0.20 \times 4186 \times 36 \;\approx\; 3.0 \times 10^4\ \text{J}. $$ (Check: $m_{\text{cold}}\,c\,(T_f - 20) = 0.30 \times 4186 \times 24 \approx 3.0 \times 10^4\ \text{J}$, the same magnitude.)（验证：$m_{\text{冷}}\,c\,(T_f - 20) = 0.30 \times 4186 \times 24 \approx 3.0 \times 10^4\ \text{J}$，大小相同。）

(d) Why the result leans toward $20\,^\circ\text{C}$为何结果偏向 $20\,^\circ\text{C}$ A1

There is more cold water ($0.30$ kg) than hot water ($0.20$ kg). The larger-mass body changes temperature less, so the equilibrium sits closer to the cold side.冷水（$0.30$ kg）比热水（$0.20$ kg）多。质量更大的物体温度变化更小，故平衡点更靠近冷端。
When both substances are the same material, the specific heat cancels and the final temperature is a mass-weighted average.当两种物质材料相同时，比热容约去，最终温度是按质量加权的平均值。 Setting $|Q_{\text{lost}}| = |Q_{\text{gained}}|$ is conservation of energy applied to a closed thermal system. With equal $c$, the result is $T_f = (m_1 T_1 + m_2 T_2)/(m_1 + m_2) = (0.20 \cdot 80 + 0.30 \cdot 20)/0.50 = 44\,^\circ\text{C}$, exactly the weighted average. The final temperature must always lie strictly between the two starting temperatures, a quick sanity check. If the two bodies were different materials (say copper and water), $c$ would not cancel and you would keep it explicit, with water's high $c$ pulling the equilibrium strongly toward the water's temperature.令 $|Q_{\text{失}}| = |Q_{\text{得}}|$ 即把能量守恒应用于封闭热系统。当 $c$ 相等时，结果为 $T_f = (m_1 T_1 + m_2 T_2)/(m_1 + m_2) = (0.20 \cdot 80 + 0.30 \cdot 20)/0.50 = 44\,^\circ\text{C}$，恰为加权平均值。最终温度必定严格介于两个初温之间，可用作快速核验。若两物体材料不同（如铜与水），$c$ 不会约去，须保留，水的高比热容会把平衡强烈拉向水的温度。
Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Latent heat + heating curve潜热与加热曲线 · SPH3U D2 [8 marks][8 分]

$0.25$ kg ice at $0\,^\circ\text{C}$ to water at $20\,^\circ\text{C}$. $L_f = 334$ kJ/kg, $c = 4186$. (a) Heat to melt. (b) Heat to warm. (c) Total. (d) Heating-curve shape and why $T$ is flat during melting.$0.25$ kg、$0\,^\circ\text{C}$ 冰至 $20\,^\circ\text{C}$ 水。$L_f = 334$ kJ/kg，$c = 4186$。(a) 熔化热。(b) 升温热。(c) 总热量。(d) 加热曲线形状及熔化时 $T$ 为何不变。

Answer:答案：  (a) $Q_1 = 83.5\ \text{kJ}$  ·  (b) $Q_2 = 20.9\ \text{kJ}$  ·  (c) $Q_{\text{tot}} \approx 104\ \text{kJ}$  ·  (d) flat plateau then a slope先平台后斜线

(a) Heat to melt the ice ($Q = mL_f$)熔化冰所需热量（$Q = mL_f$） M1·A1

$$ Q_1 \;=\; mL_f \;=\; 0.25 \times 334\,000 \;=\; 83\,500\ \text{J} \;=\; 83.5\ \text{kJ}. $$

(b) Heat to warm the meltwater ($Q = mc\Delta T$)将融水升温所需热量（$Q = mc\Delta T$） M1·A1

$$ Q_2 \;=\; mc\Delta T \;=\; 0.25 \times 4186 \times (20 - 0) \;=\; 20\,930\ \text{J} \;=\; 20.9\ \text{kJ}. $$

(c) Total heat required所需总热量 A1

$$ Q_{\text{tot}} \;=\; Q_1 + Q_2 \;=\; 83.5 + 20.9 \;=\; 104.4\ \text{kJ} \;\approx\; 104\ \text{kJ}. $$

(d) Shape of the heating curve, and why $T$ stays constant during melting加热曲线形状，以及熔化时 $T$ 为何不变 A1·A1·A1

Stage 1 (melting): a horizontal plateau at $0\,^\circ\text{C}$, because heat goes into breaking the bonds of the solid lattice (increasing potential energy), not into raising the average kinetic energy. Stage 2 (warming): a rising straight line from $0$ to $20\,^\circ\text{C}$, because all the heat now raises the kinetic energy and hence the temperature. The slope of stage 2 is $1/(mc)$ per unit heat. Temperature is constant during a phase change because it measures average kinetic energy, which does not change while bonds are being broken.阶段 1（熔化）：在 $0\,^\circ\text{C}$ 处为水平平台，因为热量用于断裂固体晶格的键（增加势能），而非提高平均动能。阶段 2（升温）：从 $0$ 升至 $20\,^\circ\text{C}$ 的上升直线，因为此时所有热量都用于提高动能进而提高温度。阶段 2 的斜率为每单位热量 $1/(mc)$。相变期间温度不变，因为温度衡量平均动能，而在断键过程中平均动能不变。
A heating curve alternates sloped segments ($Q = mc\Delta T$) with flat plateaux ($Q = mL$).加热曲线由斜线段（$Q = mc\Delta T$）与平台段（$Q = mL$）交替构成。 Sloped portions are sensible heat: the temperature changes and you use $Q = mc\Delta T$. Flat portions are latent heat: the temperature holds while a phase change happens and you use $Q = mL$. Here the melting plateau alone ($83.5$ kJ) costs four times the heat needed to warm the resulting water ($20.9$ kJ), which surprises many students. The deep reason, emphasised by NGSS HS-PS3-2, is the kinetic-versus-potential energy split: temperature tracks kinetic energy, but phase changes store energy as molecular potential energy. Sketching the curve and labelling which equation governs each segment is the reliable exam strategy.斜线段是显热：温度变化，使用 $Q = mc\Delta T$。平台段是潜热：相变进行时温度保持不变，使用 $Q = mL$。此处仅熔化平台（$83.5$ kJ）就是将所得水升温（$20.9$ kJ）所需热量的四倍，这令许多学生意外。其深层原因（NGSS HS-PS3-2 所强调）是动能与势能之分：温度跟踪动能，而相变将能量储存为分子势能。画出曲线并标注每段所遵循的方程，是可靠的应试策略。
Q8HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Combined gas law联合气体定律 · Physics 11 [8 marks][8 分]

$V_1 = 3.0$ L, $P_1 = 1.0 \times 10^5$ Pa, $T_1 = 27\,^\circ\text{C}$, compressed and heated to $V_2 = 1.0$ L, $T_2 = 127\,^\circ\text{C}$. (a) Kelvin. (b) $P_2$. (c) Which effect dominates. (d) Ideal-gas assumption.$V_1 = 3.0$ L，$P_1 = 1.0 \times 10^5$ Pa，$T_1 = 27\,^\circ\text{C}$，压缩并加热至 $V_2 = 1.0$ L，$T_2 = 127\,^\circ\text{C}$。(a) 开尔文。(b) $P_2$。(c) 哪个效应占主导。(d) 理想气体假设。

Answer:答案：  (a) $300\,\text{K},\ 400\,\text{K}$  ·  (b) $P_2 = 4.0 \times 10^5\ \text{Pa}$  ·  (c) compression dominates压缩占主导  ·  (d) negligible molecular volume / forces分子体积与作用力可忽略

(a) Convert temperatures to kelvin换算温度为开尔文 A1

$$ T_1 = 27 + 273 = 300\,\text{K}; \qquad T_2 = 127 + 273 = 400\,\text{K}. $$

(b) Combined gas law for the new pressure用联合气体定律求新压力 M1·M1·A1·A1

$$ \frac{P_1 V_1}{T_1} \;=\; \frac{P_2 V_2}{T_2} \;\Longrightarrow\; P_2 \;=\; P_1 \cdot \frac{V_1}{V_2} \cdot \frac{T_2}{T_1}. $$ $$ P_2 \;=\; (1.0 \times 10^5) \times \frac{3.0}{1.0} \times \frac{400}{300} \;=\; (1.0 \times 10^5) \times 3.0 \times 1.333 \;=\; 4.0 \times 10^5\ \text{Pa}. $$

(c) Which effect dominates哪个效应占主导 A1·A1

The volume drop to one-third multiplies pressure by $3$ (compression raises $P$). The temperature rise from $300$ to $400$ K multiplies pressure by $4/3$ (heating raises $P$). Both push $P$ up; the compression factor ($\times 3$) is larger than the heating factor ($\times 1.33$), so compression dominates.体积降至三分之一使压力乘以 $3$（压缩使 $P$ 升高）。温度从 $300$ 升至 $400$ K 使压力乘以 $4/3$（加热使 $P$ 升高）。两者都使 $P$ 升高；压缩因子（$\times 3$）大于加热因子（$\times 1.33$），故压缩占主导。

(d) State an ideal-gas assumption写出一个理想气体假设 A1

The gas molecules have negligible volume compared with the container, and intermolecular forces are negligible (no condensation over this range).气体分子体积相对容器可忽略，分子间作用力可忽略（在此范围内不发生冷凝）。
The combined gas law factorises into a volume ratio and a temperature ratio, each acting independently on the pressure.联合气体定律可分解为体积比与温度比，两者各自独立地作用于压力。 Writing $P_2 = P_1 (V_1/V_2)(T_2/T_1)$ separates the two physical effects so you can see which dominates. Note the temperature ratio uses kelvin: using $T_2/T_1 = 127/27 \approx 4.7$ (Celsius) would be a serious error, badly overstating the heating effect. The volume ratio, by contrast, can stay in litres because the units cancel. This problem rewards organising the calculation as a product of clean ratios rather than computing $n$ explicitly, which is unnecessary when the amount of gas is fixed.写成 $P_2 = P_1 (V_1/V_2)(T_2/T_1)$ 可把两个物理效应分开，便于看出哪个占主导。注意温度比须用开尔文：若用 $T_2/T_1 = 127/27 \approx 4.7$（摄氏）将是严重错误，会大幅夸大加热效应。相比之下，体积比可保留升，因为单位约去。本题的诀窍是把计算组织为几个干净比值的乘积，而无需显式计算 $n$，在气体量固定时这是不必要的。
Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 First law of thermodynamics热力学第一定律 · HS-PS3-1 [8 marks][8 分]

$\Delta U = Q - W$. (a) $Q = +650$ J, gas does $250$ J work. (b) $300$ J added at constant volume. (c) gas releases $500$ J, surroundings do $200$ J on it. (d) Why (c) conserves energy.$\Delta U = Q - W$。(a) $Q = +650$ J，气体做 $250$ J 功。(b) 恒容下加入 $300$ J。(c) 气体放出 $500$ J，外界对其做 $200$ J 功。(d) (c) 为何与能量守恒一致。

Answer:答案：  (a) $\Delta U = +400\ \text{J}$  ·  (b) $\Delta U = +300\ \text{J}$  ·  (c) $\Delta U = -300\ \text{J}$  ·  (d) work input partly offsets heat loss输入的功部分抵消了热损失

(a) Heat added, gas does work吸热并对外做功 M1·A1

$Q = +650$ J (added), $W = +250$ J (done by the gas):$Q = +650$ J（吸热），$W = +250$ J（气体对外做功）： $$ \Delta U \;=\; Q - W \;=\; 650 - 250 \;=\; +400\ \text{J}. $$

(b) Constant volume means no work恒容意味着不做功 M1·A1

At constant volume the gas does no work on its surroundings, so $W = 0$. With $Q = +300$ J:恒容时气体不对外界做功，故 $W = 0$。$Q = +300$ J： $$ \Delta U \;=\; Q - W \;=\; 300 - 0 \;=\; +300\ \text{J}. $$

(c) Heat released, work done on the gas (mind the signs)放热且外界做功（注意符号） M1·M1·A1

Heat released: $Q = -500$ J. Surroundings do $200$ J of work compressing the gas, so the work done by the gas is $W = -200$ J:放热：$Q = -500$ J。外界做 $200$ J 功压缩气体，故气体对外做的功 $W = -200$ J： $$ \Delta U \;=\; Q - W \;=\; (-500) - (-200) \;=\; -500 + 200 \;=\; -300\ \text{J}. $$

(d) Consistency with energy conservation与能量守恒的一致性 A1

The gas lost $500$ J as heat but gained $200$ J from the work done on it. The net loss of internal energy is therefore $500 - 200 = 300$ J, exactly $\Delta U = -300$ J. No energy is created or destroyed; it is redistributed between heat and work.气体以热量形式损失 $500$ J，但从外界做功获得 $200$ J。因此内能净损失为 $500 - 200 = 300$ J，恰为 $\Delta U = -300$ J。没有能量被创造或消灭；它只是在热与功之间重新分配。
In $\Delta U = Q - W$, $W$ is the work done by the gas: compression makes $W$ negative, heat release makes $Q$ negative.在 $\Delta U = Q - W$ 中，$W$ 是气体对外做的功：压缩使 $W$ 为负，放热使 $Q$ 为负。 The single most common error in first-law problems is a sign slip. Pin down two conventions before substituting: (1) $Q > 0$ when heat is added to the gas, $Q < 0$ when the gas releases heat; (2) $W > 0$ when the gas expands and does work on the surroundings, $W < 0$ when the surroundings compress the gas. Part (c) double-trap: both $Q$ and $W$ are negative, and the formula subtracts $W$, so two minus signs combine: $-(-200) = +200$. The constant-volume case in (b) is the cleanest reminder that $W = 0$ whenever volume is fixed (a rigid sealed container), so all added heat becomes internal energy. This bookkeeping is exactly the energy accounting NGSS HS-PS3-1 asks for.第一定律问题中最常见的错误是符号出错。代入前先确定两个约定：(1) 气体吸热时 $Q > 0$，气体放热时 $Q < 0$；(2) 气体膨胀对外界做功时 $W > 0$，外界压缩气体时 $W < 0$。(c) 是双重陷阱：$Q$ 与 $W$ 都为负，而公式要减去 $W$，故两个负号叠加：$-(-200) = +200$。(b) 的恒容情形是最清晰的提醒：只要体积固定（刚性密封容器），$W = 0$，故全部吸收的热量都转为内能。这种核算正是 NGSS HS-PS3-1 所要求的能量账目。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模与应用 · 详解Universal applications · 30 marks通用应用题 · 共 30 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §3 + §4 Calorimetry with melting含熔化的量热法 · SPH3U D3 [9 marks][9 分]

$0.040$ kg ice at $0\,^\circ\text{C}$ into $0.50$ kg water at $25\,^\circ\text{C}$, insulated. $L_f = 334$ kJ/kg, $c = 4186$. (a) Heat to melt. (b) Heat water can give up; confirm all melts. (c) Final $T$. (d) Real-world deviation.$0.040$ kg、$0\,^\circ\text{C}$ 冰投入 $0.50$ kg、$25\,^\circ\text{C}$ 水，绝热。$L_f = 334$ kJ/kg，$c = 4186$。(a) 熔化热。(b) 水能放出的热量；确认全部熔化。(c) 最终 $T$。(d) 真实偏差。

Answer:答案：  (a) $Q_{\text{melt}} = 13\,360\ \text{J}$  ·  (b) $Q_{\text{avail}} = 52\,325\ \text{J} > Q_{\text{melt}}$  ·  (c) $T_f \approx 17.2\,^\circ\text{C}$  ·  (d) heat leaks / cup absorbs heat热量泄漏 / 杯子吸热

(a) Heat to melt all the ice ($Q = mL_f$)熔化全部冰的热量（$Q = mL_f$） M1·A1

$$ Q_{\text{melt}} \;=\; mL_f \;=\; 0.040 \times 334\,000 \;=\; 13\,360\ \text{J}. $$

(b) Heat the warm water can release cooling to $0\,^\circ\text{C}$温水冷却至 $0\,^\circ\text{C}$ 可放出的热量 M1·A1

$$ Q_{\text{avail}} \;=\; m_w c (25 - 0) \;=\; 0.50 \times 4186 \times 25 \;=\; 52\,325\ \text{J}. $$ Since $52\,325 > 13\,360$, the warm water supplies more than enough heat to melt all the ice, so all the ice melts and the final state is liquid water above $0\,^\circ\text{C}$.由于 $52\,325 > 13\,360$，温水提供的热量足以熔化全部冰，故冰全部熔化，最终状态为高于 $0\,^\circ\text{C}$ 的液态水。

(c) Final equilibrium temperature最终平衡温度 M1·M1·A1·A1

Energy balance: heat lost by warm water $=$ heat to melt ice $+$ heat to warm the meltwater from $0$ to $T_f$.能量平衡：温水失去的热量 $=$ 熔化冰的热量 $+$ 将融水从 $0$ 升至 $T_f$ 的热量。 $$ m_w c (25 - T_f) \;=\; m_{\text{ice}} L_f + m_{\text{ice}} c\,(T_f - 0). $$ $$ 0.50 \times 4186 \times (25 - T_f) \;=\; 13\,360 + 0.040 \times 4186 \times T_f. $$ $$ 52\,325 - 2093\,T_f \;=\; 13\,360 + 167.4\,T_f \;\Longrightarrow\; 38\,965 \;=\; 2260.4\,T_f \;\Longrightarrow\; T_f \;\approx\; 17.2\,^\circ\text{C}. $$

(d) Why a real measurement differs真实测量为何不同 A1

Real cups are not perfectly insulated: heat leaks to the surroundings, and the cup itself absorbs some heat, so the measured final temperature would be slightly lower than $17.2\,^\circ\text{C}$.真实杯子并非完全绝热：热量会泄漏到环境中，杯子本身也吸收部分热量，故实测最终温度会略低于 $17.2\,^\circ\text{C}$。
In a melt-and-mix problem, first check whether all the ice melts before writing the equilibrium equation.在熔化与混合问题中，先确认冰是否全部熔化，再写平衡方程。 The crucial pre-step (part b) is comparing the heat needed to melt the ice against the heat the warm water can supply. If $Q_{\text{avail}} < Q_{\text{melt}}$, the ice does not fully melt and the final temperature is exactly $0\,^\circ\text{C}$ (a mixture of ice and water). Here $Q_{\text{avail}} > Q_{\text{melt}}$, so we proceed to a full equilibrium calculation that includes warming the meltwater. Forgetting the $m_{\text{ice}} c\,T_f$ term (warming the melted ice up to $T_f$) is the most common error and would wrongly give a higher $T_f$. The ice has a double cooling effect: it absorbs latent heat to melt, then the cold meltwater absorbs sensible heat to warm up.关键的前置步骤（b）是比较熔化冰所需的热量与温水能提供的热量。若 $Q_{\text{avail}} < Q_{\text{melt}}$，冰不会完全熔化，最终温度恰为 $0\,^\circ\text{C}$（冰水混合）。此处 $Q_{\text{avail}} > Q_{\text{melt}}$，故进行包含将融水升温的完整平衡计算。遗漏 $m_{\text{ice}} c\,T_f$ 项（把融化的冰升温至 $T_f$）是最常见的错误，会错误地得到偏高的 $T_f$。冰有双重降温作用：先吸收潜热熔化，随后冷的融水再吸收显热升温。
Q11MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Gay-Lussac (applied)盖-吕萨克定律（应用） · Physics 11 [9 marks][9 分]

Sealed rigid can: $1.0 \times 10^5$ Pa at $27\,^\circ\text{C}$, heated to $177\,^\circ\text{C}$, volume fixed. (a) Kelvin. (b) Which law. (c) New pressure. (d) Safety implication.密封刚性罐：$1.0 \times 10^5$ Pa、$27\,^\circ\text{C}$，加热至 $177\,^\circ\text{C}$，体积固定。(a) 开尔文。(b) 何种定律。(c) 新压力。(d) 安全含义。

Answer:答案：  (a) $300\,\text{K},\ 450\,\text{K}$  ·  (b) Gay-Lussac's law盖-吕萨克定律  ·  (c) $P_2 = 1.5 \times 10^5\ \text{Pa}$  ·  (d) pressure rise can rupture the can压力上升可使罐爆裂

(a) Convert temperatures to kelvin换算温度为开尔文 A1

$$ T_1 = 27 + 273 = 300\,\text{K}; \qquad T_2 = 177 + 273 = 450\,\text{K}. $$

(b) Identify the gas law确定气体定律 A1

Volume is constant and the amount of gas is fixed, so Gay-Lussac's law applies: $P_1/T_1 = P_2/T_2$.体积不变且气体量固定，故适用盖-吕萨克定律：$P_1/T_1 = P_2/T_2$。

(c) New pressure新压力 M1·M1·A1·A1

$$ P_2 \;=\; P_1 \frac{T_2}{T_1} \;=\; (1.0 \times 10^5) \times \frac{450}{300} \;=\; (1.0 \times 10^5)\times 1.5 \;=\; 1.5 \times 10^5\ \text{Pa}. $$

(d) Safety implication安全含义 A1·A1·A1

The internal pressure rises by $50\%$ (to $1.5$ times atmospheric) just from heating, with nowhere for the gas to expand. If the can is heated further, the pressure keeps climbing in proportion to absolute temperature until it exceeds the can's strength and it bursts. This is why aerosol and sealed containers carry "do not incinerate / do not store above $50\,^\circ\text{C}$" warnings: the constant-volume pressure rise is unavoidable.仅因加热，内部压力就上升 $50\%$（达到大气压的 $1.5$ 倍），而气体无处膨胀。若继续加热，压力会随绝对温度成比例持续攀升，直到超过罐体强度而爆裂。这就是气雾罐和密封容器标注"切勿焚烧 / 切勿存放于 $50\,^\circ\text{C}$ 以上"的原因：恒容下的压力上升不可避免。
At constant volume, pressure is directly proportional to absolute temperature: $P \propto T$ (kelvin).恒容时，压力与绝对温度成正比：$P \propto T$（开尔文）。 Gay-Lussac's law follows from $PV = nRT$ with $V$ and $n$ fixed. The kelvin requirement is critical here: heating from $27$ to $177\,^\circ\text{C}$ looks like a six-fold temperature increase in Celsius ($177/27$), but in kelvin it is only $450/300 = 1.5\times$. Using Celsius would predict a wildly wrong pressure. Microscopically, raising the temperature increases the average molecular speed, so molecules strike the walls both harder and more often, raising the pressure. The real-world hook (do not incinerate) makes this a favourite for BC Physics 11 and AP energy-conversion contexts.盖-吕萨克定律由 $PV = nRT$ 在 $V$、$n$ 固定时得出。此处开尔文要求至关重要：从 $27$ 加热到 $177\,^\circ\text{C}$ 在摄氏看来像是六倍升温（$177/27$），但用开尔文只有 $450/300 = 1.5$ 倍。用摄氏会预测出严重错误的压力。从微观看，升温使分子平均速率增大，故分子撞击器壁既更猛烈又更频繁，从而提高压力。"切勿焚烧"这一实际背景使本题成为卑诗物理 11 与 AP 能量转化情境中的常见题型。
Q12HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §6 + §7 Heat engine + Carnot limit热机与卡诺上限 · HS-PS3-3 / HS-PS3-4 [12 marks][12 分]

Turbine: $Q_H = 9000$ J, $W = 2700$ J per cycle, $T_H = 600$ K, $T_C = 300$ K. (a) Actual efficiency. (b) $Q_C$. (c) Carnot efficiency. (d) Compare; which law bars $100\%$. (e) Raise $T_H$ to $750$ K: new Carnot limit and trade-off.轮机：每循环 $Q_H = 9000$ J，$W = 2700$ J，$T_H = 600$ K，$T_C = 300$ K。(a) 实际效率。(b) $Q_C$。(c) 卡诺效率。(d) 比较；哪条定律禁止 $100\%$。(e) 将 $T_H$ 提高到 $750$ K：新卡诺上限与权衡。

Answer:答案：  (a) $\eta = 30\%$  ·  (b) $Q_C = 6300\ \text{J}$  ·  (c) $\eta_{\text{Carnot}} = 50\%$  ·  (d) below Carnot; second law低于卡诺；第二定律  ·  (e) $60\%$

(a) Actual thermal efficiency ($\eta = W/Q_H$)实际热效率（$\eta = W/Q_H$） M1·A1

$$ \eta \;=\; \frac{W}{Q_H} \;=\; \frac{2700}{9000} \;=\; 0.30 \;=\; 30\%. $$

(b) Heat rejected ($Q_C = Q_H - W$)排出热量（$Q_C = Q_H - W$） M1·A1

$$ Q_C \;=\; Q_H - W \;=\; 9000 - 2700 \;=\; 6300\ \text{J}. $$

(c) Carnot (maximum) efficiency卡诺（最大）效率 M1·A1·A1

$$ \eta_{\text{Carnot}} \;=\; 1 - \frac{T_C}{T_H} \;=\; 1 - \frac{300}{600} \;=\; 1 - 0.50 \;=\; 0.50 \;=\; 50\%. $$

(d) Comparison and the governing law比较与支配定律 A1·A1

The actual efficiency ($30\%$) is below the Carnot limit ($50\%$), as it must be: no real engine can exceed the Carnot value. The second law of thermodynamics forbids $100\%$ efficiency, because some heat $Q_C$ must always be rejected to the cold reservoir.实际效率（$30\%$）低于卡诺上限（$50\%$），这是必然的：任何实际热机都不能超过卡诺值。热力学第二定律禁止 $100\%$ 效率，因为总有部分热量 $Q_C$ 必须排向冷源。

(e) Raising $T_H$ to $750$ K将 $T_H$ 提高到 $750$ K M1·A1·A1

$$ \eta_{\text{Carnot}}' \;=\; 1 - \frac{300}{750} \;=\; 1 - 0.40 \;=\; 0.60 \;=\; 60\%. $$ Raising $T_H$ lifts the theoretical ceiling from $50\%$ to $60\%$, but running a hotter reservoir demands stronger, more heat-resistant materials and more fuel, so the efficiency gain is bought with higher engineering cost and stress.提高 $T_H$ 将理论上限从 $50\%$ 提升到 $60\%$，但运行更高温的热源需要更坚固、更耐热的材料和更多燃料，故效率的提升是以更高的工程成本和应力为代价换来的。
Real efficiency $\le$ Carnot efficiency $= 1 - T_C/T_H$; the gap is mandated by the second law, not by poor engineering.实际效率 $\le$ 卡诺效率 $= 1 - T_C/T_H$；这一差距由第二定律决定，而非工程缺陷。 Three efficiency ideas combine here. First, $\eta = W/Q_H = 1 - Q_C/Q_H$ defines the actual efficiency from the energy flows. Second, energy conservation (first law) gives $Q_C = Q_H - W$. Third, the Carnot formula $1 - T_C/T_H$ sets the upper bound using only the reservoir temperatures in kelvin. The actual engine ($30\%$) sits below its Carnot ceiling ($50\%$) because of friction, turbulence, and finite-rate (irreversible) heat transfer. Even a perfect, frictionless engine could not exceed $50\%$ here. The only ways to raise the ceiling are a hotter source or a colder sink, which is exactly why power plants run very high-temperature steam, and why this links to Ontario D3.12 on nuclear-plant thermodynamics.这里融合了三个效率概念。第一，$\eta = W/Q_H = 1 - Q_C/Q_H$ 由能量流定义实际效率。第二，能量守恒（第一定律）给出 $Q_C = Q_H - W$。第三，卡诺公式 $1 - T_C/T_H$ 仅用热源的开尔文温度设定上限。实际热机（$30\%$）低于其卡诺上限（$50\%$），原因是摩擦、湍流和有限速率（不可逆）的热传递。即使是完美、无摩擦的热机在此也无法超过 $50\%$。提高上限的唯一途径是更热的源或更冷的汇，这正是发电厂使用极高温蒸汽的原因，也与安大略 D3.12 关于核电站热力学的内容相联。