PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 18 marksAP 风格选择题 + 安/卑省考短答 · 共 18 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. State units in every answer. Use $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$ and convert temperatures to kelvin in every gas-law calculation. No calculator on Q1-Q2; calculator permitted on Q3-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。每道题都要写出单位。取 $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$，所有气体定律计算须将温度换算为开尔文。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。
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]
A cup holds $0.20$ kg of boiling water at $100\,^\circ\text{C}$; a bathtub holds $80$ kg of warm water at $40\,^\circ\text{C}$. Which statement is correct?一个杯子盛有 $0.20$ kg、$100\,^\circ\text{C}$ 的沸水；一个浴缸盛有 $80$ kg、$40\,^\circ\text{C}$ 的温水。下列哪句话正确？
(A)The cup has the higher temperature; the bathtub has the greater total thermal energy.杯子温度更高；浴缸的总热能更多。
(B)The cup has both the higher temperature and the greater thermal energy.杯子温度更高，热能也更多。
(C)Temperature and thermal energy are the same quantity, so both are larger in the cup.温度与热能是同一量，故杯子两者都更大。
(D)The bathtub has the higher temperature because it has more water.浴缸水更多，故温度更高。
Energy from the Sun reaches the Earth across $\approx 150$ million km of near-vacuum. By which mode of heat transfer does it travel?来自太阳的能量穿越约 $1.5$ 亿 km 的近真空到达地球。它通过哪种热传递模式传播？
A laboratory cools a gas sample from $27\,^\circ\text{C}$ to $-73\,^\circ\text{C}$.某实验室将一份气体样品从 $27\,^\circ\text{C}$ 冷却至 $-73\,^\circ\text{C}$。
(a)Convert both temperatures to kelvin.将两个温度换算为开尔文。[2]
(b)State the size of the temperature change in both kelvin and degrees Celsius.写出温度变化的大小，分别用开尔文和摄氏度表示。[1]
(c)Explain why the ideal gas law requires kelvin, not Celsius.解释为何理想气体定律要求用开尔文而非摄氏。[1]
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]
How much heat is needed to raise the temperature of a $0.50$ kg aluminium block ($c = 900\ \text{J kg}^{-1}\text{K}^{-1}$) by $60\,^\circ\text{C}$?将一块 $0.50$ kg 的铝块（$c = 900\ \text{J kg}^{-1}\text{K}^{-1}$）升温 $60\,^\circ\text{C}$ 需要多少热量？
A gas is held at a constant temperature. Its initial state is $P_1 = 1.0 \times 10^5$ Pa and $V_1 = 2.0$ L. The gas is then compressed to $V_2 = 0.50$ L.一份气体保持恒温。初始状态为 $P_1 = 1.0 \times 10^5$ Pa，$V_1 = 2.0$ L。随后将气体压缩至 $V_2 = 0.50$ L。
(a)State which gas law applies and why.写出适用的气体定律及理由。[1]
(b)Find the new pressure $P_2$.求新压力 $P_2$。[3]
(c)Explain microscopically why the pressure changed as it did.从微观角度解释压力为何如此变化。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 32 marksAP 衔接简答题 + 荣誉级 · 共 32 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Identify the governing equation before substituting values. State units in every final answer. Use $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$, $L_f = 334\ \text{kJ kg}^{-1}$ (water), and $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ where needed. Convert temperatures to kelvin in every gas-law step. Calculator permitted on Q6-Q9.每一步推理都要写出。代入数值前先注明所用方程。每个最终答案都要写单位。需要时取 $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$、$L_f = 334\ \text{kJ kg}^{-1}$（水）和 $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$。所有气体定律步骤须将温度换算为开尔文。Q6-Q9 可用计算器。
In an insulated container, $0.20$ kg of water at $80\,^\circ\text{C}$ is mixed with $0.30$ kg of water at $20\,^\circ\text{C}$. Assume no heat is lost to the container or surroundings. Use $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$.在一个绝热容器中，将 $0.20$ kg、$80\,^\circ\text{C}$ 的水与 $0.30$ kg、$20\,^\circ\text{C}$ 的水混合。假设无热量散失至容器或环境。取 $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$。
(a)State the principle that determines the final temperature.写出决定最终温度的原理。[1]
(b)Find the final equilibrium temperature.求最终平衡温度。[4]
(c)Find the heat transferred from the hot water to the cold water.求从热水传递到冷水的热量。[2]
(d)Explain why the final temperature is closer to $20\,^\circ\text{C}$ than to $80\,^\circ\text{C}$.解释为何最终温度更接近 $20\,^\circ\text{C}$ 而非 $80\,^\circ\text{C}$。[1]
A $0.25$ kg block of ice at $0\,^\circ\text{C}$ is heated until it becomes liquid water at $20\,^\circ\text{C}$. Use $L_f = 334\ \text{kJ kg}^{-1}$ and $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$.一块 $0.25$ kg、$0\,^\circ\text{C}$ 的冰被加热，直至变为 $20\,^\circ\text{C}$ 的液态水。取 $L_f = 334\ \text{kJ kg}^{-1}$，$c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$。
(a)Find the heat needed to melt the ice at $0\,^\circ\text{C}$.求在 $0\,^\circ\text{C}$ 熔化冰所需的热量。[2]
(b)Find the heat needed to warm the resulting water from $0$ to $20\,^\circ\text{C}$.求将所得的水从 $0$ 升至 $20\,^\circ\text{C}$ 所需的热量。[2]
(c)Find the total heat required.求所需的总热量。[1]
(d)On a temperature-versus-heat graph for this process, describe the shape of each stage and explain why the temperature stays constant during melting.在该过程的温度-热量图上，描述各阶段的形状，并解释熔化期间温度为何保持不变。[3]
A fixed amount of gas occupies $V_1 = 3.0$ L at $P_1 = 1.0 \times 10^5$ Pa and $T_1 = 27\,^\circ\text{C}$. It is compressed and heated to $V_2 = 1.0$ L and $T_2 = 127\,^\circ\text{C}$.一定量气体在 $P_1 = 1.0 \times 10^5$ Pa、$T_1 = 27\,^\circ\text{C}$ 时占据 $V_1 = 3.0$ L。将其压缩并加热至 $V_2 = 1.0$ L、$T_2 = 127\,^\circ\text{C}$。
(a)Convert both temperatures to kelvin.将两个温度换算为开尔文。[1]
(b)State the combined gas law and find the new pressure $P_2$.写出联合气体定律并求新压力 $P_2$。[4]
(c)Identify which two effects (volume change and temperature change) each push the pressure up or down, and which dominates.指出体积变化与温度变化各自使压力上升还是下降，以及哪一个占主导。[2]
(d)State one assumption built into treating this gas as ideal.写出将该气体视为理想气体所隐含的一个假设。[1]
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]
Use the first law of thermodynamics in the form $\Delta U = Q - W$, where $Q$ is heat added to the gas and $W$ is work done by the gas.使用热力学第一定律的形式 $\Delta U = Q - W$，其中 $Q$ 为系统吸收的热量，$W$ 为系统对外做的功。
(a)A gas absorbs $650$ J of heat and expands, doing $250$ J of work on a piston. Find $\Delta U$.气体吸收 $650$ J 热量并膨胀，对活塞做 $250$ J 的功。求 $\Delta U$。[2]
(b)In a separate process, $300$ J of heat is added to a gas held at constant volume (a rigid sealed container). Find $\Delta U$, and justify the value of $W$.在另一过程中，向保持恒容的气体（刚性密封容器）加入 $300$ J 热量。求 $\Delta U$，并说明 $W$ 的取值理由。[2]
(c)A gas releases $500$ J of heat while the surroundings do $200$ J of work compressing it. Find $\Delta U$, paying careful attention to signs.气体释放 $500$ J 热量，同时外界对其做 $200$ J 的功将其压缩。注意符号，求 $\Delta U$。[3]
(d)Explain how part (c) is consistent with energy conservation even though the gas lost heat.解释 (c) 中尽管气体放热，结果为何仍与能量守恒一致。[1]
PART III · MODELING / APPLIED第三部分 · 建模与应用Universal applications · 30 marks通用应用题 · 共 30 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define symbols (with units) at the start of each question. State the governing equation before substituting. Convert temperatures to kelvin in every gas-law and Carnot step. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用方程。所有气体定律与卡诺步骤须将温度换算为开尔文。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
A student drops $0.040$ kg of ice at $0\,^\circ\text{C}$ into $0.50$ kg of water at $25\,^\circ\text{C}$ in an insulated cup. Use $L_f = 334\ \text{kJ kg}^{-1}$ and $c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$.一名学生将 $0.040$ kg、$0\,^\circ\text{C}$ 的冰投入绝热杯中 $0.50$ kg、$25\,^\circ\text{C}$ 的水里。取 $L_f = 334\ \text{kJ kg}^{-1}$，$c_{\text{water}} = 4186\ \text{J kg}^{-1}\text{K}^{-1}$。
(a)Find the heat needed to melt all the ice at $0\,^\circ\text{C}$.求在 $0\,^\circ\text{C}$ 熔化全部冰所需的热量。[2]
(b)Find the maximum heat the warm water can give up as it cools to $0\,^\circ\text{C}$, and confirm that all the ice melts.求温水冷却至 $0\,^\circ\text{C}$ 时最多能放出的热量，并确认冰全部熔化。[2]
(c)Find the final equilibrium temperature of the mixture.求混合物的最终平衡温度。[4]
(d)State one reason a real measurement would give a slightly different result.写出真实测量结果会略有不同的一个原因。[1]
A sealed rigid steel can holds air at $1.0 \times 10^5$ Pa and $27\,^\circ\text{C}$. It is left near a fire and heats up to $177\,^\circ\text{C}$. The can's volume does not change.一个密封的刚性钢罐内装有 $1.0 \times 10^5$ Pa、$27\,^\circ\text{C}$ 的空气。它被放在火堆附近，升温至 $177\,^\circ\text{C}$。罐的体积不变。
(a)Convert both temperatures to kelvin.将两个温度换算为开尔文。[1]
(b)State which gas law applies for constant volume.写出恒容条件下适用的气体定律。[1]
(c)Find the new pressure inside the can.求罐内的新压力。[4]
(d)Explain the safety implication and why warning labels say "do not incinerate."解释其安全含义，以及警告标签为何标注"切勿焚烧"。[3]
A steam turbine absorbs $Q_H = 9000$ J of heat from a hot reservoir each cycle and produces $W = 2700$ J of mechanical work. The hot reservoir is at $T_H = 600$ K and the cold reservoir is at $T_C = 300$ K.一台蒸汽轮机每循环从热源吸收 $Q_H = 9000$ J 热量并产生 $W = 2700$ J 机械功。热源温度 $T_H = 600$ K，冷源温度 $T_C = 300$ K。
(a)Find the actual thermal efficiency of the turbine.求轮机的实际热效率。[2]
(b)Find the heat $Q_C$ rejected to the cold reservoir each cycle.求每循环排向冷源的热量 $Q_C$。[2]
(c)Find the Carnot (maximum theoretical) efficiency for these reservoir temperatures.求在该热源温度下的卡诺（理论最大）效率。[3]
(d)Compare the actual and Carnot efficiencies, and state which law forbids reaching $100\%$.比较实际效率与卡诺效率，并写出哪条定律禁止达到 $100\%$。[2]
(e)An engineer proposes raising $T_H$ to $750$ K. Find the new Carnot limit and explain the trade-off in one sentence.一名工程师提议将 $T_H$ 提高到 $750$ K。求新的卡诺上限，并用一句话解释其权衡。[3]
🇺🇸 US NGSS美国 NGSSHS-PS3-1 · HS-PS3-2 · HS-PS3-3 · HS-PS3-4
🇨🇦 Ontario安大略SPH3U Strand D · D2 · D3 · D3.12
🇨🇦 British Columbia不列颠哥伦比亚Physics 11: heat, thermal equilibrium, gas laws, efficiency物理 11：热、热平衡、气体定律、效率
Full Syllabus Map lives in ../Study Guides/Unit_11_Thermodynamics_and_Heat.html. Note: Alberta Physics 20/30 has no thermodynamics unit, so no AB region chip appears here; this set serves NGSS, ON SPH3U Strand D, BC Physics 11, and AP/IB preparation.完整大纲对照表见 ../Study Guides/Unit_11_Thermodynamics_and_Heat.html。注：阿尔伯塔 Physics 20/30 没有热力学单元，故此处不设阿省地区标签；本题集服务于 NGSS、安大略 SPH3U D 单元、卑诗物理 11 以及 AP/IB 准备。