题型分析 — Temperature

Question Type 1: 温标与测温性质

如何识别

问 thermometric property 的例子，或解释为什么某种温度计不能测量 thermodynamic temperature。

标准解题方法

概念要点

Thermometric property: a physical property that varies with temperature. Examples: density of liquid, volume of gas at constant pressure, resistance of metal, e.m.f. of thermocouple.

Thermodynamic temperature: independent of the property of any particular substance.

评分标准

评分

• B1/B2: 列举测温性质（每个1分）
• B1: depends on properties of real substance
• B1: 0 °C is not absolute zero

完整原题

Example 1 — 9702_w22_qp_41 (5 marks):

(a) State two other physical properties of materials, apart from the density of a liquid, that can be used for measuring temperature. (c)(i) Explain why the thermometer does not provide a direct measurement of thermodynamic temperature. (ii) Name the type of substance for which $T$ is proportional to the product of pressure and volume.

Details

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MS:

• B1: resistance of a metal
• B1: volume of a gas at constant pressure / e.m.f. of a thermocouple
• B1: depends on properties of a real substance
• B1: 0 °C is not absolute zero
• B1: ideal gas

Example 2 — 9702_s24_qp_41 (4 marks):

(a)(i) State the magnitude and unit of absolute zero on the thermodynamic temperature scale. (ii) Explain why temperature measured using a laboratory liquid-in-glass thermometer does not give a direct measurement of thermodynamic temperature. (b)(iii) Suggest a type of thermometer suitable for measuring a rapidly changing temperature.

Details

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MS:

• B1: 0 K
• B1: -273 °C
• B1: depends on property of liquid / depends on property of a real substance
• B1: thermocouple

常见陷阱

注意

• 列举测温性质时不要重复
• "ideal gas" 是 thermodynamic temperature 的标准物质
• 温度计测量需要达到热平衡（需要时间）

Question Type 2: 比热容计算

如何识别

给出加热功率、质量、温度变化，求比热容。

标准解题方法

方法

1. $Q = Pt$（电能转化为热能）
2. $Q = mc\Delta\theta$
3. 联立求解
4. 注意热损失对结果的影响

评分标准

评分

• C1: $Q = Pt$ 或 $Q = mc\Delta\theta$
• C1: $Q_{\text{lost}} = Q_{\text{gained}}$
• A1: 数值正确

完整原题

Example 1 — 9702_s23_qp_41 (8 marks):

A beaker of mass 42 g and specific heat capacity 0.84 J g$^{-1}$ K$^{-1}$ contains 180 g of liquid. An electrical heater of power 810 W heats the liquid. The temperature rises from 22.5 °C to 35.0 °C in 28 s. Determine the specific heat capacity of the liquid.

Details

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MS:

• C1: $Q = Pt = 810 \times 28$
• C1: $= 22680$ J
• C1: $Q = m_l c_l \Delta\theta + m_b c_b \Delta\theta$
• C1: $22680 = 180 \times c_l \times 12.5 + 42 \times 0.84 \times 12.5$
• A1: $c_l = 10.1 - 0.352 = 9.8$ J g$^{-1}$ K$^{-1}$ (需约 4 s.f.)
• B1: 最终答案带单位

Example 2 — 9702_w22_qp_41 (4 marks):

A mercury thermometer initially at 23.0 °C is inserted into water at 37.4 °C. Mass of water = 18.7 g, specific heat capacity of water = 4.18 J g$^{-1}$ K$^{-1}$. Mass of mercury = 6.94 g, specific heat capacity of mercury = 0.140 J g$^{-1}$ K$^{-1}$. The glass has negligible heat capacity. Calculate the final steady temperature.

Details

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MS:

• C1: $Q = mc\Delta\theta$
• C1: $Q_{\text{lost by water}} = Q_{\text{gained by mercury}}$
• C1: $18.7 \times 4.18 \times (37.4 - T) = 6.94 \times 0.140 \times (T - 23.0)$
• A1: $T = 37.1$ °C

常见陷阱

注意

• 容器也吸热！
• 温度变化 $\Delta\theta$ 在 K 和 °C 中数值相同
• 注意单位一致：J g$^{-1}$ K$^{-1}$ 还是 J kg$^{-1}$ K$^{-1}$
• 热损失导致计算值偏小

Question Type 3: 比潜热

如何识别

涉及相变（熔解、汽化）的热量计算。

标准解题方法

方法

1. 相变时温度不变
2. $Q = mL$
3. 如包含升温和相变，分两步

评分标准

评分

• B1: definition: energy per unit mass (during change of state)
• C1: $Q = mL$
• A1: 数值正确

完整原题

Example 1 — 9702_s22_qp_42 (4 marks):

(a) Define specific latent heat of vaporisation. (b) The specific latent heat of vaporisation of water at atmospheric pressure is $2.3 \times 10^6$ J kg$^{-1}$. Calculate the thermal energy required to vaporise 0.015 kg of water at 100 °C.

Details

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MS:

• M1: energy per unit mass
• A1: during change of state (at constant temperature)
• C1: $Q = mL$
• C1: $= 0.015 \times 2.3 \times 10^6$
• A1: $= 3.45 \times 10^4$ J

常见陷阱

注意

• 相变时温度不变
• 区分 fusion（固液）和 vaporisation（液气）
• 定义题必须包含 "during change of state" 或 "at constant temperature"