# 热力学第二定律

> AP 物理 2 · 第2单元：热力学
> 来源: https://www.owlsprep.com/zh/study/ap-physics-2-u2-second-law-of-thermodynamics/

本模块涵盖热力学第二定律的等价表述、熵变计算、热机的最大卡诺效率，以及符合AP物理2考纲要求的自发过程判断规则。

**先修:** [热力学第一定律和能量守恒](https://www.owlsprep.com/zh/study/ap-physics-2-u2-first-law-of-thermodynamics/); 热量、功和绝对温度的定义; 循环过程的理想气体性质

## 学习目标

- 阐述热力学第二定律的多种等价表述
- 计算系统和环境的熵变
- 计算热机效率和最大卡诺效率
- 判断过程是否符合热力学第二定律

## 热力学第二定律核心导论

热力学第一定律只要求能量守恒，但它无法解释为什么某些过程（比如热量从低温物体流向高温物体，或者破碎的杯子自动复原）永远不会自发发生。热力学第二定律填补了这一空白，定义了所有物理过程的自然方向。

在AP物理2考试中，该内容占总分的2-4%，既会出现在选择题（自发性概念题）中，也会作为热机相关大题的子问题出现在自由解答题中。

**热力学第二定律** — 定义自发过程允许方向的基本物理定律，指出对于任何物理过程，宇宙的总熵永不减少。

## 熵与热力学第二定律的熵表述

**熵** — 描述系统粒子和能量的可能微观排布（微观态）数量的状态函数，直观上对应系统的无序或混乱程度。

*记法:* S

对于任何恒温下的可逆过程（等温过程），熵变可简化为:

$$\Delta S = \frac{Q}{T}$$

where $Q$ is the total heat transferred to the system, and $T$ is the constant absolute temperature. The most general statement of the second law, in terms of total entropy change of the universe (system + surroundings), is:

$$\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \geq 0$$

Spontaneous processes have $\boxed{\Delta S_{\text{univ}} > 0}$, ideal reversible processes have $\boxed{\Delta S_{\text{univ}} = 0}$, and no process can have $\boxed{\Delta S_{\text{univ}} < 0}$. Because entropy is a state function, $\Delta S$ only depends on start and end states, not the path taken.

**例题:** 0.80 mol of an ideal gas undergoes a reversible isothermal compression at 290 K. The gas releases 1800 J of heat to the surroundings during the compression. Calculate $\Delta S_{\text{sys}}$, $\Delta S_{\text{surr}}$, $\Delta S_{\text{univ}}$, and state if this process violates the second law.

1. First, define signs: the system releases heat, so $Q_{\text{sys}} = -1800$ J, and $T = 290$ K for both system and surroundings.
2. Calculate $\Delta S_{\text{sys}}$:
3. $$\Delta S_{\text{sys}} = \frac{Q_{\text{sys}}}{T} = \frac{-1800}{290} \approx -6.21 \text{ J/K}$$
4. The surroundings gain 1800 J, so $Q_{\text{surr}} = +1800$ J, so:
5. $$\Delta S_{\text{surr}} = \frac{+1800}{290} \approx +6.21 \text{ J/K}$$
6. Calculate total entropy change:
7. $$\Delta S_{\text{univ}} = -6.21 + 6.21 = 0 \text{ J/K}$$
8. A total entropy change of zero is allowed for an ideal reversible process, so this process does not violate the second law.

> **考试提示:** When calculating entropy change of the surroundings, always explicitly flip the sign of $Q$ from the system. If the system gains heat, the surroundings lose it, so their entropy changes in the opposite direction.

## Heat Engines and Macroscopic Statements

Two common macroscopic statements of the second law describe practical devices:

- **Kelvin-Planck Statement**: No heat engine can convert 100% of input heat to useful work in a cycle.
- **Clausius Statement**: Heat cannot spontaneously flow from a cooler body to a hotter body without work input.

A heat engine operates between two thermal reservoirs: a hot reservoir at $T_H$ that supplies heat $Q_H$, and a cold reservoir at $T_C$ that accepts waste heat $Q_C$. For a full cycle, the engine returns to its original state so $\Delta U_{\text{engine}} = 0$. By the first law, net work output is:

$$W = Q_H - |Q_C|$$

Efficiency $e$ of the engine is the ratio of useful work output to heat input:

$$e = \frac{W}{Q_H} = 1 - \frac{|Q_C|}{Q_H}$$

**例题:** A prototype heat engine absorbs 1500 J of heat from a hot reservoir at 600 K, does 500 J of work, and expels 1000 J of waste heat to a cold reservoir at 300 K. Is this engine allowed by the second law?

1. Calculate entropy change of each reservoir: the hot reservoir loses 1500 J, so:
2. $$\Delta S_H = \frac{-1500}{600} = -2.5 \text{ J/K}$$
3. The cold reservoir gains 1000 J, so:
4. $$\Delta S_C = \frac{+1000}{300} \approx +3.33 \text{ J/K}$$
5. The engine completes a full cycle, so its entropy change $\Delta S_{\text{engine}} = 0$, because entropy is a state function.
6. Total entropy change of the universe:
7. $$\Delta S_{\text{univ}} = -2.5 + 3.33 + 0 = +0.83 \text{ J/K}$$
8. Since $\Delta S_{\text{univ}} > 0$, this engine is allowed by the second law.

> **考试提示:** For any full-cycle process, the entropy change of the engine itself is always zero—don't accidentally add a non-zero entropy change for the engine to the total.

## Maximum Carnot Efficiency

The most efficient possible heat engine operating between two fixed temperatures $T_H$ and $T_C$ is the ideal reversible Carnot engine. The Carnot efficiency sets an absolute upper limit that no real engine can exceed, per the second law. The formula is:

$$e_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$$

Critical note: $T_H$ and $T_C$ *must* be absolute temperatures in Kelvin, not Celsius. 100% efficiency ($e=1$) is only possible if $T_C = 0$ K, which is physically impossible, consistent with the second law.

**例题:** A natural gas power plant has a combustion chamber temperature of 1100°C and discharges waste heat to cooling towers at 40°C. What is the maximum possible efficiency this plant can achieve?

1. Convert temperatures to Kelvin:
2. $$T_H = 1100 + 273 = 1373 \text{ K}, \quad T_C = 40 + 273 = 313 \text{ K}$$
3. Substitute into the Carnot efficiency formula:
4. $$e_{\text{max}} = 1 - \frac{313}{1373}$$
5. Calculate the final result:
6. $$\frac{313}{1373} \approx 0.228, \quad e_{\text{max}} = 1 - 0.228 = 0.772 = 77.2\%$$
7. This matches expectations: real plants typically achieve 50-60% efficiency, which is below the Carnot limit.

> **考试提示:** Always convert Celsius temperatures to Kelvin before plugging into the Carnot efficiency formula. Using Celsius directly will often give you a negative or impossible efficiency, which is an immediate red flag.

## AP Style Concept Check

**概念自测**

Test your understanding of core second law rules:

1. Which of the following processes is allowed by the second law of thermodynamics?

   - A process with $\Delta S_{\text{sys}} = -10$ J/K and $\Delta S_{\text{surr}} = +8$ J/K
   - A heat engine operating between 500 K and 200 K with an efficiency of 65%
   - A process that decreases the total entropy of the universe
   - An ice cube freezing in a 0°C freezer, where the freezer uses work input to move heat from the ice to the room

   *解析:* Option A has total $\Delta S_{\text{univ}} = -2 < 0$ (violates second law). Option B's maximum Carnot efficiency is 60%, so 65% exceeds the allowed limit. Option C directly violates the second law. Option D is allowed: the entropy increase of the surroundings (room) outweighs the entropy decrease of the ice, so total $\Delta S_{\text{univ}} > 0$.

**例题:** 1.00 mol of an ideal gas undergoes a spontaneous irreversible free expansion into a vacuum, doubling its volume at constant temperature 300 K. (a) What is the change in internal energy of the gas? Justify. (b) Calculate the entropy change of the system. (c) Calculate total entropy change and state if the process is spontaneous.

1. (a) For an ideal gas, internal energy depends only on temperature. The expansion is constant temperature, so $\Delta T = 0$, meaning $\Delta U = 0$. No work is done in free expansion, so the first law confirms $Q=0$.
2. (b) Entropy is a state function, so $\Delta S$ equals the change for a reversible isothermal expansion between the same states. For reversible expansion, $Q = nRT \ln(V_2/V_1)$, so $\Delta S_{\text{sys}} = Q/T = nR \ln(V_2/V_1)$:
3. $$\Delta S_{\text{sys}} = (1.00)(8.31)(\ln 2) \approx 5.76 \text{ J/K}$$
4. (c) No heat is exchanged with the surroundings, so $Q_{\text{surr}} = 0$, $\Delta S_{\text{surr}} = 0$. Total entropy change:
5. $$\Delta S_{\text{univ}} = 5.76 + 0 = +5.76 \text{ J/K}$$
6. Since $\Delta S_{\text{univ}} > 0$, the process is spontaneous, as expected.

## 常见错误

- **错误做法:** Using Celsius temperatures directly in the Carnot efficiency formula instead of converting to Kelvin
  - 原因: Most problems give temperatures in Celsius, and students forget the formula requires absolute temperature
  - 正确做法: Always add 273 to any Celsius temperature before plugging into the Carnot efficiency formula
- **错误做法:** Claiming a process with $\Delta S_{\text{sys}} < 0$ is impossible per the second law
  - 原因: Students memorize "entropy increases" but forget this applies to the total entropy of the universe, not just the system
  - 正确做法: Always calculate $\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$ before judging if a process is allowed
- **错误做法:** Calculating heat engine efficiency as $e = W/Q_C$ instead of $e = W/Q_H$
  - 原因: Students mix up input heat from the hot reservoir and waste heat to the cold reservoir
  - 正确做法: Remember: efficiency is work you get out divided by heat you put in, and input heat always comes from the hot reservoir ($Q_H$)
- **错误做法:** Forgetting to flip the sign of $Q$ when calculating $\Delta S_{\text{surr}}$
  - 原因: Students carry the system's $Q$ sign directly over to the surroundings, flipping the sign of the total entropy change
  - 正确做法: Explicitly write $Q_{\text{surr}} = -Q_{\text{sys}}$ before calculating $\Delta S_{\text{surr}}$ for any process
- **错误做法:** Claiming a reversible Carnot engine can have 100% efficiency
  - 原因: Students confuse reversibility with zero total entropy change, but forget the temperature requirement
  - 正确做法: Remember even ideal Carnot engines only reach 100% efficiency if $T_C = 0$ K, which is physically impossible, so all real engines have $e < 1$

## 速查表

| Category | Formula | Notes |
| --- | --- | --- |
| Entropy change (reversible isothermal) | $\Delta S = \frac{Q}{T}$ | $Q$ = heat added to system, $T$ = absolute temperature (K) |
| Second Law (entropy form) | $\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \geq 0$ | $>0$ = spontaneous, $=0$ = ideal reversible |
| Heat engine net work | $W = Q_H - \|Q_C\|$ | Full cycle: $\Delta U_{\text{engine}} = 0$ |
| Heat engine efficiency | $e = \frac{W}{Q_H} = 1 - \frac{\|Q_C\|}{Q_H}$ | Always $e < 1$ for real engines |
| Maximum (Carnot) efficiency | $e_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$ | Temperatures must be in Kelvin, upper limit for all engines |
| Entropy change for full cycle | $\Delta S_{\text{cycle}} = 0$ | Entropy is a state function, returns to starting value |
| Clausius Statement | N/A | No device can transfer heat cold to hot with no work input |

## 下一步

The second law of thermodynamics is a foundational concept that extends beyond thermodynamics to all areas of physics, explaining why all real processes are irreversible and why perpetual motion machines are impossible. After mastering this sub-topic, you have completed the core thermodynamics content for AP Physics 2 Unit 2, and are ready to move on to applied topics. Understanding the second law also helps you contextualize real-world energy conversion systems like power plants and refrigerators, which are common FRQ contexts on the AP exam, so be sure to practice calculating entropy changes and Carnot efficiency before test day.

- [Unit 2: Thermodynamics Overview](https://www.owlsprep.com/zh/study/ap-physics-2-u2-overview/)
- [熵](https://www.owlsprep.com/zh/study/ap-physics-2-u2-entropy/)
- [电力、电场与电势](https://www.owlsprep.com/zh/study/ap-physics-2-u3-overview/)

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