Thermal equilibrium · 热平衡
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| heat/hiːt/ | 热量 | rè liàng |
| thermal energy/ˈθɜːml ˈenədʒi/ | 热能 | rè néng |
| thermal equilibrium/ˈθɜːml ˌiːkwɪˈlɪbrɪəm/ | 热平衡 | rè píng héng |
| net flow/net fləʊ/ | 净流动 | jìng liú dòng |
A spoon in hot soup
- Put a cold spoon in hot soup: the spoon warms, the soup cools — until they match.
- Energy moved from the hot thing to the cold thing.
- That one-way flow is the key idea of temperature.
热汤里的勺子
- 把一把冷勺子放进热汤:勺子变热,汤变凉——直到两者一样。
- 能量从热的东西转移到了冷的东西。
- 这种单向的流动是温度的核心思想。
Thermal equilibrium route · 热平衡路径
Watch energy transfer until two objects reach the same temperature. · 观察能量传递直到两个物体达到相同温度。
Heat 热量 flows hot to cold
- Heat (thermal energy 热能) flows from higher to lower temperature.
- It keeps flowing until the temperatures are equal.
热量从热流向冷
- 热量(heat)(热能)从 较高 温度流向 较低 温度。
- 它持续流动,直到温度相等。
Heat (thermal energy) naturally flows from: · 热量（热能）自然从：
Energy always flows down the temperature difference, from hot to cold, until they are equal. · 能量总是顺着温差流动，从热到冷，直到两者相等。
Thermal equilibrium 热平衡
- When two things reach the same temperature, they are in thermal equilibrium.
- There is then no net flow 净流动 of energy between them.
A thermal (infrared) camera turns temperature into colour: the hot fries glow bright orange, while the cold drink stays dark
热平衡
- 当两个物体达到 相同温度 时,它们处于 热平衡(thermal equilibrium)。
- 此时它们之间 没有净流动 的能量。
热成像(红外)相机把温度变成颜色:热薯条发出亮橙色,冷饮保持深色
At thermal equilibrium there is no net flow of energy between two bodies. · 在热平衡状态下，两个物体之间没有净能量流动。
Equal temperatures mean no net flow — particles still exchange energy, but with no overall transfer. · 温度相等意味着没有净流动——粒子仍在交换能量，但总体上传递量为零。
Two objects in thermal equilibrium are at the same ____. · 处于热平衡的两个物体具有相同的 ____。
Same temperature is exactly the condition for thermal equilibrium. · 相同的温度正是热平衡的条件。
Match each term to the definition the examiner marks. · 将每个术语与考官标记的定义匹配。
Thermal equilibrium is defined by NO NET transfer, not by no transfer at all. Energy still crosses in both directions; the two flows simply match. · 热平衡是由“无净传递”定义的，而非完全不传递。能量仍双向穿过；两股流量仅仅相互抵消。
Temperature is not energy
- Temperature decides the direction of heat flow — not how much energy a body holds.
- A tiny spark at $1000\ °\text{C}$ holds far less energy than a warm swimming pool.
温度不是能量
- 温度决定热流动的 方向——而不是物体含有多少能量。
- $1000\ °\text{C}$ 的一颗小火花所含的能量远少于一个温暖的游泳池。
Temperature tells you: · 温度告诉你：
Temperature decides the direction of heat flow; the amount of energy also depends on mass and material. · 温度决定热流方向；能量的多少还取决于质量和物质种类。
A hot spark always contains more thermal energy than a warm swimming pool. · 一颗炽热的火花总是比一个温水游泳池含有更多的热能。
No — the spark is at a higher temperature, but the huge pool holds far more thermal energy overall. · 不——火花温度更高，但巨大的游泳池总体上储存着多得多的热能。
A bucket of warm water contains more internal energy than a red-hot nail, so it is at a higher temperature. · 一桶温水含有的内能比一根红热的铁钉多，所以它的温度更高。
Temperature is not energy. The bucket holds far more internal energy in total, yet energy flows from the nail to the water, which is what temperature actually decides. · 温度不是能量。水桶总共含有多得多的内能，但能量是从铁钉流向水的，这才是温度实际决定的。
Saying it the exam way
- "Two objects in thermal equilibrium are at the same temperature, so there is no net transfer of thermal energy between them." Both halves score.
- Energy still passes both ways between them — at equal rates, so the net flow is zero.
- A thermometer works by reaching thermal equilibrium with what it touches, which is why it must be left long enough to settle.
用考试的方式说
- "处于热平衡的两个物体 温度相同,所以它们之间 没有热能的净转移。"两个部分都得分。
- 能量仍然在它们之间 双向 传递——速率相等,所以净流动为零。
- 温度计的工作原理是与它接触的物体达到热平衡,所以必须放置足够长的时间让它稳定下来。
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Two objects are in thermal equilibrium. Which statements are true? Select all that apply. · 两个物体处于热平衡。哪些陈述是正确的？选择所有适用的选项。
Equilibrium means equal temperature and zero net flow. Energy still passes both ways at equal rates, and the two objects can hold very different amounts of energy. · 平衡意味着温度相等且净流量为零。能量仍以相等的速率双向传递，且两个物体可以储存截然不同的能量量。
Worked example: mixing hot and cold water
$0.20\ \text{kg}$ of water at $80\ °\text{C}$ is poured into $0.50\ \text{kg}$ of water at $20\ °\text{C}$ in an insulated cup. Find the final temperature $T$. Take $c = 4200\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$.
- Energy lost by the hot water = energy gained by the cold water, since none escapes.
- $0.20 \times 4200 \times (80 - T) = 0.50 \times 4200 \times (T - 20)$; the $4200$ cancels.
- $16 - 0.20T = 0.50T - 10$, so $0.70T = 26$ and $T = 37\ °\text{C}$.
- Check: the answer lies between $20$ and $80$, and nearer the cold water because there is more of it. Write each $\Delta T$ as a positive number on its own side.
例题:热水和冷水混合
把 $0.20\ \text{kg}$、$80\ °\text{C}$ 的水倒入绝热杯中 $0.50\ \text{kg}$、$20\ °\text{C}$ 的水里。求末温度 $T$。取 $c = 4200\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$。
- 热水失去的能量 = 冷水获得的能量,因为没有能量逃逸。
- $0.20 \times 4200 \times (80 - T) = 0.50 \times 4200 \times (T - 20)$;$4200$ 消去。
- $16 - 0.20T = 0.50T - 10$,所以 $0.70T = 26$,$T = 37\ °\text{C}$。
- 检查: 答案在 $20$ 和 $80$ 之间,更接近冷水,因为冷水更多。每个 $\Delta T$ 在自己那一边写成 正 数。
$0.30\ \text{kg}$ of water at $90\ °\text{C}$ is mixed with $0.60\ \text{kg}$ of water at $15\ °\text{C}$ in an insulated container. What is the final temperature, in °C? · 将 $0.30\ \text{kg}$ 温度为 $90\ °\text{C}$ 的水与 $0.60\ \text{kg}$ 温度为 $15\ °\text{C}$ 的水混合在绝热容器中。最终温度是多少 °C？
$0.30(90 - T) = 0.60(T - 15)$: $27 - 0.30T = 0.60T - 9$, so $0.90T = 36$ and $T = 40\ °\text{C}$. · $0.30(90 - T) = 0.60(T - 15)$：$27 - 0.30T = 0.60T - 9$，所以 $0.90T = 36$ 和 $T = 40\ °\text{C}$。
200 g of water at 80 C is mixed with 300 g at 20 C. What is the final temperature, in C? · 将 200 g 80°C 的水与 300 g 20°C 的水混合。最终温度是多少 °C？
Energy lost equals energy gained: 200(80 - T) = 300(T - 20), so T = 44 C. The specific heat capacity cancels because both are water. · 损失的能量等于获得的能量：200(80 - T) = 300(T - 20)，解得 T = 44°C。比热容被约去因为两者都是水。
Worked example: the beaker takes its share
An $810\ \text{W}$ heater warms $120\ \text{g}$ of a liquid in a $42\ \text{g}$ glass beaker from $25\ °\text{C}$ to $80\ °\text{C}$ in $21\ \text{s}$. The specific heat capacity of glass is $0.84\ \dfrac{\text{J}}{\text{g}\,\text{K}}$. Find $c$ for the liquid.
- Energy supplied: $Q = Pt = 810 \times 21 = 17\,000\ \text{J}$.
- Energy into the beaker: $42 \times 0.84 \times 55 = 1940\ \text{J}$ — it warms through the same $55\ \text{K}$.
- Energy into the liquid: $17\,010 - 1940 = 15\,070\ \text{J}$.
- Specific heat capacity: $c = \dfrac{15\,070}{120 \times 55} = 2.3\ \dfrac{\text{J}}{\text{g}\,\text{K}}$.
- Check: the beaker is in thermal equilibrium with the liquid throughout, so it ends at the same temperature and must be counted. Forgetting it gives $2.6$, which looks plausible and is wrong.
例题:烧杯也分走一份
一个 $810\ \text{W}$ 的加热器在 $21\ \text{s}$ 内把 $42\ \text{g}$ 玻璃烧杯中的 $120\ \text{g}$ 液体从 $25\ °\text{C}$ 加热到 $80\ °\text{C}$。玻璃的比热容为 $0.84\ \dfrac{\text{J}}{\text{g}\,\text{K}}$。求液体的 $c$。
- 供给的能量: $Q = Pt = 810 \times 21 = 17\,000\ \text{J}$。
- 进入烧杯的能量: $42 \times 0.84 \times 55 = 1940\ \text{J}$——它升高了同样的 $55\ \text{K}$。
- 进入液体的能量: $17\,010 - 1940 = 15\,070\ \text{J}$。
- 比热容: $c = \dfrac{15\,070}{120 \times 55} = 2.3\ \dfrac{\text{J}}{\text{g}\,\text{K}}$。
- 检查: 烧杯全程与液体处于热平衡,所以它最终温度相同,必须计入。忘记它会得到 $2.6$,看起来合理却是错的。
Equal temperature is not equal energy: a hot spark and a warm pool. At equilibrium the net flow is zero, not all flow. In any mixing calculation, count every object whose temperature changes — the container included — and use temperature changes in the same units on both sides.
温度相等不等于能量相等:热火花与温水池。平衡时 净 流动为零,而不是所有流动都停止。在任何混合计算中,计入 每一个 温度改变的物体——包括容器——并在两边用同样单位的温度 变化量。
You've got it
- heat flows from higher to lower temperature
- thermal equilibrium: same temperature, no net transfer of thermal energy
- temperature sets the direction of flow, not the amount of energy; mixing problems: energy lost = energy gained, container included
你掌握了
- 热量从 较高温度流向较低温度
- 热平衡:温度相同,没有热能的 净 转移
- 温度决定流动的 方向,而不是能量的多少;混合问题:失去的能量 = 获得的能量,包括容器