Temperature scales · 温标
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| thermometer/θɜːˈmɒmɪtə/ | 温度计 | wēn dù jì |
| thermocouple/ˈθɜːməkʌpl/ | 热电偶 | rè diàn ǒu |
| thermodynamic/ˌθɜːməʊdaɪˈnæmɪk/ | 热力学 | rè lì xué |
| kelvin/ˈkelvɪn/ | 开尔文 | kāi'ěrwén |
| absolute zero/ˈæbsəluːt ˈzɪərəʊ/ | 绝对零度 | jué duì líng dù |
| Celsius/ˈselsɪəs/ | 摄氏度 | shè shì dù |
How a thermometer 温度计 works
- A thermometer needs something that changes with temperature.
- A liquid expands, a metal's resistance rises, a gas swells.
- Read that change and you read the temperature.
温度计是怎么工作的
- 温度计需要某种随温度 变化 的东西。
- 液体膨胀,金属的电阻升高,气体胀大。
- 读出这种变化,你就读出了温度。
Temperature scale lab · 温标实验室
Kelvin index = Celsius index + 2.73 · 开尔文索引 = 摄氏索引 + 2.73
Slide Celsius temperature and see the Kelvin scale shift by 273. · 滑动摄氏温度并观察开尔文刻度如何平移 273。
Kinds of thermometer
- Anything that changes repeatably with temperature works:
- expansion of a liquid, resistance of a metal, e.m.f. of a thermocouple 热电偶, volume of a gas.
A thermometer measures temperature on a defined scale
各种温度计
- 任何随温度 可重复 变化的东西都可以:
- 液体 的膨胀、金属的 电阻、热电偶(thermocouple)的 电动势、气体的 体积。
温度计在规定的标度上测量温度
Select all · 所有 the properties that can be used to measure temperature. · 选择 所有 可用于测量温度的属性。
Any property that changes repeatably with temperature works. The fixed colour of the case does not. · 任何随温度可重复变化的属性均可使用。外壳固定的颜色不行。
The absolute scale
- The thermodynamic 热力学 (absolute) scale uses the kelvin 开尔文 (K) and depends on no single substance.
- Absolute zero 绝对零度 ($0\ \text{K}$) is the lowest possible temperature — least internal energy. Nothing is colder.
绝对温标
- 热力学(thermodynamic)(绝对) 温标使用 开尔文(kelvin)(K),不依赖于任何单一物质。
- 绝对零度(absolute zero)($0\ \text{K}$)是可能的最低温度——内能最小。没有比它更冷的。
Absolute zero (0 K) is: · 绝对零度 (0 K) 是：
At 0 K a system has its least possible internal energy; nothing can be made colder. · 在 0 K 时，系统具有最小的可能内能；无法变得更冷。
Kelvin and Celsius 摄氏度
- $T/\text{K} = \theta/°\text{C} + 273.15$. So $0\ °\text{C} = 273\ \text{K}$.
- A kelvin and a Celsius degree are the same size, so a change of $1\ \text{K} = 1\ °\text{C}$.
- Always use kelvin in gas-law calculations.
Extrapolating the pressure-temperature line back to zero pressure gives absolute zero, about -273 degC
开尔文与摄氏度
- $T/\text{K} = \theta/°\text{C} + 273.15$。所以 $0\ °\text{C} = 273\ \text{K}$。
- 一开尔文和一摄氏度 大小相同,所以 变化量 $1\ \text{K} = 1\ °\text{C}$。
- 气体定律计算中总是用 开尔文。
把压强–温度直线外推到压强为零,得到绝对零度约 −273 °C
What is $27\ °\text{C}$ in kelvin? · $27\ °\text{C}$ 等于多少开尔文？
$T/\text{K} = \theta/°\text{C} + 273 = 27 + 273 = 300\ \text{K}$.
A temperature rise of 1 K is the same as a rise of 1 °C. · 升高 1 K 等同于升高 1 °C。
The two scales have the same size of degree — only their zero points differ (by 273.15). · 两个刻度的分度值大小相同——仅零点不同（相差 273.15）。
In gas-law calculations, the temperature must be in: · 在气体定律计算中，温度必须是：
Gas laws use absolute temperature — kelvin. Using °C gives wrong answers. · 气体定律使用绝对温度——开尔文。使用 °C 会得到错误答案。
A temperature is 27 degrees Celsius. What is it in kelvin? · 某温度是 27 摄氏度。换算成开尔文是多少？
T = 27 + 273 = 300 K. A temperature DIFFERENCE, though, is the same number in both scales: a rise of 27 C is a rise of 27 K. · T = 27 + 273 = 300 K。不过，一个温度差值在两个刻度上是相同的数值：升高 27 C 等同于升高 27 K。
Why the thermodynamic scale is special
- It does not depend on the property of any particular substance — it is defined from the behaviour of an ideal gas, with absolute zero and the triple point of water as its fixed points.
- A liquid-in-glass thermometer assumes its liquid expands uniformly with temperature. That is only approximately true, so it agrees with the thermodynamic scale exactly only at the points where it was calibrated.
- That is the exam's "explain why a liquid-in-glass thermometer does not measure thermodynamic temperature".
热力学温标为什么特殊
- 它 不 依赖于任何特定物质的性质——它由理想气体的行为定义,以绝对零度和水的三相点为固定点。
- 液体温度计 假设 它的液体随温度均匀膨胀。这只是近似成立,所以它只在校准过的那些点上与热力学温标完全一致。
- 这就是考试题"解释为什么液体温度计测的不是热力学温度"的答案。
What makes the thermodynamic scale special? Select all · 所有 that apply. · 什么使热力学温标特别？选择 所有 适用的选项。
The unit is a consequence, not the defining feature. A mercury thermometer's scale depends on how mercury happens to expand; this one depends on nothing of the sort. · 单位是结果而非定义特征。水银温度计的刻度取决于水银如何膨胀；而这个温标与此无关。
Choosing a thermometer
- Platinum resistance: resistance varies linearly and repeatably with temperature over a wide range — very accurate. But it has a large thermal capacity, so it is slow and cannot follow a rapidly changing temperature.
- Thermocouple: two metal junctions give an e.m.f.; tiny, fast, wide range, measures at a point — ideal for a changing temperature.
- Liquid in glass: cheap and simple; limited range, slow, and reads only where the bulb is.
选择温度计
- 铂电阻: 在很宽的范围内电阻随温度 线性 且可重复地变化——非常精确。但它的热容量大,所以 反应慢,跟不上快速变化的温度。
- 热电偶: 两个金属接点产生电动势;很小、快、范围宽、测的是一个点——最适合变化的温度。
- 液体温度计: 便宜简单;范围有限、慢,而且只能读出球泡所在处的温度。
Match each thermometer to its key feature. · 将每种温度计与其关键特性匹配。
A large thermal capacity makes the resistance thermometer accurate but slow; a thermocouple's tiny junction responds fast; the absolute scale is defined from ideal-gas behaviour, not a material. · 较大的热容量使电阻温度计准确但响应慢；热电偶微小的接点响应快；绝对温标由理想气体行为定义，而非某种材料。
Match each thermometer to the property it uses. · 将每种温度计与其使用的属性匹配。
A thermocouple has a small thermal capacity and responds fast, which is why it is chosen for a rapidly changing temperature. · 热电偶的热容量小且响应快，因此被选用于测量快速变化的温度。
Worked example: finding absolute zero, then converting
A fixed volume of gas has pressure $100\ \text{kPa}$ at $0\ °\text{C}$ and $136.6\ \text{kPa}$ at $100\ °\text{C}$. Then: a gas is heated from $27\ °\text{C}$ to $742\ °\text{C}$.
- Gradient: $\dfrac{136.6 - 100}{100} = 0.366\ \dfrac{\text{kPa}}{°\text{C}}$; pressure is a straight line in temperature.
- Extrapolate to zero pressure: $\theta = -\dfrac{100}{0.366} = -273\ °\text{C}$ — absolute zero.
- Convert: $27\ °\text{C} = 300\ \text{K}$; $742\ °\text{C} = 1015\ \text{K}$.
- Temperature change: $742 - 27 = 715\ °\text{C} = 715\ \text{K}$ — a difference is the same number on both scales.
- Check: absolute zero is where the pressure of an ideal gas would fall to zero; the gas would liquefy long before, which is why the point is found by extrapolation, not by cooling.
例题:先找绝对零度,再换算
定容气体在 $0\ °\text{C}$ 时压强为 $100\ \text{kPa}$,在 $100\ °\text{C}$ 时为 $136.6\ \text{kPa}$。然后:一份气体从 $27\ °\text{C}$ 加热到 $742\ °\text{C}$。
- 斜率: $\dfrac{136.6 - 100}{100} = 0.366\ \dfrac{\text{kPa}}{°\text{C}}$;压强随温度是一条直线。
- 外推到压强为零: $\theta = -\dfrac{100}{0.366} = -273\ °\text{C}$——绝对零度。
- 换算: $27\ °\text{C} = 300\ \text{K}$;$742\ °\text{C} = 1015\ \text{K}$。
- 温度变化: $742 - 27 = 715\ °\text{C} = 715\ \text{K}$——差值在两种温标上是同一个数。
- 检查: 绝对零度是理想气体压强将降到零的温度;真实气体早就液化了,所以这个点是通过 外推 而不是通过冷却找到的。
A gas is heated from $20\ °\text{C}$ to · 到 $95\ °\text{C}$. What is the temperature change, in K? · 气体从 $20\ °\text{C}$ 加热到 $95\ °\text{C}$。温度变化是多少开尔文 (K)？
A change is the same number on both scales: $95 - 20 = 75\ °\text{C} = 75\ \text{K}$. Adding 273 would be wrong here. · 两个温标上相同的数值是：$95 - 20 = 75\ °\text{C} = 75\ \text{K}$。在此处加 273 是错误的。
Add $273$ to convert a Celsius temperature to kelvin, but a temperature difference is the same number on both scales — do not add $273$ to a $\Delta T$. A kelvin temperature can never be negative. And absolute zero is the temperature of minimum internal energy, not zero energy — "the molecules stop moving" is not an accepted answer.
把摄氏 温度 换算成开尔文要加 $273$,但温度 差 在两种温标上是同一个数——不要给 $\Delta T$ 加 $273$。开尔文温度永远不可能为负。而且绝对零度是内能 最小 的温度,不是能量为零——"分子停止运动"不是被接受的答案。
Absolute zero is the temperature at which a substance has its ____ internal energy. · 绝对零度是指物质具有 ____ 内能时的温度。
Not zero energy — the minimum possible. That is why "the molecules stop moving" is not accepted. · 不是零能量——而是可能的最小值。这就是为什么“分子停止运动”的说法不被接受。
You've got it
- a thermometer uses any property that changes repeatably with temperature; the thermodynamic scale depends on no substance
- absolute zero $= 0\ \text{K} = -273.15\ °\text{C}$, the lowest possible temperature (minimum internal energy)
- $T/\text{K} = \theta/°\text{C} + 273$; a difference is the same in K and °C; use kelvin in gas laws
你掌握了
- 温度计利用任何随温度 可重复 变化的性质;热力学温标 不 依赖于任何物质
- 绝对零度 $= 0\ \text{K} = -273.15\ °\text{C}$,可能的最低温度(内能最小)
- $T/\text{K} = \theta/°\text{C} + 273$;差值 在 K 和 °C 中相同;气体定律用 开尔文