Specific heat capacity and latent heat · 比热容与潜热
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| latent heat/ˈleɪtənt hiːt/ | 潜热 | qián rè |
| specific heat capacity/spəˈsɪfɪk hiːt kəˈpæsɪti/ | 比热容 | bǐ rè róng |
| change of state/tʃeɪndʒ ɒv steɪt/ | 状态变化 | zhuàng tài biàn huà |
| specific latent heat/spəˈsɪfɪk ˈleɪtənt hiːt/ | 比潜热 | bǐ qián rè |
| fusion/ˈfjuːʒn/ | 熔化 | róng huà |
| vaporisation/ˌveɪpəraɪˈzeɪʃn/ | 汽化 | qì huà |
Slow to boil, slow to cool
- A metal spoon heats up in seconds; a pan of water takes minutes.
- And boiling water stays stuck at $100\ °\text{C}$ even as it bubbles away.
- Two ideas explain this: heat capacity and latent heat 潜热.
烧得慢,冷得慢
- 金属勺几秒钟就热了;一锅水要几分钟。
- 而沸腾的水即使不断冒泡,也停在 $100\ °\text{C}$。
- 两个概念解释了这些:热容 和 潜热(latent heat)。
Specific heat capacity 比热容
- $c$ = energy to raise 1 kg by 1 K: $Q = mc\Delta T$.
- The exam definition: "the energy required per unit mass to raise the temperature by one kelvin (or one degree)".
- Water's $c \approx 4200\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$ is high — it heats and cools slowly.
Water's specific heat capacity towers over metals' — why it heats and cools so slowly
比热容
- $c$ = 使 1 kg 升高 1 K 所需的能量:$Q = mc\Delta T$。
- 考试定义:"单位质量 升高 一开尔文(或一度)所需的能量"。
- 水的 $c \approx 4200\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$ 很高——它升温和降温都慢。
水的比热容远远高于金属——这就是它升温和降温都慢的原因
Energy to heat it: E = mcΔT · 加热所需能量:E = mcΔT
Pick a material, set the mass and the temperature rise, and read the energy. Water needs far more energy than the metals. · 选一种材料,设置质量和温度升高,读出能量。水比金属需要多得多的能量。
Specific heat capacity · 比热容
Q = mcΔT
The heat needed is proportional to the temperature rise — the gradient depends on mass and the material's specific heat capacity. · 所需热量与温升成正比——斜率取决于质量和材料的比热容。
How much energy raises $2.0\ \text{kg}$ of water by $10\ \text{K}$? (Use $c = 4200\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$.) · 把 $2.0\ \text{kg}$ 的水升高 $10\ \text{K}$ 需要多少能量?(取 $c = 4200\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$。)
$Q = mc\Delta T = 2.0 \times 4200 \times 10 = 84000\ \text{J}$. · $Q = mc\Delta T = 2.0 \times 4200 \times 10 = 84000\ \text{J}$。
Water has a high specific heat capacity compared with most metals. · 与大多数金属相比,水有很高的比热容。
Water ~4200 vs aluminium ~900, copper ~385 J/(kg·K) — water needs much more energy per kelvin. · 水约 4200,而铝约 900、铜约 385 J/(kg·K)——水每开尔文需要多得多的能量。
The heating curve
- Heat steadily and the temperature rises — except during a change of state 状态变化.
- There it stays constant: the energy goes into breaking bonds, not raising temperature.
加热曲线
- 稳定加热,温度上升——除了在 状态变化(change of state) 期间。
- 那时温度保持 恒定:能量用于打破分子间的键,而不是升高温度。
While a solid is melting, its temperature: · 当固体正在熔化时,它的温度:
During a state change the energy breaks bonds rather than raising temperature, so it stays constant. · 在状态变化期间,能量破坏键而不是升高温度,所以它保持恒定。
Specific latent heat 比潜热
- $L$ = energy to change the state of 1 kg at constant temperature: $Q = mL$ — that last phrase is part of the definition.
- Fusion 熔化 $L_{\text{f}}$ (melting); vaporisation 汽化 $L_{\text{v}}$ (boiling).
- $L_{\text{v}} > L_{\text{f}}$ for two reasons: boiling must separate the molecules completely, and the vapour does work pushing back the atmosphere as it expands.
比潜热
- $L$ = 在 恒定温度 下使 1 kg 改变状态所需的能量:$Q = mL$——最后那个短语是定义的一部分。
- 熔化(fusion) $L_{\text{f}}$;汽化(vaporisation) $L_{\text{v}}$。
- $L_{\text{v}} > L_{\text{f}}$ 有两个原因:沸腾必须 完全 分开分子,而且蒸气膨胀时要 做功推开大气。
How much energy melts $0.10\ \text{kg}$ of ice at $0\ °\text{C}$? (Use $L_{\text{f}} = 3.34 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$.) · 熔化 $0\ °\text{C}$ 的 $0.10\ \text{kg}$ 冰需要多少能量?(取 $L_{\text{f}} = 3.34 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$。)
$Q = mL_{\text{f}} = 0.10 \times 3.34 \times 10^{5} = 3.34 \times 10^{4}\ \text{J}$. · $Q = mL_{\text{f}} = 0.10 \times 3.34 \times 10^{5} = 3.34 \times 10^{4}\ \text{J}$。
The latent heat of vaporisation is larger than that of fusion because: · 汽化潜热大于熔化潜热,因为:
Melting only loosens the bonds; boiling separates the particles completely and the vapour expands against the air. · 熔化只是松开键;沸腾把粒子完全分开,而且蒸气对着空气膨胀。
Match each term to the definition the examiner marks. · 把每个术语与评分认可的定义配对。
"At constant temperature" is what separates latent heat from heating, and leaving it out loses the mark even though the formula is right. · "在温度不变时"正是把潜热与升温区分开的说法,漏掉它就丢分,尽管公式是对的。
Multi-step problems
- Warming through a state change splits into steps.
- Use $Q = mc\Delta T$ on each sloped part, and $Q = mL$ at each flat plateau — then add them up.
Split the heating into sloped (mcΔT) and flat (mL) stages, then add the energies
多步骤问题
- 经过状态变化的加热要分成几步。
- 在每段斜线上用 $Q = mc\Delta T$,在每段平台上用 $Q = mL$——然后把它们加起来。
把加热分成斜线(mcΔT)和平台(mL)两种阶段,再把能量相加
During a change of state, the energy is $Q = m$____ (not $mc\Delta T$). · 在状态变化期间,能量是 $Q = m$____(不是 $mc\Delta T$)。
A phase change is at constant temperature, so use $Q = mL$ there; use $mc\Delta T$ for the sloped (temperature-changing) parts. · 相变在恒温下进行,所以那里用 $Q = mL$;倾斜的(温度变化的)部分用 $mc\Delta T$。
Worked example: an ice cube in a drink
A $37.0\ \text{g}$ ice cube at $0.0\ °\text{C}$ is dropped into $250\ \text{g}$ of water at $24.0\ °\text{C}$ in an insulated beaker. Find the final temperature. Take $L_{\text{f}} = 3.34 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$ and $c = 4200\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$.
- Melt the ice first: $mL = 0.037 \times 3.34 \times 10^{5} = 12\,400\ \text{J}$, at $0\ °\text{C}$.
- Then warm the meltwater from $0$ to $T$: $0.037 \times 4200 \times T = 155\,T$.
- The warm water supplies both, cooling from $24$ to $T$: $0.250 \times 4200 \times (24 - T) = 25\,200 - 1050\,T$.
- Balance: $12\,400 + 155\,T = 25\,200 - 1050\,T$, so $1205\,T = 12\,800$ and $T = 10.6\ °\text{C}$.
- Check: the answer is well above $0$ (all the ice melted) and below $24$. Leaving out the melting step gives $21\ °\text{C}$ — the commonest error.
例题:饮料里的冰块
把一块 $37.0\ \text{g}$、$0.0\ °\text{C}$ 的冰块放入绝热烧杯中 $250\ \text{g}$、$24.0\ °\text{C}$ 的水里。求末温度。取 $L_{\text{f}} = 3.34 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$,$c = 4200\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$。
- 先把冰熔化: $mL = 0.037 \times 3.34 \times 10^{5} = 12\,400\ \text{J}$,在 $0\ °\text{C}$。
- 再把融水加热 从 $0$ 到 $T$:$0.037 \times 4200 \times T = 155\,T$。
- 温水提供这两部分,从 $24$ 降到 $T$:$0.250 \times 4200 \times (24 - T) = 25\,200 - 1050\,T$。
- 平衡: $12\,400 + 155\,T = 25\,200 - 1050\,T$,所以 $1205\,T = 12\,800$,$T = 10.6\ °\text{C}$。
- 检查: 答案远高于 $0$(冰全部熔化)且低于 $24$。漏掉熔化这一步会得到 $21\ °\text{C}$——最常见的错误。
$20\ \text{g}$ of ice at $0\ °\text{C}$ is added to $200\ \text{g}$ of water at $30\ °\text{C}$ in an insulated cup. Take $L_{\text{f}} = 3.3 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$ and $c = 4200\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$. What is the final temperature, in °C? · 把 $20\ \text{g}$、$0\ °\text{C}$ 的冰加入绝热杯中 $200\ \text{g}$、$30\ °\text{C}$ 的水里。取 $L_{\text{f}} = 3.3 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$,$c = 4200\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$。末温度是多少(单位 °C)?
Melting: $0.020 \times 3.3 \times 10^{5} = 6600\ \text{J}$. Then $6600 + 0.020 \times 4200 \times T = 0.200 \times 4200 \times (30 - T)$, so $6600 + 84T = 25200 - 840T$, giving $T = 20.1\ °\text{C}$. · 熔化:$0.020 \times 3.3 \times 10^{5} = 6600\ \text{J}$。然后 $6600 + 0.020 \times 4200 \times T = 0.200 \times 4200 \times (30 - T)$,即 $6600 + 84T = 25200 - 840T$,得 $T = 20.1\ °\text{C}$。
Put the energy stages for turning ice at -10 C into steam in order. · 把 -10 C 的冰变成水蒸气的能量阶段按顺序排列。
Each stage uses a different formula and a different constant. Using mc(delta T) across a change of state, or mL for a temperature rise, is the standard error. · 每个阶段用不同的公式和不同的常量。在相变过程中用 mc(delta T),或用 mL 去算升温,是标准错误。
How much energy melts 0.20 kg of ice at 0 C? (latent heat of fusion 3.3e5 J/kg) Give the answer in kJ. · 熔化 0 C 的 0.20 kg 冰需要多少能量?(熔化潜热 3.3e5 J/kg)用 kJ 作答。
E = mL = 0.20 x 3.3e5 = 6.6e4 J = 66 kJ, all at 0 C with no temperature change at all. There is no delta T in this stage. · E = mL = 0.20 x 3.3e5 = 6.6e4 J = 66 kJ,全程在 0 C、温度完全不变。这一阶段没有 delta T。
Worked example: heating a block at constant pressure
A $2.0\ \text{kg}$ metal block is heated by $100\ \text{K}$ at atmospheric pressure ($1.0 \times 10^{5}\ \text{Pa}$) and expands by $1.0 \times 10^{-6}\ \text{m}^{3}$. It absorbs $7.8 \times 10^{4}\ \text{J}$ of thermal energy.
- Work done by the block on the air: $p\Delta V = 1.0 \times 10^{5} \times 1.0 \times 10^{-6} = 0.10\ \text{J}$ — so the work done on the block is $-0.10\ \text{J}$ (it expands, pushing the air away).
- First law: the rise in internal energy is $q + W = 7.8 \times 10^{4} - 0.1 \approx 7.8 \times 10^{4}\ \text{J}$.
- Specific heat capacity: $c = \dfrac{q}{m\Delta T} = \dfrac{7.8 \times 10^{4}}{2.0 \times 100} = 390\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$ — copper.
- Check: for a solid the expansion work is tiny, so heating at constant pressure and at constant volume give almost the same $c$. For a gas the difference is large, which is the point of the next topic.
例题:恒压下加热金属块
一个 $2.0\ \text{kg}$ 的金属块在大气压($1.0 \times 10^{5}\ \text{Pa}$)下升温 $100\ \text{K}$,体积膨胀了 $1.0 \times 10^{-6}\ \text{m}^{3}$。它吸收了 $7.8 \times 10^{4}\ \text{J}$ 的热能。
- 金属块对空气做的功: $p\Delta V = 1.0 \times 10^{5} \times 1.0 \times 10^{-6} = 0.10\ \text{J}$——所以 对 金属块做的功是 $-0.10\ \text{J}$(它膨胀,把空气推开)。
- 第一定律: 内能的增加是 $q + W = 7.8 \times 10^{4} - 0.1 \approx 7.8 \times 10^{4}\ \text{J}$。
- 比热容: $c = \dfrac{q}{m\Delta T} = \dfrac{7.8 \times 10^{4}}{2.0 \times 100} = 390\ \dfrac{\text{J}}{\text{kg}\,\text{K}}$——铜。
- 检查: 对固体来说膨胀功极小,所以恒压加热和恒容加热得到的 $c$ 几乎相同。对 气体 差别很大,这正是下一主题的要点。
$Q = mc\Delta T$ has no term for a change of state — at a plateau use $Q = mL$ with $\Delta T = 0$. Keep the mass in kg when $c$ is in $\dfrac{\text{J}}{\text{kg}\,\text{K}}$. And a definition needs its qualifiers: "per unit mass" and "per unit temperature rise" for $c$; "per unit mass" and "at constant temperature" for $L$.
$Q = mc\Delta T$ 中 没有 状态变化的项——在平台上用 $Q = mL$,且 $\Delta T = 0$。当 $c$ 的单位是 $\dfrac{\text{J}}{\text{kg}\,\text{K}}$ 时,质量用 kg。而且定义需要限定词:$c$ 要"单位质量"和"单位温度升高";$L$ 要"单位质量"和"恒定温度下"。
Which phrases must appear in a full definition of specific latent heat of vaporisation? Select all that apply. · 完整定义 汽化比潜热 时必须出现哪些短语?选出所有正确的。
Latent heat is energy per unit mass to change state at constant temperature. "Per kelvin" belongs to specific heat capacity, where the temperature does change. · 潜热是恒定温度下单位质量改变状态所需的能量。"每开尔文"属于比热容,那里温度是变化的。
Measuring $c$ and $L$ in the lab
- Electrical heating: $Q = VIt$ (or $Pt$) from an immersion heater; plot temperature against time and use the gradient, so heat loss can be corrected.
- For $L_{\text{v}}$: heat the liquid at its boiling point and measure the mass boiled away in a set time; repeat at a second power so that the heat loss, the same in both runs, subtracts out.
- Sources of error: energy lost to the surroundings and to the container, and evaporation before boiling — lag the apparatus and take readings only once the temperature is steady.
在实验室测量 $c$ 和 $L$
- 电加热: 用浸入式加热器,$Q = VIt$(或 $Pt$);画出温度随时间的图并用斜率,这样可以修正热损失。
- 测 $L_{\text{v}}$: 在沸点加热液体,测量一定时间内沸腾掉的质量;换一个功率重复一次,两次相同的热损失就被抵消了。
- 误差来源: 散失到周围环境和容器的能量,以及沸腾前的蒸发——给装置保温,只在温度稳定后读数。
You've got it
- $Q = mc\Delta T$ to change temperature (energy per unit mass per kelvin); water's $c$ is high
- during a state change the temperature is constant: $Q = mL$ (energy per unit mass at constant temperature)
- $L_{\text{v}} > L_{\text{f}}$ (boiling separates the molecules fully and does work against the atmosphere); melt first, then warm, in a mixing problem
你掌握了
- $Q = mc\Delta T$ 改变温度(单位质量每开尔文的能量);水的 $c$ 很高
- 状态变化 期间温度恒定:$Q = mL$(恒定温度下单位质量的能量)
- $L_{\text{v}} > L_{\text{f}}$(沸腾完全分开分子 并且 要对大气做功);混合问题中先熔化,再升温