| English | 意味 |
|---|---|
| latent heat/ˈleɪtənt hiːt/ | 潜熱 |
| specific heat capacity/spəˈsɪfɪk hiːt kəˈpæsɪti/ | 比熱容 |
| change of state/tʃeɪndʒ ɒv steɪt/ | 状態変化 |
| specific latent heat/spəˈsɪfɪk ˈleɪtənt hiːt/ | 比潜熱 |
| fusion/ˈfjuːʒn/ | 融解 |
| vaporisation/ˌveɪpəraɪˈzeɪʃn/ | 蒸発 |
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.
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
Energy to heat it: E = mcΔT
Pick a material, set the mass and the temperature rise, and read the energy. Water needs far more energy than the metals.
比熱容
Q = mcΔT
必要な熱量は 温度上昇に比例する — 勾配は質量と物質の比熱容に依存する。
How much energy raises $2.0\ \text{kg}$ of water by $10\ \text{K}$? (Use $c = 4200\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$.)
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$Q = mc\Delta T = 2.0 \times 4200 \times 10 = 84000\ \text{J}$.
Water has a high specific heat capacity compared with most metals.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Water ~4200 vs aluminium ~900, copper ~385 J/(kg·K) — water needs much more energy per kelvin.
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.
While a solid is melting, its temperature:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
During a state change the energy breaks bonds rather than raising temperature, so it stays constant.
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.
How much energy melts $0.10\ \text{kg}$ of ice at $0\ °\text{C}$? (Use $L_{\text{f}} = 3.34 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$.)
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$Q = mL_{\text{f}} = 0.10 \times 3.34 \times 10^{5} = 3.34 \times 10^{4}\ \text{J}$.
The latent heat of vaporisation is larger than that of fusion because:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Melting only loosens the bonds; boiling separates the particles completely and the vapour expands against the air.
Match each term to the definition the examiner marks.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
"At constant temperature" is what separates latent heat from heating, and leaving it out loses the mark even though the formula is right.
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
During a change of state, the energy is $Q = m$____ (not $mc\Delta T$).
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A phase change is at constant temperature, so use $Q = mL$ there; use $mc\Delta T$ for the sloped (temperature-changing) parts.
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.
$20\ \text{g}$ 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?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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 $T = 20.1\ °\text{C}$.
Put the energy stages for turning ice at -10 C into steam in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
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.
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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.
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.
$Q = mc\Delta T$ 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 $L$.
Which phrases must appear in a full definition of specific latent heat of vaporisation? Select all that apply.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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.
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.
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