Specific heat capacity and latent heat
| English | Chinese | Pinyin |
|---|---|---|
| latent heat | 潜热 | qián rè |
| specific heat capacity | 比热容 | bǐ rè róng |
| change of state | 状态变化 | zhuàng tài biàn huà |
| specific latent heat | 比潜热 | bǐ qián rè |
| fusion | 熔化 | róng huà |
| vaporisation | 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.
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.
Specific heat capacity
Q = mcΔT
The heat needed is proportional to the temperature rise — the gradient depends on mass and the material's specific heat capacity.
How much energy raises $2.0\ \text{kg}$ of water by $10\ \text{K}$? (Use $c = 4200\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$.)
$Q = mc\Delta T = 2.0 \times 4200 \times 10 = 84000\ \text{J}$.
Water has a high specific heat capacity compared with most metals.
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:
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 $L_{\text{f}} = 3.34 \times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$.)
$Q = mL_{\text{f}} = 0.10 \times 3.34 \times 10^{5} = 3.34 \times 10^{4}\ \text{J}$.
The latent heat of vaporisation is larger than that of fusion because:
Melting only loosens the bonds; boiling separates the particles completely and the vapour expands against the air.
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 $mc\Delta T$).
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?
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 $T = 20.1\ °\text{C}$.
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.
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