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: $Q = mc\Delta T$.
- 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: $Q = mL$.
- Fusion 熔化 $L_{\text{f}}$ (melting); vaporisation 汽化 $L_{\text{v}}$ (boiling). $L_{\text{v}} > L_{\text{f}}$ — boiling must break all the bonds and push back the air.
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.
You've got it
- $Q = mc\Delta T$ to change temperature; water's $c$ is high
- during a state change the temperature is constant: $Q = mL$
- $L_{\text{v}} > L_{\text{f}}$ (boiling breaks all the bonds and does work expanding)