Specific heat capacity and latent heat
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.
Specific heat capacity
- $c$ = 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.
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.
- 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
- $L$ = energy to change the state of 1 kg at constant temperature: $Q = mL$.
- Fusion $L_{\text{f}}$ (melting); vaporisation $L_{\text{v}}$ (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.
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)