Energy of Phase Changes
Introduced| English | Meaning |
|---|---|
| heat of fusion/hiːt ɒv ˈfjuːʒn/ | The energy to melt a solid. |
| heat of vaporization/hiːt ɒv ˌveɪpəraɪˈzeɪʃn/ | The energy to boil a liquid. |
The flat spots on the graph
- Heat ice steadily and its temperature climbs — then stalls at $0$ degrees.
- All the heat goes into melting, not warming.
- Only once it is fully liquid does the temperature rise again.
- Those flat spots hide a big energy cost.
Temperature pauses
- During a phase change, the temperature stays constant.
- Added heat breaks the attractions rather than speeding particles up.
- So a heating curve has flat plateaus.
Heating curve: the temperature stays flat while a phase changes
The temperature stays constant during a phase change.
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Heat breaks attractions instead of raising the temperature.
On a heating curve, a phase change appears as a flat ____.
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The temperature is constant, giving a flat plateau.
The hidden heats
- The heat of fusion 熔化热 is the energy to melt a solid.
- The heat of vaporization 汽化热 is the energy to boil a liquid.
- Vaporization takes far more energy than fusion.
For water, which usually requires more energy?
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The heat of vaporization is much larger than the heat of fusion.
Warming versus changing
- Warming within one phase uses $q = mc\Delta T$.
- Changing phase uses $q = m \times$ (heat of fusion or vaporization).
- A full heating curve combines both.
The heating curve
Temperature stays flat while a substance melts or boils, as latent heat breaks the forces between particles.
To warm a liquid without changing its phase, you use...
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Within a single phase, use $q = mc\Delta T$.
Why does steam burn worse than boiling water at the same temperature?
- Steam releases its large heat of vaporization as it condenses on skin.
- That extra energy makes the burn much worse.
Steam burns worse than boiling water because it also releases its...
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Condensing steam gives up a large heat of vaporization to the skin.
You should use $q = mc\Delta T$ during a phase change.
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With $\Delta T = 0$, use the latent heat instead.
The temperature is constant during a phase change — the heat goes into breaking attractions, so do not use $q = mc\Delta T$ there ($\Delta T = 0$). Use the latent heat instead. Vaporization needs much more energy than fusion. And the plateaus on a heating curve mark the phase changes.
During a phase change the temperature holds constant while heat breaks attractions — the plateaus on a heating curve. The heat of fusion melts a solid and the heat of vaporization boils a liquid (much larger). Warming within a phase uses $q = mc\Delta T$; changing phase uses the latent heat.