Heating Curve & Phase Change Essentials
Phase changes and latent heat
- Phase changes: solid ↔ liquid ↔ gas; includes solid→liquid (melting), liquid→gas (vaporization), and reverse processes (freezing, condensation)
- During a phase change, temperature does not change; the absorbed/released heat goes into rearranging particles, not warming/cooling
- Latent heats:
- Fusion (solid ⇄ liquid): Q=mLf
- Vaporization (liquid ⇄ gas): Q=mLv
- For heating/cooling within a single phase: Q=mcΔT (c is specific heat)
- Common water values:
- Lf≈334 Jg−1
- Lv≈2260 Jg−1
- cw≈4.18 Jg−1 ∘C−1
Heating and cooling curves
- Axes: X = input heat, Y = temperature
- Single-phase regions: temperature rises/falls (slopes)
- Phase-change plateaus: constant temperature while phase change occurs
- Melting at 0°C for water (solid → liquid)
- Boiling at 100°C for water (liquid → gas)
- During melting/boiling, the added heat goes into changing phase, not increasing temperature
- Two-phase regions exist during partial melting or partial vaporization (mixtures of phases)
- The cooling curve is the reverse of the heating curve (same plateaus, different direction)
Water heating curve: key sequence (ice to steam)
- Start: solid ice at a low temperature (e.g., −40°C)
- Heat to 0°C: solid ice warms until melting point
- Melting plateau at 0°C: solid → liquid; heat input goes into fusion
- Liquid water heats from 0°C to 100°C
- Boiling plateau at 100°C: liquid → steam; heat input goes into vaporization
- Steam heats above 100°C or, upon cooling, steam condenses back through the same plateaus
- States at points:
- Below 0°C: solid
- During 0°C plateau: solid–liquid mixture
- Between plateaus: liquid
- At 100°C plateau: liquid–gas mixture
- Above 100°C: gas
Practical burn comparison: 100°C steam vs 100°C water
- Both at 100°C, but steam can transfer more heat to skin due to latent heat release during condensation
- Heat transfer per mass (conceptual):
- Hot water burn: Q<em>extwater=mc</em>w(100−37)
- Steam burn: Q<em>extsteam=mL</em>v+mcw(100−37)
- Since L<em>v≫c</em>w×(100−37), steam releases more energy when it condenses and cools on skin
- Latent heat is the key factor in the greater severity of steam burns
Other examples and problem types
- Ethanol: apply the same approach with ethanol’s L<em>v and its own c and L</em>f
- General problem approach:
- Identify phase-change steps
- Assign the appropriate heat equation to each step
- Sum the heats of all steps
Exam-type problem: steam to ice at 0°C (three steps)
- Given: 175 g steam at 100°C condenses, then cools to 0°C, then freezes to 0°C ice
- Steps (identify and write equations):
1) Condense steam: Q<em>1=mL</em>v
- 100°C steam → 100°C water
2) Cool water: Q<em>2=mc</em>w(100−0) - 100°C water → 0°C water
3) Freeze at 0°C: Q<em>3=mL</em>f - 0°C water → 0°C ice
- Total heat lost: Q=Q<em>1+Q</em>2+Q3
- Numerical values (approx):
- Lv≈2260 Jg−1
- cw≈4.18 Jg−1∘C−1
- Lf≈334 Jg−1
- Mass: m=175 g
- Example total (rounded): approximately 5.27×105 J≈527 kJ
- Key exam skill: decompose into steps, apply the correct equation to each, and sum
Quick recall tips
- Latent heat governs heat during phase changes; temperature stays constant during these steps
- Use Q=mL<em>f for fusion, Q=mL</em>v for vaporization, and Q=mcΔT for heating/cooling within one phase
- On a heating/cooling curve, read the substance’s state at any point by the plateau presence and phase regions
- For burns, steam is more dangerous than hot water because of the additional heat from condensation (latent heat)
- In problems, list steps (e.g., H1, H2, H3) and apply the corresponding equations to each step