Thermal Properties of Matter · Lesson 4 of 8
Changes of State, Latent Heat and Phase Diagrams
“Explain phase changes using heating curves and pressure–temperature diagrams, then calculate latent and sensible energy in single- and multistage calorimetry.”
• Interpret sloping and horizontal regions of heating curves for a pure substance at fixed pressure. • Explain pressure-dependent melting, boiling, sublimation and triple points. • Use latent heat with a complete phase and energy inventory. • Solve ice–water, hot-metal-on-ice and ice-to-steam energy balances.
Heating does not always raise temperature
Warm ice below 0 °C at ordinary atmospheric pressure and its temperature rises until melting begins. During slow equilibrium melting of pure ice at fixed pressure, the ice–water mixture remains near its melting temperature while energy is supplied. When all the ice has become liquid, further energy raises the water temperature. At boiling, energy again changes the relative amounts of two phases without the equilibrium temperature rising. Only after all liquid is vaporized does further heating warm the vapour.
On a temperature-versus-energy graph, a sloping single-phase segment has slope 1/(mc). Different slopes show different specific heats. A plateau’s width is the energy required for that phase conversion, not a temperature interval. If absorbed heater power is constant and losses negligible, energy is proportional to time, so the same pattern appears on a temperature-versus-time graph. Unequal power or significant losses prevent reading time widths directly as latent-energy ratios.
Energy during a phase change alters molecular arrangement and intermolecular potential energy rather than merely increasing the thermal motion associated with rising temperature. The constant-temperature statement assumes a pure substance, fixed pressure and conditions close to equilibrium. Mixtures, changing pressure, superheating and experimental delays can distort the simple plateaus. Melting and boiling temperatures are not universal numbers independent of conditions.
Pressure changes the phase boundaries
At ordinary pressure, ice is less dense and occupies more volume than the same mass of water. Increasing pressure favours the smaller-volume liquid phase, so it lowers ice’s melting temperature near ordinary conditions. Most substances expand on melting and respond with the opposite sign. Pressure effects must therefore be linked to the volume difference between the actual phases, not memorized as one rule for all melting.
In the weighted-wire demonstration, melting beneath the wire allows it to move downward while water above can refreeze when the local pressure is relieved. The slab can remain joined rather than simply breaking into two pieces. Conditions must permit energy transfer for melting and freezing. Pressure melting helps explain this demonstration, but it is not a complete general explanation of the slippery surface under skates: frictional heating and surface effects can also contribute.
Boiling occurs when vapour bubbles can persist within the liquid against the surrounding pressure. Raising that pressure raises the boiling temperature; lowering it lowers the boiling temperature. Water therefore boils below about 100 °C at high altitude and cooks many foods more slowly there. A pressure cooker permits a higher liquid temperature and faster cooking. The higher temperature, rather than pressure alone acting on the food, supplies the basic thermal explanation.
Small bubbles first seen during heating can be dissolved air escaping. Later, vapour bubbles formed at the hot base may collapse in colder liquid above. Sustained boiling occurs when bubbles survive and reach the surface. Water vapour itself is invisible; the white cloud often called steam is a mist of condensed liquid droplets. In a suitable pressure-controlled demonstration, condensing the vapour above warm water lowers its pressure and can restart boiling below 100 °C. This requires apparatus designed for the pressure conditions.
Read a pressure–temperature phase diagram
A phase diagram has pressure on the vertical axis and temperature on the horizontal axis. A point within a labelled region represents the stable bulk phase under those conditions. Moving across a boundary changes phase; moving along a boundary allows two phases to coexist in equilibrium. The solid–vapour boundary describes sublimation, the direct conversion between solid and vapour without a liquid stage. Iodine and dry ice provide familiar examples under suitable conditions.
Three coexistence lines meet at the triple point, where solid, liquid and vapour can coexist. Water’s triple-point temperature is approximately 273.16 K, or 0.01 °C, and its pressure about 611.657 Pa, roughly 0.00604 atm. The water fusion boundary slopes toward lower temperature as pressure increases in this range. For carbon dioxide the triple point is near −56.6 °C and 5.11 atm, and its fusion boundary slopes the other way. At 1 atm CO₂ has no stable liquid region: dry ice sublimes instead of melting into a puddle.
The liquid–vapour boundary terminates at the critical point: above that temperature, raising pressure alone does not produce the ordinary liquid–vapour transition. The diagram’s endpoint is near 374 °C for water and 31.1 °C for CO₂. These diagrams are qualitative here; the relative sizes and slopes are not numerical predictions. For example, a horizontal path at 1 atm through the CO₂ diagram crosses the solid–vapour boundary but never a stable liquid region.
Latent heat measures phase-conversion energy
Latent heat is a mass-specific conversion energy, not a temperature and not an extra specific heat to multiply by a temperature rise. Fusion changes solid to liquid; vaporization changes liquid to vapour. The reverse process releases energy. For water at approximately 1 atm, use Lf ≈ 3.33 × 10⁵ J kg⁻¹ and Lv ≈ 22.6 × 10⁵ J kg⁻¹ unless a problem supplies a different rounded value. Much more energy is required to vaporize a kilogram than to melt it.
| Substance | Melting point / °C | Lf / (10⁵ J kg⁻¹) | Boiling point / °C | Lv / (10⁵ J kg⁻¹) |
|---|---|---|---|---|
| Ethanol | −114 | 1.0 | 78 | 8.5 |
| Gold | 1063 | 0.645 | 2660 | 15.8 |
| Lead | 328 | 0.25 | 1744 | 8.67 |
| Mercury | −39 | 0.12 | 357 | 2.7 |
| Nitrogen | −210 | 0.26 | −196 | 2.0 |
| Oxygen | −219 | 0.14 | −183 | 2.1 |
| Water | 0 | 3.33 | 100 | 22.6 |
These are representative transition data near standard pressure. Nitrogen and oxygen boil far below room temperature, whereas metals such as gold require very high temperatures. The different latent heats reflect different phase energy changes; their magnitudes are not inferred from melting temperature alone. When steam at 100 °C condenses, it first releases latent energy; the resulting water can release additional energy while cooling. Equal masses of water and steam initially at 100 °C therefore need not transfer equal energy to a cooler object.
Problem
Mix 0.15 kg of ice at 0 °C with 0.30 kg of water at 50 °C in a calorimeter of negligible heat capacity. All ice melts and the final temperature is 6.7 °C. Use cwater = 4186 J kg⁻¹ K⁻¹.
- 1.The initially warm water loses 0.30 × 4186 × (50 − 6.7) = 54376.14 J.
- 2.The ice requires 0.15Lf to melt. Its meltwater then warms from 0 °C to 6.7 °C, requiring 0.15 × 4186 × 6.7 = 4206.93 J.
- 3.Equate loss and gain: 54376.14 = 0.15Lf + 4206.93. Hence Lf = 334461.4 J kg⁻¹ ≈ 3.34 × 10⁵ J kg⁻¹.
- 4.Ignoring the warming of the meltwater would falsely count all transferred energy as melting energy and overestimate Lf.
Track the state before adding energy terms
Start with an inventory: initial temperature and state, mass in each phase, desired final temperature and state. List each warming, melting, vaporizing or cooling stage in physical order. Use mcΔT only within a phase and mL only during conversion. If the available energy is insufficient to finish melting, the final state can be an ice–water mixture at the transition temperature; do not automatically assume all ice melts.
Problem
Find the energy needed for 3.0 kg of ice at −12 °C to become steam at 100 °C at approximately 1 atm. Use cice = 2100, cwater = 4186 J kg⁻¹ K⁻¹, Lf = 3.35 × 10⁵ and Lv = 2.256 × 10⁶ J kg⁻¹.
- 1.Warm the ice to 0 °C: Q1 = 3 × 2100 × 12 = 75600 J.
- 2.Melt it: Q2 = 3 × 335000 = 1005000 J.
- 3.Warm the resulting water to 100 °C: Q3 = 3 × 4186 × 100 = 1255800 J.
- 4.Vaporize it at 100 °C: Q4 = 3 × 2256000 = 6768000 J.
- 5.The total is 9104400 J ≈ 9.1 MJ. Vaporization is the largest contribution. No steam-warming term is needed because the desired steam is already at its formation temperature.
Problem
A 2.5 kg copper block at 500 °C is placed on a large ice slab at 0 °C. Estimate the maximum melted mass if all available block energy melts ice. Use ccopper = 390 J kg⁻¹ K⁻¹ and Lf = 3.35 × 10⁵ J kg⁻¹.
- 1.The large ice reservoir holds the final contact system at about 0 °C while ice remains.
- 2.Copper cooling releases 2.5 × 390 × 500 = 487500 J.
- 3.Set mLf = 487500 to obtain m ≈ 1.455 kg, approximately 1.5 kg.
- 4.This is a maximum within the model: environmental loss or warming of other bodies leaves less energy for melting.
Problem
In an idealized calculation, a 30 kg body cools from 101 °F to 98 °F in 20 min solely by additional water evaporation. Model its specific heat as 1 cal g⁻¹ K⁻¹ and use Lv = 580 cal g⁻¹. Find the average additional evaporated mass per minute.
- 1.The temperature difference is 3 °F = 3 × 5/9 = 5/3 K. Do not add the Fahrenheit offset to a temperature interval.
- 2.The modelled energy decrease is 30000 × 1 × 5/3 = 50000 cal.
- 3.The total additional evaporation is 50000/580 ≈ 86.2 g, giving about 4.31 g min⁻¹.
- 4.This is an intentionally simplified physics energy balance, not a complete account of physiological heat exchange.
Quick Quiz
Quiz
During equilibrium melting of pure ice at fixed pressure, added energy chiefly
The phase proportions change while the equilibrium coexistence temperature stays approximately fixed.
Why does increased pressure lower ice’s melting point near ordinary conditions?
Pressure favours the smaller-volume phase, which is liquid water in this case.
Why does dry ice at 1 atm sublime rather than form a stable liquid?
The stable liquid region begins above the CO₂ triple-point pressure near 5.11 atm.
Which expression gives the energy to melt mass m already at its melting point?
Melting at the transition conditions requires the latent heat of fusion, not an invented temperature interval.
Equal masses of steam and water at 100 °C cool to the same lower water temperature. Steam can release more energy because it
Both can release water-cooling energy, but steam additionally condenses.
Practice Problems
- Sketch a temperature-versus-absorbed-energy curve for pure water initially as ice at −10 °C and finally as steam at 110 °C at 1 atm. Label both plateaus and explain all three single-phase slopes without assigning unsupported numerical widths.
- Find the energy to convert 0.200 kg of ice at −10 °C into liquid water at 20 °C. Use cice = 2100, cwater = 4186 J kg⁻¹ K⁻¹ and Lf = 3.34 × 10⁵ J kg⁻¹; assume no loss.
- Mix 0.100 kg of ice at 0 °C with 0.400 kg of water at 30 °C in a negligible-capacity insulated vessel. Use Lf = 3.34 × 10⁵ and cwater = 4186 in SI units. First test whether all the ice melts, then calculate the final temperature or remaining ice.
- Compare the energy released by 0.050 kg of steam at 100 °C and by 0.050 kg of water at 100 °C when each finally becomes water at 30 °C. Use Lv = 2.256 × 10⁶ and cwater = 4186 in SI units.
- Use phase-boundary reasoning to explain three observations: a pressure cooker allows hotter liquid water, a lake surface can freeze while water survives below, and CO₂ at 1 atm has no stable liquid region. Distinguish the density explanation from the boiling-pressure explanation.
Key Takeaways
• A phase-change plateau represents changing phase proportions at fixed pressure, not an absence of energy transfer. • Pressure shifts coexistence temperatures; water’s melting boundary has an unusual sign because ice is less dense. • Use mcΔT for warming within one phase and mL for conversion; inventory every stage. • Check the available energy before assuming that all ice melts or all liquid vaporizes.