riaology
Unit 02 · Lesson 01 of 09

What happens during a phase change?

Temperature can pause while energy keeps entering.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do
  • One substance, several phases.
  • Saturation ties pressure to temperature.
  • The plateau is an energy conversion.
Before you touch the controls

Predict whether temperature keeps rising when a liquid begins boiling.

An experiment you can change

Heat a kilogram of water

Predict the direction of change, move one control, then explain the response using the equation.

Water heated at constant pressure: temperature 99.60591861133764 degrees Celsius, vapor mass fraction 0.4727535065890169Heat in99.61°C · vapor mass: 47.3%Heat a kilogram of water. Heat added (kJ) versus Temperature (°C). Selected result 99.6059 °C. Two phases coexist: heat changes vapor mass fraction while saturation temperature stays fixed.00100.221,400200.442,800Temperature (°C)Heat added (kJ)
T=99.606 ∘C,x=0.47275T=99.606\,{}^\circ\mathrm C,\quad x=0.47275

Two phases coexist: heat changes vapor mass fraction while saturation temperature stays fixed.

Model assumptions and units

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Q=m(h2−h1),h=hf+xhfgQ=m(h_2-h_1),\qquad h=h_f+xh_{fg}

Explore the saturation dome

Water saturation dome: temperature 100 degrees Celsius, quality 0.25, specific volume 0.41874774186596714 cubic meters per kilogram; horizontal volume axis is logarithmic.0.0010.010.1110Temperature (°C)Specific volume (m³/kg), logarithmic axisLiquidMixtureVaporpsat=101.42 kPa,v=0.41875 m3/kgp_{sat}=101.42\,\mathrm{kPa},\quad v=0.41875\,\mathrm{m^3/kg}

The tie line is straight in volume, but the position looks nonlinear because the volume axis is logarithmic. Saturation properties are calculated using IAPWS-IF97.

Saturated water at 100 kPa · IAPWS-IF97
Phaseh (kJ/kg)u (kJ/kg)v (m³/kg)
Liquid · 99.606°C417.436417.3320.001043
Vapor · same temperature2,674.952,505.5471.694023
01

One substance, several phases.

A pure substance has a fixed chemical composition. Liquid water and water vapor can form a pure-substance mixture because their chemistry is the same. Air can often be modeled as a pure substance while its composition stays uniform, but a liquid-air mixture can have different phase compositions.

02

Saturation ties pressure to temperature.

At a given pressure, equilibrium boiling occurs at a particular saturation temperature. Compressed liquid is colder than that temperature; superheated vapor is hotter. In the two-phase region pressure and temperature are linked, so they do not provide two independent properties.

T=Tsat(p)T=T_{sat}(p)
03

The plateau is an energy conversion.

At constant pressure, heat into a closed system with only boundary work equals its enthalpy increase. During vaporization the energy changes phase proportions rather than temperature. The saturated-vapor enthalpy exceeds saturated-liquid enthalpy by the latent enthalpy.

Q=m(h2−h1),hfg=hg−hfQ=m(h_2-h_1),\quad h_{fg}=h_g-h_f
04

Read the dome carefully.

The left edge is saturated liquid and the right edge saturated vapor. Inside are mixtures; above the critical point the liquid–vapor distinction disappears. The diagram below uses a logarithmic volume axis because vapor volume can dwarf liquid volume.

A worked example

One kilogram, halfway through boiling.

  1. At 100 kPa, use the displayed saturation table: Tsat≈99.606 ∘CT_{sat}\approx99.606\,{}^\circ\mathrm C.
  2. Half the mass vaporized means x=0.5x=0.5, so h=hf+0.5hfgh=h_f+0.5h_{fg}.
  3. Starting from saturated liquid, Q=0.5mhfg≈1128.8 kJQ=0.5mh_{fg}\approx1128.8\,\mathrm{kJ}. The temperature stays at saturation during this step.
Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Heat without a temperature rise

Heat 1 kg water at 100 kPa from 20°C by 1400 kJ. Find its final temperature and inspect the vapor mass fraction.

Hint

First pay the sensible heat needed to reach saturation; the remaining energy vaporizes liquid.

Explore the visual solution
Water heated at constant pressure: temperature 99.60591861133764 degrees Celsius, vapor mass fraction 0.4727535065890169Heat in99.61°C · vapor mass: 47.3%Heat a kilogram of water. Heat added (kJ) versus Temperature (°C). Selected result 99.6059 °C. Two phases coexist: heat changes vapor mass fraction while saturation temperature stays fixed.00100.221,400200.442,800Temperature (°C)Heat added (kJ)
T=99.606 ∘C,x=0.47275T=99.606\,{}^\circ\mathrm C,\quad x=0.47275

Two phases coexist: heat changes vapor mass fraction while saturation temperature stays fixed.

Model assumptions and units

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Q=m(h2−h1),h=hf+xhfgQ=m(h_2-h_1),\qquad h=h_f+xh_{fg}

Controls explore variations. The steps below retain the original question’s values.

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Q=m(h2−h1),h=hf+xhfgQ=m(h_2-h_1),\qquad h=h_f+xh_{fg}
2 · Compare

Reach the start of boiling

Supply 200 kJ to the same kilogram. Find temperature. Explain why a quality value is not a valid descriptor yet.

Hint

Compare the heat supplied with 4.18(Tsat − 20).

Explore the visual solution
Water heated at constant pressure: temperature 67.8468899521531 degrees Celsius, vapor mass fraction 0Heat in67.85°C · vapor mass: 0%Heat a kilogram of water. Heat added (kJ) versus Temperature (°C). Selected result 67.8469 °C. Compressed liquid: heat raises temperature.00100.221,400200.442,800Temperature (°C)Heat added (kJ)
T=67.847 ∘C,quality undefined (single phase)T=67.847\,{}^\circ\mathrm C,\quad \text{quality undefined (single phase)}

Compressed liquid: heat raises temperature.

Model assumptions and units

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Q=m(h2−h1),h=hf+xhfgQ=m(h_2-h_1),\qquad h=h_f+xh_{fg}

Controls explore variations. The steps below retain the original question’s values.

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Q=m(h2−h1),h=hf+xhfgQ=m(h_2-h_1),\qquad h=h_f+xh_{fg}
3 · Challenge the model

Pass the vapor endpoint

Supply 2800 kJ under the model assumptions. Find the final temperature and identify which heat-capacity approximation controls this last stage.

Hint

Subtract both the sensible-liquid heat and latent heat before using vapor heat capacity.

Explore the visual solution
Water heated at constant pressure: temperature 200.43962302897887 degrees Celsius, vapor mass fraction 1Heat in200.44°C · vapor mass: 100%Heat a kilogram of water. Heat added (kJ) versus Temperature (°C). Selected result 200.4396 °C. Superheated vapor: temperature rises again. Vapor heat capacity here is an approximation.00100.221,400200.442,800Temperature (°C)Heat added (kJ)
T=200.44 ∘C,quality undefined (single phase)T=200.44\,{}^\circ\mathrm C,\quad \text{quality undefined (single phase)}

Superheated vapor: temperature rises again. Vapor heat capacity here is an approximation.

Model assumptions and units

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Q=m(h2−h1),h=hf+xhfgQ=m(h_2-h_1),\qquad h=h_f+xh_{fg}

Controls explore variations. The steps below retain the original question’s values.

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Q=m(h2−h1),h=hf+xhfgQ=m(h_2-h_1),\qquad h=h_f+xh_{fg}
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