# Meaning of temperature during phase transition

Microscopically, temperature is related to the average kinetic energy of a system in thermal equilibrium. Both LIQUID water and GASEOUS steam can exist (independently) at a temperature of 100 degree Celsius (say during boiling). But surely, steam has higher kinetic energy than liquid water. How then, can they be said to have the same temperature?

• Temperature is, by definition of the second law of thermodynamics, that which drives heat transfer. If there is no heat transfer between two co-existing phases, then they have the same temperature. You don't need any microscopic theory for this and, to be honest, in most cases attempts to explain thermodynamic phenomena microscopically are failures to understand and properly apply the macroscopic theory correctly. Keep in mind that statistical mechanics and thermodynamics have EXACTLY the same physical content. If you can describe one thing with one, then you can also use the other. Jan 14 '16 at 16:03
• Temperature is generally related to the average kinetic energy of the atoms in a system, but the problem here is that you seem to be assuming an equivalence, and that's not true for systems as disparate as water and steam. All that you've demonstrated with your water and steam example is that this equivalence cannot be taken too far. It is, for example, possible to have systems with negative temperatures, which clearly doesn't make any sense if one tries to somehow equate temperature with kinetic energy.
– user93237
Jan 14 '16 at 17:43
• So going by the "second law definition" of temperature, does it mean that if I take two separate systems-one with steam at 100 degree Celsius and the other with water at 100 degree Celsius, each system having diathermal boundaries, will no heat exchange will take place? But surely, "microscopically", the molecules on the steam side of the whole system are bombarding the wall with more kinetic energy than the liquid water molecules on the other side (less K.E. due to H-bonding) ? Jan 15 '16 at 2:33

## 2 Answers

There are some intricate concepts here that I can only begin to explain.

To start with - in liquid water there are strong intermolecular forces. In order to water to turn into steam, these forces have to be overcome. The work done is the latent heat of evaporation. But the question is - does it follow that the molecules in steam at 100°C have higher kinetic energy than molecules in water? Or is the difference just in the intermolecular bonds?

The specific heat of water at 100C is 4.219 kJ/kg/K; for steam, the corresponding number is 1.890 kJ/kg/K. That's very strange, but explained here. Basically, as water is heated and the molecules move faster, more and more of the hydrogen bonds are broken; this requires a fair bit of energy - it accounts for more than half the heat capacity of liquid water.

However, both in liquid water and in steam there are five degrees of freedom (two rotation, three translation) which store the kinetic energy. And there's no reason to believe that they have a different amount of energy / motion when they are at the same temperature. That doesn't mean they don't have different energy content - but you specifically asked about kinetic energy content.

Because steam has latent heat in it. Latent heat is required to break the attractions in water and convert it to a gaseous phase. So gaseous steam seems to have a higher kinetic energy.

EDIT: The earlier answer was unclear and its presentation was misleading. The energy of the system comprises of potential energy terms, which are attractions between the water molecules, and kinetic energy terms. Now the difference between gaseous and water steam is although they are both at same temperature, gaseous steam has latent heat added to it which breaks the attractions , i.e the potential energy terms. Temperature is the measure of the average kinetic energy and therefore it remains the same. Steam doesn't have a higher kinetic energy but a higher total energy.

• So temperature is not a measure of average kinetic energy? Jan 14 '16 at 16:26
• @garyp you're right, the answer was misleading. i edited it now. thanks Jan 14 '16 at 16:54
• That's what I thought. (whew!). Your "total energy" is the thermodynamic "internal energy", I think. Jan 14 '16 at 16:59
• @garyp precisely it is the internal energy. They are equivalent. Jan 14 '16 at 17:19