It should be 0 as internal energy is a function of temperature, i.e: U (T)
At the boiling point, there is no change in temperature, and hence DU= 0
The change in enthalpy is dH = TdS + VdP. Since the pressure P is the same for the the vapor and the liquid, the change in enthalpy is just TdS, which is just the latent heat.
The heat needed to melt one gram of a solid at its melting point is the heat of fusion.
I think you mean vaporization. Vaporization is the phase change of matter from a liquid to a gas (vapor). Vaporization can occur through evaporation or boiling.
Boiling
Evaporation is a type of vaporization of a liquid that occurs only on the surface of a liquid. The other type of vaporization is boiling, which, instead, occurs on the entire mass of the liquid.
Molar heat of fusion: the heat (enthalpy, energy) needed to transform a solid in liquid (expressed in kJ/mol). Molar heat of vaporization: the heat (enthalpy, energy) needed to transform a liquid in gas (expressed in kJ/mol).
Latent Heat of Evaporation, or Evaporation Enthalpy. It is given in units of energy over unit of mass, i.e., KJ/Kg.
Heat of vaporization or enthalpy of vaporization. It is the additional energy, per unit mass, required after vaporization temperature (boiling point) is reached, to accomplish the change in state, from liquid to gas.
Temperature doesn't give the whole picture when you talk about boiling. A more useful property to talk about is enthalpy. Enthalpy is the energy held by the water. Prior to the boiling point, enthalpy and temperature both rise linearly. At the boiling point, temperature stops rising, but enthalpy continues to rise until it becomes steam. If you were to continue adding energy to the steam, it's temperature would rise again. The amount of energy that must be added to water to get it from water just at the boiling point to steam is the latent heat of vaporization and is equal to the enthalpy rise discussed in the previous paragraph. The latent heat of vaporization and the temperature where boiling will occur are dependant on the pressure.
The change in enthalpy is dH = TdS + VdP. Since the pressure P is the same for the the vapor and the liquid, the change in enthalpy is just TdS, which is just the latent heat.
Vaporization
Because it is closer to its enthalpy of vaporization (ΔvH), at boiling point the liquid molecules have highest kinetic energy so they easily escape from liquid as vapours.
The heat needed to melt one gram of a solid at its melting point is the heat of fusion.
Two types of vaporization are evaporation and boiling.
OK.With entalpy od vaporization and temperature of vaporization is very easy to calculate entropy of vaporization of etanol.So the equation to calculate this is:Delta_S=-Delta_H/TbWhere:Delta_S= Entropy of vaporizationDelta_H=Entalpy of vaporizationTb= Normal Boiling point temperatureSo the Delta_S become:Delta_S=-(-109000.8)/(78.5+273)Delta_S=310.1 J.mol-1.K-1
Evaporation is vaporization under the boiling point.
Boiling.