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Liquids - Latent Heat of Evaporation

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The input energy required to change the state from liquid to vapor at constant temperature is called the latent heat of vaporization. When a liquid vaporize at the normal boiling point the temperature of the liquid will not rise beyond the temperature of the boiling point.

The latent heat of vaporization is the amount of

"heat required to convert a unit mass of a liquid into vapor without a change in temperature".

Liquids - Latent Heat of Evaporation
ProductLatent Heat of Evaporation*)
- he -
(kJ/kg)(Btu/lb)
Acetic acid 402 173
Acetone 518 223
Alcohol 896 385
Alcohol, ethyl (ethanol) 846 364
Alcohol, methyl (methanol alcohol, wood alcohol, wood naphtha or wood spirits) 1100 473
Alcohol, propyl 779 335
Ammonia 1369 589
Aniline 450 193
Benzene 390 168
Bromine 193 83
Carbon bisulphide 160
Carbon dioxide 574 247
Carbon disulphide 351 151
Carbon tetrachloride 194 83
Chlorine 293
Chloroform 247 106
Decane 263 113
Dodecane 256 110
Ether 377 162
Ethylene glycol 800 344
Trichlorofluoromethane refrigerant R-11 180 77
Dichlorodifluoromethane refrigerant R-12 165 71
Chlorodifluoromethane refrigerant R-22 232 100
Glycerine 974 419
Helium 21 9
Heptane 318 137
Hexane 365 157
Hydrogen 461 198
Iodine 164 71
Kerosene 251 108
Mercury 295 127
Methyl chloride 406
Nitrogen 199 86
Octane 298 128
Oxygen 214 92
Propane 428 184
Propylene 342 147
Propylene glycol 914 393
Sulphur 1510 650
Sulfur dioxide 164
Toluene 351 151
Turpentine 293 126
Water 2256 970.4
  • 1 kJ/kg = 0.43 Btu/lbm = 0.24 kcal/kg

*) The latent heats of evaporation are based on fluid boiling point temperatures at atmospheric pressure.

.

Evaporation Heat

The heat required to evaporate a fluid can be calculated as:

q = he m                                                             (1)

where

q = evaporation heat (kJ, Btu)

he = evaporation heat (kJ/kg, Btu/lb)

m = mass of liquid (kg, lb)

Example - Calculate heat required to evaporate 10 kg of water

The latent heat of evaporation for water is 2256 kJ/kg at atmospheric pressure and 100oC. The heat required to evaporate 10 kg can be calculated as

q = (2256 kJ/kg) (10 kg)

   = 22560 kJ

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