How many calories are required to evaporate 1 gram of boiling water?
Also, how many calories of heat are lost by water freezing at 0 degrees celsius?
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December 11th, 2010 at 8:34 am
This is a two step process.
1) heat the water to boiling point
2) phase change from liquid to gas
Q= cm(dtT)
Q= mL, where L is the latent heat
c-water= 4.184 j/g
c-vapor = 2258 j/g
1) Q = 4.184j/g * 1g*100degrees
2) Q = 1*2258
418.4j+2258j = 2,676.4j or 639.2 cal
1j=4.18 cal