Posted by justin on Sunday, August 7, 2011 at 5:32am.
1 liter of water has a mass of 1.00 kg
Energy added in one hour =
3600 s * 350 J/s = 1.26*10^6 J
Energy required to heat all of the water to 100C (the boiling point) =
1 kg*60 K*4200 J/kgK = = 2.52*10^5 J
Energy available to boil water =
1.26*10^6 - 0.252*10^6 = 0.998*10^6 J
Amount that boils away
= 0.998*10^6 J/2.26*10^6 J/kg
= 0.44 kg
The amount left will be 0.56 kg.
Notye: The units of the latent heat of vaporization should be J/kg, NOT J/kg*K
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