physics
posted by justin .
. A water heater can has a power of 350 W. It is placed in a container filled with 1 litres of water with a temperature of 40 ÂșC. The heater is then allow to run for 1 hour. How much water will there be left in the container after 1 hour? (Specific heat capacity of water = 4200 J/kg K and specific latent heat of vaporization = 2260 000 J/kg K)

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