Question

how much heat energy is required to vaporize a 1 g ice cube at 0 C. The heat of fusion of ice is 3.36 x 10^5 J/kg . The heat of vaporization of water is 2.26 x 10^6 J/Kg. ( specific heat of water = 4200 J/Kg C )

Answer 1

Answer:

Q = M * (Cf +  C * 100 + Cv)

Cf and Cf are heats of fusion and vaporization and C is the heat required to heat mass M of water 1 deg

Q = .001 kg ( 3.36 * E5 + 100 deg * 4200 + 2.26 * E6) J

Q = .001 kg ( 3.36 J / kg + 4.2 J / kg + 22.6 J /kg) * 10E5

Q = .001 kg * 30.2 * 10E5 J / kg = 3020 J