Question

part A How much heat energy, in kilojoules, is required to convert 69.0 g of ice at −18.0 ∘C to water at  25.0 ∘C ? Part B How long would it take for 1.50 mol of water at 100.0 ∘C to be converted completely into steam if heat were added at a constant rate of 22.0 J/s ?

Specific heat of ice: sice=2.09 J/(g⋅∘C)

Specific heat of liquid water: swater=4.18 J/(g⋅∘C)

Enthalpy of fusion (H2O(s)→H2O(l)): ΔHfus=334 J/g

Enthalpy of vaporization (H2O(l)→H2O(g)): ΔHvap=2250 J/g

Answer 1

A) To heat the ice to 0°C:
69g x 2.09J/g/°C x 18°C ΔT = 2596 J = 2.596 kJ.

To melt the ice at 0°C = 69g x 334J/g = 23046 J = 23.046 kJ.

To heat the water to 25°C = 69g x 4.184J/g/°C x 25°C ΔT = 7217 J = 7.217 kJ.

Total heat required = 2.596 kJ + 23.046 kJ + 7.217kJ = 32.859 kJ.

B) 1.5mol water = 27g
latent heat of vaporisation of water at 100°C to steam at 100°C = 2258kJ/kg
we have 27g, So heat required is 2258 X 27/1000 = 60.966 kJ
Heat is supplied at 22.0 J/sec
It will take 60966/22 = 2771 seconds (or) 46.18 minutes.