At 1 atm, how much energy is required to heat 83.0 g of H2O(s) at –12.0 °C to H2O(g) at 115.0 °C?
Specific heat capacity, ice: 2.108 kJ/kg-K
Specific heat capacity, water: 4.187 kJ/kg-K
Specific heat capacity, water vapor: 1.996 kJ/-kgK
Heat of fusion of water: 334 J/g
Heat of vaporization of water: 2.26 kJ/g
There are two phase changes occurring. Water in its different phases has a different heat capacity. Then the phase changes take energy as well.
(83 g ice)(12 °C)(2.108 kJ/kg-K) +
(83 g ice)(334 J/g) +
(83 g water)(100 °C)(4.187 kJ/kg-K) +
(83 g water)(2.26 KJ/g) +
(83 g steam)(15 °c)(1.996 kJ/kg-K) = 254 KJ
Note the relatively high heat of vaporization of water. It takes a lot of energy to make steam.