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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?

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1 Answer

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.