How much energy (in kJ) is required to heat 100.0 g of water from 0.00 C to 100.0 C? ∆Hfus = 6.02 kJ/mol ∆Hvap = 44.0 kJ/mol Cwater = 4.184 J/gK

Question

How much energy (in kJ) is required to heat 100.0 g of water from 0.00 C to 100.0 C?

∆H_fus = 6.02 kJ/mol

∆H_vap = 44.0 kJ/mol

C_water = 4.184 J/gK

Answer 1

To heat water from liquid water at 0 to 100°C is just Q = mc_w(100-0) (adjust to J to kJ)

To go from ice to steam would add mΔH_f + mΔH_v

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