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Answered

Calculate the enthalpy change, ΔH, for the process in which 14.5g of water is converted from liquid at 10.5∘C to vapor at 25.0∘C .


For water, ΔHvap = 44.0kJ/mol at 25.0∘C and Cs = 4.18 J/(g⋅∘C) for H2O(l).

Verified Answer :

Edufirst

You need to calculate two stages: 1) heating of liquid water from 10.5 °C to 25.0° C, and 2) vaporization of water at 25.0°C.

1) Heating of liquid water

ΔH = mCsΔT = 14.5 g (1kg / 1000g) 4.18 J/(kg°C) (25.0°C - 10.5°) = 0.878845 J

2) Heating of vaporization

You need to pass the mass in grams to moles, that is why is divide by the molar mass of water, 18 g/mol.

ΔH = n ΔH vap = [14.5g / 18 g/mol] * 44.0 J/mol = 35.44 J

3) Total enthalpy change

ΔH = 0.879 J + 35.44 J =  36.32 J

Answer: 36.32 J

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