Question

Calculate the enthalpy change, ΔH, for the process in which 46.8 g of water is converted from liquid at 0.7 ∘C to vapor at 25.0 ∘C . For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and Cs = 4.18 J/(g⋅∘C) for H2O(l).

Answer 1

Mass of water = 46.8 g

Molar mass of water = 18.02 g/mol

Moles of water = mass/ molar mass = 46.8g/( 18.02 g/mol) = 2.597 mol

Change in temperature = 25-0.7 = 24.3 ^oC

Heat capacity of water = 4.184 J/g.^oC

Heat absorbed by water to raise the temperature from 0.7^oC to 25.0^oC = Mass*heat capacity*change in temperature = 46.8 g*4.18 J/g.^oC* 24.3 ^oC = 4753.66J = 4.754kJ

Heat absorbed to vaporize water = moles* latent heat of vaporization = 2.597mol*44 kJ/mol = 114.27kJ

Thus, total heat absorbed = 4.754kJ+114.27kJ = 119.03 kJ