Calculate the enthalpy change, ΔH, for the process in which 30.1 g of water is converted from liquid at 10.1 ∘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).

You can split the process in two parts:

1) heating the liquid water from 10.1 °C to 25.0 °C, and

2) vaporization of liquid water at constant temperature of 25.0 °C.

For the first part, you use the formula ΔH = m*Cs*ΔT

ΔH = 30.1g * 4.18 j / (g°C) * (25.0°C - 10.1°C) = 1,874 J

For the second part, you use the formula ΔH = n*ΔHvap

Where n is the number of moles, which is calculated using the mass and the molar mass of the water:

n = mass / [molar mass] = 30.1 g / 18.0 g/mol = 1.67 mol

=> ΔH = 1.67 mol * 44,000 J / mol = 73,480 J

3) The enthalpy change of the process is the sum of both changes:

ΔH total = 1,874 J + 73,480 J = 75,354 J

Answer: 75,354 J