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28 October, 20:10

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).

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  1. 28 October, 20:33
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    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
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