The heat of solution of potassium acetate (KC2H3O2) in water is -15.3 kJ/mol.
If 18.8 g of KC2H3O2(s) is dissolved in 655 mL of water that is initially at 25.8 °C, what will be the final temperature (in degrees Celcius) of the resulting aqueous solution?
(Assume no heat exchange with the surroundings.)

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  1. q = -15.3 kJ/mol x (18.8/molar mass KC2H3O2) = approx - 3.03 kJ or 3030 J.
    Then -3030 = mass H2O x specific heat H2O x Tfinal-Tinitial)
    Substitute the numbers and solve for Tfinal

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  2. (18.8 g KC2H3O2) / (98.1423 g KC2H3O2/mol) x (15.3 kJ/mol) = 2.931 kJ = 2931 J

    (2931 J) / (4.186 J/g·°C) / (655 g + 18.8 g) + 25.8 °C = 26.84°C = 26.8°C

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