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Compute the change in temperature when 1.1 g of sodium chloride is dissolved in 200.0 mL of water initially at 28 C. The heat capacity of water is 4.184 J/mL/K. Neglect the heat capacity of the NaCl.

Species ΔfH (kJ/mol)
NaCl(s) -411.2

  • Thermochemistry -

    DH = delta H.
    DHsoln = DHsolvation - DHlatticeenergy
    DHsoln = -407.3 -(-411.2) = 3.9 kJ/mol
    The + means it is an endothermic reaction and the water will become cooler; we are extracting heat from the water. How much? 3.9 kJ/mol x (1000 J/kJ) x 1.1 g NaCl x (1 mol NaCl/molar mass NaCl) = q and since we are extracting heat it is -q.
    Then -q = mass H2O x specific heat H2O x (Tfinal-Tinitial) and solve for Tfinal.

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