q = CM∆T CALORIMETRY NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)

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q = CM∆T CALORIMETRY NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) In a coffee cup calorimeter 100.0 mL of NaOH is mixed with 100.0 mL of 1.0 molar HCl. Both solutions were at 24.6 C, with the temperature after the reactiojn at 31.3 C. Assuming all the solutions have a density of 1.0 grams/ml, and a heat capacity of 4.18 J/C * g calculate the enthalpy of neutralization for the reaction. Assume 100 efficiency. NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) q = CM∆T

q = CM∆T 200.0mL x 1.0g/ml = 200.0g q = (4.18 J/gC)(200g)(31.3-24.6) Q= -5.6 X 103 J, -5.6 kJ