The solution is as follows:
Since the mass of the calorimeter is given, let's take the heat effects of the calorimeter as negligible.
Compute for sensible heat from 24.60°C (297.6 K) to 298 K. The heat capacity of water is 4.18 J/g·K.
ΔH₁ = ∫[(0.1 mol/L AgNO₃)(0.1 L)(169.87 g/mol AgNO₃)(4.18 J/g·K)dT + (0.2 mol/L NaCl)(0.1 L)(58.44 g/mol NaCl)(4.18 J/g·K)dT]
Take the integral from limits 297.6 K to 298.8 K(25.30°C).
ΔH₁ = 4,289.12 J
Compute for the heat of reaction at room temperature:
ΔH₂ = ∑(Heat of formation of products*stoichiometric coefficient) - ∑(Heat of formation of reactants*stoichiometric coefficient)
ΔH₂ = [(-127 kJ/mol AgCl)(1 mol) + (-467 kJ/mol NaNO₃)(1 mol)] - [(-123.02 kJ/mol AgNO₃)(1 mol) + (−407.27 kJ/mol NaCl)(1 mol)]
ΔH₂ = -63.71 kJ or -63,710 J
Thus,
ΔHrxn = ΔH₁ + ΔH₂ = 4,289.12 J + -63,710 J
ΔHrxn = 59,420.88 J or 59.42 kJ
