posted by william on .
Consider the dissolution of CaCl2.
CaCl2(s) Ca2+(aq) + 2 Cl-(aq) ΔH = -81.5 kJ
A 13.0 g sample of CaCl2 is dissolved in 134 g of water, with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/°C·g.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
I assume that is 81.5 kJ/mol. If so then we have 13g x (1mol/110.98g) = ? mol and
81,500 J/mol x ?mol = q
Solve for Tfinal. I get approximately 40 C but I didn't check my figures.