Question

In the following experiment, a coffee-cup calorimeter containing 100. mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 2.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution, ΔHsoln, of CaCl2 is −82.8 kJ/mol. The specific heat of water is CS=4.184 J/(g−K

Answer 1

given the mass of CaCl2 = 2.00 g

molecular mass of CaCl2 = 111.0 g/mol

Hence moles of CaCl2 = 2.00 g / 111.0 g/mol = 0.01802 mol

Given ΔHsoln = - 82.8 KJ/mol (exothermic)

Hence the amount of heat liberated by 2.00 g of CaCl2 = (- 82.8 KJ/mol) x 0.01802 mol = 1.492 KJ = 1492 J

Hence heat accepted by water = 1492 J

mass of water = 100.0 mL x 1g/mL = 100 g

Let the final temperature be 'T' DegC

=> mxSxT = 1492 J

=> 100 g x 4.184 J/(g−C) x (T - 23.0) = 1492 J

=> T - 23.0 = 3.566

=> T = 26.566 DegC (answer)