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 7.10 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.

Express your answer with the appropriate units.

Answer 1

Mass of CaCl₂ = 7.10g

Moles of CaCl₂ = Mass of CaCl₂ / Molar Mass of CaCl₂

Moles of CaCl₂ = 7.10g / 111g/mol

Moles of CaCl₂ = 0.064 mol

Heat of solution per mol of CaCl₂ = -82.8kJ

Heat of solution for 0.064 mol of CaCl₂ = -82.8 x 0.064 kJ

Heat of solution for 0.064 mol of CaCl₂ = -5.2992kJ = -5299.2J [Since 1kJ = 1000J]

Mass of water = Density of water x Volume of water

Mass of water = 1g/ml x 100ml

Mass of water = 100g

Heat = Mass x Heat Capacity x T

5229.2J = 100g x 4.184J/g⁰C x T

T = 12.665⁰C

Final Temperature = 23⁰C + 12.665⁰C = 35.665⁰C