q = -82.8 kJ/mol x (3.40/molar mass CaCl2) = about -2.5 kJ = about -2500 J.
Then -2500 = mass H2O x specific heat H2O x (T2-T1) and solve for T2. I get approximately 28 or so.
In the following experiment, a coffee-cup calorimeter containing 100mL of H2 is used. The initial temperature of the calorimeter is 23.0 C . If 3.40g CaCl2 of is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution of CaCl2 is -82.8 kJ/mol.
2 answers
28.9 degrees C
5 answers