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 2.60 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.

Answer 1

Apply

-Qloss = Qgain

since CaCl2 Hsln is negative, this must be exothermic, so expect Qgain to be cup and Qlost the reaction

Qreaction = n*HRxn

n CaCl2 = mass/MW = 2.6/110.98 = 0.02342 mol of CaCl2 present

Qreaction = n*HRxn = 0.02342 *(-82.8) = -1.93917 kJ

so

Qgain -(Qsystem) = -(-1.93917) = 1.93917 kJ

now...

Qwater --> 100 mL --> 100 g --> 0.1 kg

Qgain = Qwater = m*C*(Tf-Ti)

1.93917 = 0.1*4.184*(Tf-23)

Tf = 1.93917 /( 0.1*4.184) + 23

Tf = 27.6347 °C