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 9.60 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

The heat of solution says that when 1 mole of CaCl2 dissolves, it will release 82.8 kJ of heat. So, you need to calculate how many moles of CaCl2 you used, and how much heat that amount of CaCl2 will release:

9.60g / 111 g/mol = 0.0865 moles

0.0856 moles X -82.8 kJ/mol = -7.16 kJ

Now, since the dissolving of CaCl2 releases heat (is exothermic) the water in the calorimeter will get warmer. You can use this equation to calculate the final temperature of the water:

q = m c (T2-T1) where m is the mass of the water, c is water's specific heat, and T2 and T1 are the final and initial temperatures of the water:

Since specific heat of water is 4.184 J/gC, convert your heat released into joules first. Then, plug things into the equation and calculate T2:

7160 J = (100 g)(4.184 J/gC)(T2-23)
T2-23 = 17.11
T2 = 40.11 C