Chem Help Please?

Question

Consider the dissolution of CaCl2.

CaCl2(s) --> Ca2+(aq) + 2 Cl-(aq) (Delta)H = -81.5 kJ

A 13.2 g sample of CaCl2 is dissolved in 119 g of water, with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/°Cg.

Answer 1

Presuming (delta)H is for one mole of CaCl2 dissolving, 13.2 g is 0.119 moles (MW = 111), so the heat given off would be 9.70 kJ. Assuming the water absorbs that heat, heat = specific heat* (delta)T * mass

(delta)T = heat/(specific heat * mass) = 9700 J/(4.18 * 132) = 17.6 degrees.

Final temperature should be 42.6 degrees Centigrade.

Answer 2

Total mass = 119+13.2 g = 132.2 g

So,

132.2 * 4.18 *(T-25) = 81500

or T-25 = 147.49

or T = 122.49 deg C