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 8.10 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.

Answer 1

molar mass of CaCl2 = 111.0 g/mol
number of mole of CaCl2 = (given mass)/(molar mass)
= (8.10/111.0) mole
= 0.0730 mole

heat releases by dissolving CaCl2 = (number of mole of CaCl2)*(delta Hsoln)
= (82.8*0.0730) KJ
= 6.044 KJ
= 6044 J

let the final temperature be T
mass of water = (volume)*(density)
= (100*1) g
= 100 g
total mass of solution = (100+8.10) g
= 108.10 g

assuming specific heat capicity of solution = assuming specific heat capicity of water
= 4.18 J/g.oC

so, use
heat releases by dissolving CaCl2 = (total mass of solution)*s*(T2-T1)
6044 = 108.10*4.18*(T-23.0)
6044 = 451.9*(T-23.0)
(T-23.0) = 13.38
T = 36.38 oC
= 36.4 oC

Answer : 36.4 oC