Question

A coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 9.00 g of CaCl2 is added to the calorimeter, what will be the final temperature (in ∘C ) of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol.

Answer 1

Mass of CaCl2 = 9.00 g

Enthalpy of solution = ΔHsoln of CaCl2 = -82.8 kJ/mol

Heat evolved 9.00 g CaCl2 =

= -6.714 kJ

Heat evolved 9.00 g CaCl2 = - 6.714 kJ

Heat absorbed by water = q = -ΔH = - (-6.714 kJ) = 6.714 kJ = 6714 J

But, q = m x c x (T2-T1)

mass of water = m = 100 mL = 100 g (density of water 1 g/ mL)

specific heat = c = 4.18 J/g. ^oC

Initial temperature = T1 = 23 ^oC

Final temperature = T2 = ?

Plug in this values in the formula

q = m x c x (T2-T1)

6714 J = 100 g x 4.18 x (T2-23)

dividing 418 on both sides

T2-23 = 6714/418 = 16

adding 23 on both sides

T2 = 23 + 16 = 39 ^oC

Final temperature = T2 = 39 ^oC