Consider the dissolution of CaCl2:

Question

CaCl2(s) -----------> Ca2+(aq) + 2Cl-(aq)
ΔH=-81.5 kJ/mol
An 11.0 gram sample of CaCl2 is dissolved in 125 g water
with both substances at 25.0 °C. Calculate the final
temperature of the solution assuming the solution has no
heat loss to the surroundings and assuming that the
solution has a specific heat of 4.18 J/g °C.

Answer 1

First, we need to find the moles of CaCl2 in the 11.0 gram sample:
moles = mass / molar mass

The molar mass of CaCl2 = 40.08 (Ca) + 2*35.45 (Cl) = 110.98 g/mol

moles = 11.0 g / 110.98 g/mol ≈ 0.0991 mol

Now, we'll find the heat released during dissolution:
q = moles * ΔH

q = 0.0991 mol * -81.5 kJ/mol = -8.077 kJ

Now we have to convert 8.077 kJ to J:
-8.077 kJ * 1000 J/kJ = -8077 J

Now we can use the formula for heat transfer in a substance:
q = mcΔT

where
m = the mass of the substance, in this case, water (125 g)
c = the specific heat of the substance (4.18 J/g °C)
ΔT = the change in temperature of the substance, which we need to find

We are solving for ΔT:
ΔT = q / mc

ΔT = -8077 J / (125 g * 4.18 J/g °C) ≈ -15.48 °C

Since the initial temperature was 25.0 °C, the final temperature will be
25.0 - 15.48 ≈ 9.52 °C