Consider the dissolution of CaCl2.

CaCl2(s) Ca2+(aq) + 2 Cl-(aq) ΔH = -81.5 kJ
A 13.0 g sample of CaCl2 is dissolved in 134 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/°C·g.

q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

I assume that is 81.5 kJ/mol. If so then we have 13g x (1mol/110.98g) = ? mol and
81,500 J/mol x ?mol = q
Solve for Tfinal. I get approximately 40 C but I didn't check my figures.

To solve this problem, we can use the principle of energy conservation. We know that the heat absorbed by the solution is equal to the heat released by the dissolution of CaCl2. The equation relating heat, mass, specific heat, and temperature change is:

q = m * c * ΔT

Where:
q = heat (in joules)
m = mass (in grams)
c = specific heat (in J/°C·g)
ΔT = temperature change (in °C)

First, let's calculate the amount of heat released by the dissolution of CaCl2:

Given:
ΔH (enthalpy change) = -81.5 kJ = -81,500 J
Mass of CaCl2 = 13.0 g

The heat released by the dissolution is equal to the enthalpy change of the reaction:

q = ΔH

Now, let's calculate the amount of heat absorbed by the solution:

Mass of water = 134 g
Specific heat of water = 4.18 J/°C·g

q = m * c * ΔT

We need to solve for ΔT, so we rearrange the equation:

ΔT = q / (m * c)

Now, let's substitute the values into the equation:

ΔT = (-81,500 J) / ((13.0 g) * (4.18 J/°C·g) + (134 g) * (4.18 J/°C·g))

ΔT ≈ -320.61 °C

Since temperature cannot be negative in this context, it's likely that the dissolution of CaCl2 will raise the solution's temperature significantly. However, the actual final temperature cannot be determined without additional information.