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## Homework Statement

Consider the dissolution of CaCl2.

CaCl2(

*s*)

*aq*) + 2 Cl-(

*aq*) Δ

*H*= -81.5 kJ

A 10.6-g sample of CaCl2 is dissolved in 109 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.

## Homework Equations

Q=mcΔT

## The Attempt at a Solution

I had thought Q=ΔH = -81.5kJ = -81.5 x 10^3 J

the mass would be 10.6+109 g

and since the solution overall has a heat capacity of 4.18 J/°Cg, I plugged the numbers in:

81.5 x 10^3 = (10.6+109)x4.18x(T_f-25)

Then I got T_f = 188 °C, which is wrong, but I don't understand why. I don't have much time until the assignment is due, so could someone please help me out? Thank you!