Answer
39.2$^{\circ}$C
Work Step by Step
You begin with 11g Ca$Cl_{2}$
Find the molecular mass by adding the atomic mass of calcium plugs two time the atomic mass of chlorine because there are two in the formula
Ca + 2(Cl) = Molar Mass
40.078 + 2(35.4527) = 110.9834g/mol
Convert the grams of Ca$Cl_{w}$ to kJ of amount of energy generated from one mole of Ca$Cl_{w}$ decomposing
11g Ca$Cl_{2}$ * $\frac{1 mol CaCl_{2}}{110.9834g}$ * $\frac{81.5 kJ}{1 mol}$ = 8.077784606 kJ
Use the equation
q = mass * specific heat * change in temperature
(4.18 is the specific heat of water)
(136 g)(4.18)(Tf - 25$^{\circ}$C) = 8077.784606 J
Tf - 25$^{\circ}$C = 14.209$^{\circ}$C
Tf = 39.2$^{\circ}$C
