Answer:
28.3 °C
Explanation:
We assume that the specific heat of the solution is equal to the specific heat of water (4.184 Jg⁻¹ °C⁻¹).
First, we find the heat released by the dissolution of CaCl₂. The grams will be converted to moles using the molar mass (110.986 g/mol), then multiplied by the molar heat of solution:
(3.00 g)(mol/110.986 g)(-82.8 kJ/mol) = -2.23812 kJ
The negative sign indicates that heat is released. Extra significant figures are included to avoid round-off errors.
The amount of heat released by the CaCl₂ dissolution is equal to the heat absorbed by the water. The equation is rearranged to solve for Δt, the temperature change of the water.
Q = mcΔt ⇒ Δt = Q/(mc)
Δt = (2.23812 kJ)(1000 J/kJ) / (100 mL)(1 g/mL)(4.184 Jg⁻¹ °C⁻¹) = 5.3 °C
We can then calculated the final temperature t₂ of the solution:
Δt = t₂ - t₁
t₂ = Δt + t₁ = 5.3 °C + 23.0 °C = 28.3 °C