In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 3.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is - 82.8 kJ/mol. Assume that the specific heat of the solution form

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