Question

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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 4.00 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 formed in the calorimeter is the same as that for pure water: Cs=4.184 J/g⋅∘C .Express your answer with the appropriate units.

Answer 1

First, we need to calculate the amount of heat released by the dissolution of 4.00 g of CaCl2:

- The molar mass of CaCl2 is 40.08 g/mol, so 4.00 g corresponds to 4.00/40.08 = 0.0998 mol.

- The heat of solution of CaCl2 is -82.8 kJ/mol, so the heat released by the dissolution of 0.0998 mol is:

- (-82.8 kJ/mol) x (0.0998 mol) = 8.25 kJ.

This heat is absorbed by the calorimeter and the water it contains, causing an increase in temperature. We can use the equation:

q = m x Cs x ΔT

where q is the heat absorbed by the solution (which we just calculated to be 8.25 kJ), m is the mass of the solution (which is the mass of the water plus the mass of the CaCl2), Cs is the specific heat of the solution (which is given to be 4.184 J/g⋅∘C), and ΔT is the change in temperature.

We can rearrange this equation to solve for ΔT:

ΔT = q / (m x Cs)

The mass of the solution is:

- 100 mL of water has a mass of 100 g (since the density of water is 1 g/mL).

- The mass of 4.00 g of CaCl2 is also 4.00 g.

- So the total mass of the solution is 100 g + 4.00 g = 104 g.

Plugging in the values, we get:

ΔT = (8.25 kJ) / (104 g x 4.184 J/g⋅∘C) = 1.94 ∘C.

Therefore, the final temperature of the solution in the calorimeter is:

23.0 ∘C + 1.94 ∘C = 24.9 ∘C.

The final temperature of the solution in the calorimeter is 24.9 ∘C.