CaCl2(s) Ca2+(aq) + 2 Cl-(aq)
H = -81.5 kJ
A 14.0 g sample of CaCl2 is dissolved in 130. 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.
I need a step by step process on how to do this.
To calculate the final temperature of the solution, we can use the principle of conservation of energy. The heat released when CaCl2 dissolves in water will be equal to the heat absorbed by the solution, causing a change in temperature. Here's a step-by-step process:
Step 1: Determine the heat released during the dissolution of CaCl2.
Given: H = -81.5 kJ
Since we have a 14.0 g sample of CaCl2, we need to convert it to moles. The molar mass of CaCl2 is:
Ca: 40.08 g/mol
Cl: 35.45 g/mol (x2 for two Cl atoms)
Molar mass of CaCl2 = 40.08 g/mol + (35.45 g/mol * 2) = 110.98 g/mol
Now, calculate moles:
moles of CaCl2 = mass / molar mass
moles of CaCl2 = 14.0 g / 110.98 g/mol = 0.1261 mol
To calculate the heat released, we multiply the moles of CaCl2 by the heat change (H):
Heat released = moles of CaCl2 x H
Heat released = 0.1261 mol x -81.5 kJ/mol = -10.27 kJ
Step 2: Determine the heat absorbed by the solution.
To calculate the heat absorbed, we use the formula:
Heat absorbed = mass of water x specific heat capacity of water x temperature change
Given:
mass of water = 130.0 g
specific heat capacity of water = 4.18 J/°C·g
Assuming the temperature change is ΔT, we can write the following equation:
-10.27 kJ = 130.0 g x 4.18 J/°C·g x ΔT
Step 3: Solve for the temperature change (ΔT).
Rearrange the equation from step 2 to isolate ΔT:
ΔT = -10.27 kJ / (130.0 g x 4.18 J/°C·g)
Step 4: Convert the units and calculate the final temperature.
Substitute the given values and solve for ΔT:
ΔT = -10.27 kJ / (130.0 g x 4.18 J/°C·g)
ΔT ≈ -19.578 °C
The negative sign indicates a decrease in temperature. The final temperature of the solution can be calculated by subtracting ΔT from the initial temperature:
Final temperature = 25.0 °C - 19.578 °C
Final temperature ≈ 5.422 °C
Therefore, the final temperature of the solution is approximately 5.422 °C.
Note the reaction produces 81.5 kJ energy per 1 mol CaCL2.
Convert that to kJ/1 g CaCl2.
q = mass x speific heat water x (Tf-Ti).
mass = 14.0
sp.h. water is in the problem.
Tf is final T
Ti is initial T.