Heat energy is released when anhydrous calcium chloride is dissolved in water.
CaCl2 (s) ¡æ CaCl2 (aq) + 83 kJ
Calculate the final temperature when 0.01 mol of calcium chloride is dissolved into 100 mL of water initially at a temperature of 18.0 degree celsius.
So how much heat is involved? That is 0.01 mole x 83 kJ/mol = 0.83 kJ or 830 J.
830 = mass H2O x specific heat H2O x (Tfinal-Tinitial).
Solve for Tfinal.
To calculate the final temperature when 0.01 mol of calcium chloride is dissolved into 100 mL of water initially at 18.0 degrees Celsius, we can use the equation for heat energy:
q = mcΔT
Where:
q = heat energy (in joules)
m = mass of the water (in grams)
c = specific heat capacity of water (4.18 J/g·°C)
ΔT = change in temperature (final temperature - initial temperature)
First, we need to calculate the mass of water:
mass = volume × density
density = 1 g/mL (density of water)
mass = 100 mL × 1 g/mL = 100 g
Next, we need to calculate the heat energy released by dissolving 0.01 mol of calcium chloride:
ΔH = 83 kJ (given in the problem)
ΔH in joules = 83 kJ × 1000 J/kJ = 83000 J
Since 0.01 mol of calcium chloride is dissolved, the heat energy released is spread over 100 g of water:
q = 83000 J / 100 g = 830 J/g
Finally, we can find the change in temperature:
q = mcΔT
830 J/g = 100 g × 4.18 J/g·°C × (final temperature - 18.0 °C)
Simplifying the equation:
830 J = 418 J/°C × (final temperature - 18.0 °C)
Divide both sides of the equation by 418 J/°C:
830 J / 418 J/°C = final temperature - 18.0 °C
1.984 = final temperature - 18.0 °C
Adding 18.0 °C to both sides of the equation:
final temperature = 1.984 + 18.0 ≈ 19.984 °C
Therefore, the final temperature when 0.01 mol of calcium chloride is dissolved into 100 mL of water initially at a temperature of 18.0 degrees Celsius is approximately 19.984 °C.
To calculate the final temperature when calcium chloride is dissolved in water, we can use the equation:
q = m * C * ΔT
where:
- q is the heat energy released (in J)
- m is the mass of the solution (in kg)
- C is the specific heat capacity of water (4.18 J/g·°C)
- ΔT is the change in temperature (in °C)
First, let's convert 100 mL of water to grams, using the density of water (1 g/mL):
100 mL * 1 g/mL = 100 g
Next, we need to calculate the mass of the solution, taking into account the mass of calcium chloride and the mass of water.
The molar mass of CaCl2 is 111 g/mol.
Given that we have 0.01 mol of calcium chloride, we can calculate the mass:
mass of CaCl2 = 0.01 mol * 111 g/mol = 1.11 g
Now we can calculate the total mass of the solution:
mass of solution = mass of water + mass of CaCl2
mass of solution = 100 g + 1.11 g = 101.11 g
Next, we need to calculate the heat energy released:
q = 83 kJ = 83,000 J
Now we can rearrange the equation q = m * C * ΔT to solve for ΔT:
ΔT = q / (m * C)
ΔT = 83,000 J / (101.11 g * 4.18 J/g·°C)
ΔT ≈ 197.98 °C
Finally, we can calculate the final temperature when 0.01 mol of calcium chloride is dissolved in 100 mL of water initially at 18.0 °C:
final temperature = initial temperature + ΔT
final temperature = 18.0 °C + 197.98 °C
final temperature ≈ 215.98 °C
So, the final temperature when 0.01 mol of calcium chloride is dissolved into 100 mL of water initially at a temperature of 18.0 °C is approximately 215.98 °C.