Consider the dissolution of CaCl2:
CaCl2(s) -----------> Ca2+(aq) + 2Cl-(aq)
ΔH=-81.5 kJ/mol
An 11.0 gram sample of CaCl2 is dissolved in 125 g water
with both substances at 25.0 °C. Calculate the final
temperature of the solution assuming the solution has no
heat loss to the surroundings and assuming that the
solution has a specific heat of 4.18 J/g °C.
First, we need to find the moles of CaCl2 in the 11.0 gram sample:
moles = mass / molar mass
The molar mass of CaCl2 = 40.08 (Ca) + 2*35.45 (Cl) = 110.98 g/mol
moles = 11.0 g / 110.98 g/mol ≈ 0.0991 mol
Now, we'll find the heat released during dissolution:
q = moles * ΔH
q = 0.0991 mol * -81.5 kJ/mol = -8.077 kJ
Now we have to convert 8.077 kJ to J:
-8.077 kJ * 1000 J/kJ = -8077 J
Now we can use the formula for heat transfer in a substance:
q = mcΔT
where
m = the mass of the substance, in this case, water (125 g)
c = the specific heat of the substance (4.18 J/g °C)
ΔT = the change in temperature of the substance, which we need to find
We are solving for ΔT:
ΔT = q / mc
ΔT = -8077 J / (125 g * 4.18 J/g °C) ≈ -15.48 °C
Since the initial temperature was 25.0 °C, the final temperature will be
25.0 - 15.48 ≈ 9.52 °C
To calculate the final temperature of the solution after dissolving CaCl2, we need to use the principle of conservation of energy. The heat gained by the water should be equal to the heat lost by the CaCl2. The equation for this principle is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
Where:
m1 = mass of CaCl2
c1 = specific heat of CaCl2
ΔT1 = change in temperature for CaCl2 (final temperature - initial temperature)
m2 = mass of water
c2 = specific heat of water (which is given as 4.18 J/g °C)
ΔT2 = change in temperature for water (final temperature - initial temperature)
First, let's calculate the heat gained by the water (m2 * c2 * ΔT2). We need to convert the mass of water from grams to kilograms to match the units of specific heat:
m2 = 125 g = 0.125 kg
Now, substituting the values into the equation, we get:
(11.0 g * c1 * ΔT1) + (0.125 kg * 4.18 J/g °C * ΔT2) = 0
Since the heat gained by the water is equal to the heat lost by the CaCl2, we can write:
11.0 g * c1 * ΔT1 = -0.125 kg * 4.18 J/g °C * ΔT2
Now, we can solve this equation for ΔT2 to find the final temperature of the solution.
ΔT2 = (11.0 g * c1 * ΔT1) / (-0.125 kg * 4.18 J/g °C)
Substituting the given values: c1 = specific heat of CaCl2 (which is not provided), and ΔT1 = final temperature - initial temperature.
Without the specific heat of CaCl2, it's not possible to calculate the final temperature of the solution. You would need the specific heat value of CaCl2 to proceed with the calculation.
To calculate the final temperature of the solution, we need to use the concept of heat transfer and the equation:
q = m * c * ΔT
where:
q is the amount of heat transferred
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
1. Calculate the heat change from the dissolution of CaCl2:
First, calculate the moles of CaCl2 using the molar mass of CaCl2:
Moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
Molar mass of CaCl2 = 40.08 g/mol (Ca) + 2 * 35.45 g/mol (Cl)
Molar mass of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 11.0 g / 110.98 g/mol
Next, calculate the heat change using the equation:
Heat Change = moles of CaCl2 * ΔH
Heat Change = (11.0 g / 110.98 g/mol) * (-81.5 kJ/mol)
(Note: The negative sign indicates energy is being released.)
2. Calculate the heat absorbed by water:
q = m * c * ΔT
q = (mass of water) * (specific heat of water) * ΔT
The mass of water is given as 125 g, and the specific heat of water is 4.18 J/g °C.
Let's assume the final temperature of the solution is T.
Thus, the temperature change is ΔT = (T - 25.0 °C).
Substitute these values into the equation:
Heat Change = (125 g) * (4.18 J/g °C) * (T - 25.0 °C)
3. Equate the heat changes:
Since there is no heat loss or gain to the surroundings, the heat released by the dissolution of CaCl2 is equal to the heat absorbed by the water:
(11.0 g / 110.98 g/mol) * (-81.5 kJ/mol) = (125 g) * (4.18 J/g °C) * (T - 25.0 °C)
4. Solve for T:
Convert the units to be consistent:
(11.0 g / 110.98 g/mol) * (-81.5 kJ/mol) = (125 g) * (4.18 J/g °C) * (T - 25.0 °C)
(-9.88 kJ) = (125 g) * (4.18 J/g °C) * (T - 25.0 °C)
Simplifying:
-9.88 kJ = 523.75 J/°C * (T - 25.0 °C)
Divide both sides by 523.75 J/°C:
-9.88 kJ / 523.75 J/°C = T - 25.0 °C
Solve for T:
T - 25.0 °C = -9.88 kJ / 523.75 J/°C
T - 25.0 °C = -0.01885 °C
Add 25.0 °C to both sides:
T = -0.01885 °C + 25.0 °C
T = 24.98 °C
So, the final temperature of the solution assuming no heat loss to the surroundings is approximately 24.98 °C.