The ΔH per mole of CaCl2 is -81.0 kJ. Calculate the final temperature of an aqueous solution if 30 g of CaCl2 are added to 100 g of water at 23oC. Assume that the specific heat of the solution is the same as that of water

To calculate the final temperature of the aqueous solution after adding CaCl2, we can use the formula:

q = m × c × ΔT

Where:
q is the heat gained or lost by the system
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature

In this case, we have CaCl2 and water. We need to calculate the heat gained or lost when CaCl2 dissolves in water.

First, calculate the heat gained or lost by CaCl2:
q1 = m1 × ΔH1

Where:
q1 is the heat gained or lost by CaCl2
m1 is the mass (in moles) of CaCl2
ΔH1 is the heat per mole of CaCl2 (-81.0 kJ/mol)

To find the mass of CaCl2 in moles (m1), we need to know the molar mass of CaCl2:
Ca: 40.08 g/mol
Cl: 35.45 g/mol × 2 = 70.90 g/mol
Total molar mass of CaCl2 = 40.08 + 70.90 = 111.98 g/mol

m1 = 30 g ÷ 111.98 g/mol = 0.268 mol

q1 = 0.268 mol × (-81.0 kJ/mol) = -21.708 kJ

Next, calculate the heat gained or lost by water:
q2 = m2 × c × ΔT

Where:
q2 is the heat gained or lost by water
m2 is the mass of water
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (final temperature - initial temperature)

Given:
m2 = 100 g
c = 4.18 J/g°C
Initial temperature = 23°C

Let's assume the final temperature of the solution is T.

q2 = 100 g × 4.18 J/g°C × (T - 23)°C

Since the specific heat capacity of the solution is assumed to be the same as that of water, we can equate the heat gained or lost by CaCl2 to the heat gained or lost by water:

q1 = q2
-21.708 kJ = 100 g × 4.18 J/g°C × (T - 23)°C

Now, we can solve for the final temperature (T):

-21.708 kJ = 418 J/°C × (T - 23)°C

Convert kJ to J:
-21.708 kJ × 1000 J/kJ = 418 J/°C × (T - 23)°C

-21708 J = 418 J/°C × (T - 23)°C

Divide both sides of the equation by 418 J/°C:

-21708 J ÷ 418 J/°C = T - 23

-51.9462 °C = T - 23

Add 23 to both sides of the equation:

T = -28.9462 °C

Therefore, the final temperature of the aqueous solution after adding 30 g of CaCl2 to 100 g of water at 23°C is approximately -28.95 °C.

To calculate the final temperature of the solution, we need to use the equation for heat transfer:

q = m * c * ΔT

Where:
q - heat transfer (in J)
m - mass of the substance (in g)
c - specific heat capacity (in J/g°C)
ΔT - change in temperature (in °C)

First, let's calculate the heat transfer from the dissolution of CaCl2:

q1 = ΔH * n

Where:
q1 - heat transfer for the dissolution of CaCl2 (in J)
ΔH - enthalpy change per mole of CaCl2 (in kJ/mol)
n - number of moles of CaCl2

Since we have the mass of CaCl2 (30 g), we need to calculate the number of moles using its molar mass:

Molar mass of CaCl2 = (1 * atomic mass of Ca) + (2 * atomic mass of Cl)
= (1 * 40.08 g/mol) + (2 * 35.45 g/mol)
= 40.08 g/mol + 70.90 g/mol
= 111.98 g/mol

n = mass / molar mass
= 30 g / 111.98 g/mol
≈ 0.268 mol

Now we can calculate q1:

q1 = ΔH * n
= -81.0 kJ/mol * 0.268 mol
≈ -21.708 kJ

We also need to calculate the heat transfer for the water:

q2 = m * c * ΔT

Where:
q2 - heat transfer for the water (in J)
m - mass of the water (in g)
c - specific heat capacity of water (4.18 J/g°C)
ΔT - change in temperature for the water

Given that the mass of the water is 100 g and the initial temperature is 23°C, but we need to find ΔT.

To find ΔT, we can use the equation:

q1 + q2 = 0

Since the q1 is negative (exothermic), the equation becomes:

q2 = -q1

We can substitute the values to solve for ΔT:

m * c * ΔT = -q1
100 g * 4.18 J/g°C * ΔT = -(-21.708 kJ * 1000 J/kJ)
100 * 4.18 * ΔT = 21708
ΔT = 21708 / (100 * 4.18)
ΔT ≈ 52°C

Finally, to find the final temperature, we add the initial temperature with ΔT:

Final temperature = Initial temperature + ΔT
Final temperature = 23°C + 52°C
Final temperature ≈ 75°C

Therefore, the final temperature of the solution is approximately 75°C.

Q=mc∆t

where
Q=heat
m= total mass
c=specific heat of solution= 4.186 x 10^-3 kilojoule/gram °C
∆t=final temp- initial temp

solve for final temperature