An 11.1-g sample of CaCl2 is dissolved in 123 g water, with both substances at 24.0°C. Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/°C·g.
To calculate the final temperature of the solution, we can use the principle of conservation of energy. The total energy gained by the water and CaCl2 must be equal to the total energy lost, assuming no heat loss to the surroundings.
The equation we can use is:
m1 * c1 * (Tf - Ti) + m2 * c2 * (Tf - Ti) = 0
Where:
m1 = mass of water
c1 = specific heat capacity of water
m2 = mass of CaCl2
c2 = specific heat capacity of CaCl2
Tf = final temperature of the solution
Ti = initial temperature of the solution
Let's plug in the values given in the problem:
m1 = 123 g (mass of water)
c1 = 4.18 J/°C·g (specific heat capacity of water)
m2 = 11.1 g (mass of CaCl2)
c2 = ?? (specific heat capacity of CaCl2)
To find the specific heat capacity of CaCl2, we need to refer to a reliable source or use a heat capacity table. Let's assume the specific heat capacity of CaCl2 is also 4.18 J/°C·g.
Now we can rewrite the equation as:
123 * 4.18 * (Tf - 24.0) + 11.1 * 4.18 * (Tf - 24.0) = 0
We can simplify the equation:
514.14 * (Tf - 24.0) + 46.338 * (Tf - 24.0) = 0
Now, let's solve for Tf:
560.478 * Tf - 13357.216 + 46.338 * Tf - 1112.512 = 0
606.816 * Tf - 14469.728 = 0
606.816 * Tf = 14469.728
Tf = 14469.728 / 606.816
Tf ≈ 23.85°C
Therefore, the final temperature of the solution, assuming no heat loss to the surroundings, is approximately 23.85°C.