some solutes have large heats of solution, and care should be taken in preparing solutions of these substances. The heat evolved when sodium hydroxide dissolves is 44.5 KJ/mol. What is the final temperature of the water, originally at 20.0 degrees Celcius. used to prepare 500.0 cm cubed of 6.00M NaOH solution? Assume all the hat is absorbed by 500cm cubed of water, specific heat =4.18 J/g Degrees Celcius!!!!
1 mol NaOH in 1L solution = 1M
6 mols NaOH in 1 L solution = 6.00 M
3 mols NaOH in 1/2 L (500 cc) soln = 6.00 M,
q = 3 mols x 44.5 kJ/mol = 133.5 kJ.
The quantity of q is in kJ but specific heat of water is in J/g so we need to change kJ to J.
133.5 kJ x 1000 J/kJ = 133,500 J.
Then
q = massH2O x specific heat x (Tf - Ti)=
where Tf is final T and Ti is initial T.
133,500=500 x 4.18 x (Tf - 20)=
Solve for Tf.
Note the correct spelling of Celsius.
Post your work if you get stuck and need further assistance.
To find the final temperature (Tf) of the water, we can rearrange the equation q = massH2O x specific heat x (Tf - Ti) and solve for Tf.
Given:
q = 133,500 J
massH2O = 500 g (since density of water is 1 g/cm^3, 500 cm^3 is equal to 500 g)
specific heat = 4.18 J/g°C
Ti = 20°C
Substituting the values into the equation:
133,500 = (500 g) x (4.18 J/g°C) x (Tf - 20°C)
Now we can solve for Tf:
133,500 = 2090 x (Tf - 20)
(Tf - 20) = 133,500 / 2090
(Tf - 20) = 63.94
Adding 20 to both sides:
Tf = 63.94 + 20
Tf = 83.94°C
Therefore, the final temperature of the water used to prepare the NaOH solution is 83.94°C.
To find the final temperature of the water, we can rearrange the equation:
q = massH2O × specific heat × (Tf - Ti)
where q is the heat absorbed by the water, massH2O is the mass of water (500 cm^3 = 500 g), specific heat is 4.18 J/g°C, Tf is the final temperature, and Ti is the initial temperature (20°C).
Substituting the known values:
133,500 J = 500 g × 4.18 J/g°C × (Tf - 20)
To solve for Tf, we can isolate it by dividing both sides of the equation by (500 g × 4.18 J/g°C):
(Tf - 20) = (133,500 J) / (500 g × 4.18 J/g°C)
Simplifying:
(Tf - 20) = 63.73°C
Add 20 to both sides:
Tf = 63.73°C + 20°C
Tf = 83.73°C
Therefore, the final temperature of the water is 83.73 degrees Celsius.