What will be the final temp. of the solution in a coffee cip calorimeter if a 50.00mL sample of 0.250 M HCL is added to a 50.00 mL sample of 0.250. The initial temp. is 19.50 °C and the Delta H is -57.2 Kj/mol NaOH
assume density of solution is 1.00 g/mL and specific heat of the solution is 4.18 J/g°c
MY TEACHER DID NOT GIVE US AN EQUATION
I KNOW
Cs is 4.18 j/g°c
Mass is 100g
q idk
I do not know T2 but the T1 is 19.50
You didn't tell us 0.250 WHAT.
First of all, since it's 0.250 M and 50 mL for both reactants, you don't really need to know the equation since both are limiting reactants, but you put that delta H is 57.2 Kj/mol of NaOH. So I'm guessing that NaOH is the other reactant, anyways HCl plus NaOH is a favorite reaction among chem teachers.
First you need to convert 57.2 kJ/mol NaOH to just joules. So since you have 50 g of NaOH*(1mol/40g)*(57.2KJ/1mol NaOH)*(1000J/1kJ) = 71500 J.
Use Q=CmdeltaT, C=4.18, m=100, so then 71500J/(4.18*100)=171.05, which is the change in T. We know that Ti is 19.5 so Tf-19.5=171.05, and that means Tf=190.55 degrees Celsius.
Also, I just noticed, the delta H for NaOH is for the standard enthalpy of formation, which is different from the enthalpy change of the reaction.
If 57.2 was meant to be the enthalpy change of the reaction, then just convert to J, and substitute 57200J instead of 71500J.
**BTW if you have the answer to the question, that would help me understand how to do the question better.
You have how many mols? That's M x L = 0.25 x 0.05 - 0.0125
The reaction releases 57,200 J/mol so in this reaction it releases 57,200 x 0.0125 = 715 J. Now, as Victor has said, substitute that into Q = mcdT
715 = 100 x 4.18 x (Tfinal - 19.5) and solve for Tfinal. I obtained about 21 C but that's just a close estimate. You need to work through it.
To calculate the final temperature of the solution in a coffee cup calorimeter, we can use the principle of heat exchange, which states that the heat gained by the solution is equal to the heat lost by the surroundings.
First, let's determine the heat gained by the solution (q_gain):
q_gain = m × Cs × ΔT
where
m = mass of the solution (in grams)
Cs = specific heat capacity of the solution (in J/g°C)
ΔT = change in temperature of the solution (T2 - T1)
Here, the mass of the solution (m) is given as 100 g, and the specific heat capacity (Cs) is given as 4.18 J/g°C. However, we don't know the change in temperature (ΔT) yet.
Next, let's determine the heat lost by the surroundings (q_loss):
q_loss = -ΔH
where
ΔH = enthalpy change (in J)
Here, the enthalpy change (ΔH) is given as -57.2 kJ/mol NaOH. However, it needs to be converted to joules to match the units with q_gain. Since there are 1000 J in 1 kJ, we can convert ΔH to joules by multiplying it by 1000:
ΔH = -57.2 kJ/mol NaOH × 1000 J/kJ = -57200 J/mol NaOH
Now, we can set up an equation to represent the heat exchange:
q_gain = q_loss
m × Cs × ΔT = -ΔH
Substituting the known values:
100 g × 4.18 J/g°C × ΔT = -57200 J/mol NaOH
Rearranging the equation to solve for ΔT:
ΔT = (-57200 J/mol NaOH) / (100 g × 4.18 J/g°C)
Now, plug in the values and calculate ΔT:
ΔT = -13.675°C
Finally, to find the final temperature (T2), we add the change in temperature (ΔT) to the initial temperature (T1):
T2 = T1 + ΔT
T2 = 19.50°C + (-13.675°C)
Calculating this, we find:
T2 ≈ 5.825°C
Therefore, the final temperature of the solution in the coffee cup calorimeter is approximately 5.825°C.