Respuesta :
Answer : The [tex]C_p[/tex] of the the calorimeter is, [tex]98.35J/^oC[/tex]
Explanation :
First we have to calculate the moles of NaOH.
[tex]\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }NaOH=5.6mole/L\times 0.0044L=0.02464mole[/tex]
conversion used for volume : (1 L = 1000 ml)
As we know that, HCl and NaOH are the strong acid and bases. So, the heat of neutralisation is -55.9 kJ/mole.
Now we have to calculate the heat released by neutralization.
As, 1 mole of heat released by neutralization = 55.9 kJ
So, 0.02464 mole of heat released by neutralization = [tex]0.02464\times 55.9kJ=1.377kJ[/tex]
Now we have to calculate the [tex]C_p[/tex] of the the calorimeter.
[tex]Q=C_p\times \Delta T[/tex]
where,
Q = heat of reaction = 1.377 kJ
[tex]C_p[/tex] = specific heat capacity of calorimeter =?
[tex]\Delta T[/tex] = change in temperature = [tex]14^oC[/tex]
Now put all the given values in this expression, we get:
[tex]1.377kJ=C_p\times 14^oC[/tex]
[tex]C_p=0.0983kJ/^oC=98.35J/^oC[/tex]
Therefore, the [tex]C_p[/tex] of the the calorimeter is, [tex]98.35J/^oC[/tex]
The Cp of the calorimeter is equal to 98.35J/ºC
Steps to get to this answer:
- First, it is necessary to know the molarity of NaOH. This will be done with the following equations:
[tex]Mole_N_a_O_H= 5.6 mol/L*0.0044L= 0.02464 mole[/tex]
- We must identify the heat released during the neutralization reaction. For this, we must consider that the neutralization of 1 mol of HCl or NaOH releases -55.9 kJ/mole. In this case, we will use the following equation:
[tex]\left \ {{1mole=-55.9 kJ} \atop {0.02464 mole=x}} \right.\\x= 55.9*0.02464 mole\\x= 1.377kJ[/tex]
- From this, we can calculate the Cp of the calorimeter and this will be done with the following equation:
[tex]Q=C_p*\Delta T\\1,377= C_p*14 \textdegree C\\C_p= \frac{1.377}{14} = 98.35J\textdegree C[/tex]
More information about calorimetry at the link:
https://brainly.com/question/15011162
