Respuesta :
Answer : The final temperature of the solution in the calorimeter is, [tex]31.6^oC[/tex]
Explanation :
First we have to calculate the heat produced.
[tex]\Delta H=\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = 82.8 kJ/mol
q = heat released = ?
m = mass of [tex]CaCl_2[/tex] = 5.10 g
Molar mass of [tex]CaCl_2[/tex] = 110.98 g/mol
[tex]\text{Moles of }CaCl_2=\frac{\text{Mass of }CaCl_2}{\text{Molar mass of }CaCl_2}=\frac{5.10g}{110.98g/mole}=0.0459mole[/tex]
Now put all the given values in the above formula, we get:
[tex]82.8kJ/mol=\frac{q}{0.0459mole}[/tex]
[tex]q=3.80kJ[/tex]
Now we have to calculate the final temperature of solution in the calorimeter.
[tex]q=m\times c\times (T_2-T_1)[/tex]
where,
q = heat produced = 3.80 kJ = 3800 J
m = mass of solution = 100 + 5.10 = 105.10 g
c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]23.0^oC[/tex]
[tex]T_2[/tex] = final temperature = ?
Now put all the given values in the above formula, we get:
[tex]3800J=105.10g\times 4.18J/g^oC\times (T_2-23.0)[/tex]
[tex]T_2=31.6^oC[/tex]
Thus, the final temperature of the solution in the calorimeter is, [tex]31.6^oC[/tex]
