Respuesta :
Answer: The final temperature is 8.13°C
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of NaOH = 13.9 g
Molar mass of NaOH = 40 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of NaOH}=\frac{13.9g}{40g/mol}=0.35mol[/tex]
- To calculate the heat released in the reaction, we use the equation:
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
[tex]q[/tex] = amount of heat released
n = number of moles = 0.35 moles
[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction = -44.4 kJ/mol
Putting values in above equation, we get:
[tex]-44.4kJ/mol=\frac{q}{0.35mol}\\\\q=(-44.4kJ/mol\times 0.35mol)=-15.54kJ[/tex]
- To calculate the final temperature, we use the equation:
[tex]q=mc\Delta T=mc(T_2-T_1)[/tex]
where,
q = heat released = -15.54 kJ = -15540 J (Conversion factor: 1 kJ = 1000 J)
m = mass of water = 250.0 g
c = specific heat capacity of water = 4.18 J/g.K
[tex]T_1[/tex] = initial temperature = [tex]23^oC=[23+273]=296K[/tex]
[tex]T_2[/tex] = final temperature = ?
Putting values in above equation, we get:
[tex]-15540J=250g\times 4.18J/g.K\times (T_2-296)\\\\T_2=281.13K[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]281.13=T(^oC)+273\\T(^oC)=8.13^oC[/tex]
Hence, the final temperature is 8.13°C
