Respuesta :
Answer : The heat energy required in total for this process was 273529.7 joules.
Solution :
The conversions involved in this process are :
[tex](1):H_2O(l)(50^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\\\(3):H_2O(g)(100^oC)\rightarrow H_2O(g)(110^oC)[/tex]
Now we have to calculate the enthalpy change.
[tex]\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})][/tex]
where,
[tex]\Delta H[/tex] = enthalpy change or heat required = ?
m = mass of water = 110.0 g
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
[tex]c_{p,g}[/tex] = specific heat of water vapor = [tex]1.84J/g^oC[/tex]
n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{110.0g}{18g/mole}=6.11mole[/tex]
[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole
Now put all the given values in the above expression, we get
[tex]\Delta H=[110.0g\times 4.184J/gK\times (100-50)^oC]+6.11mole\times 40670J/mole+[110.0g\times 1.84J/gK\times (110-100)^oC][/tex]
[tex]\Delta H=273529.7J[/tex]
Therefore, the heat energy required in total for this process was 273529.7 joules.
