Answer : The amount of energy absorbed is, 81.2 kJ
Explanation :
The process involved in this problem are :
[tex](1):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)[/tex]
The expression used will be:
[tex]Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{vap}][/tex]
where,
[tex]Q[/tex] = heat required for the reaction = ?
m = mass of liquid = 30.3 g
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization = [tex]40.79kJ/mol=\frac{40790J/mol}{18.02g/mol}=2263.6J/g[/tex]
Now put all the given values in the above expression, we get:
[tex]Q=[30.3g\times 4.18J/g^oC\times (100-0)^oC]+[30.3g\times 2263.6J/g][/tex]
[tex]Q=81252.48J=81.2kJ[/tex]
Therefore, the amount of energy absorbed is, 81.2 kJ
