Question

The molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar enthalpy of fusion of ice is 6.009 kJ/mol. The molar mass of water is 18.02 g/mol. How much energy is absorbed when 30.3 g of liquid water boils

Answer 1

Answer : The amount of energy absorbed is, 81.2 kJ

Explanation :

The process involved in this problem are :

$(1):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{vap}]$

where,

$Q$ = heat required for the reaction = ?

m = mass of liquid = 30.3 g

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

$\Delta H_{vap}$ = enthalpy change for vaporization = $40.79kJ/mol=\frac{40790J/mol}{18.02g/mol}=2263.6J/g$

Now put all the given values in the above expression, we get:

$Q=[30.3g\times 4.18J/g^oC\times (100-0)^oC]+[30.3g\times 2263.6J/g]$

$Q=81252.48J=81.2kJ$

Therefore, the  amount of energy absorbed is, 81.2 kJ