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* c_(p,l)* (T_(final)-T_(initial))]+[m* \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=(40790J/mol)/(18.02g/mol)=2263.6J/g`

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

`Q=[30.3g* 4.18J/g^oC* (100-0)^oC]+[30.3g* 2263.6J/g]`

`Q=81252.48J=81.2kJ`

Therefore, the amount of energy absorbed is, 81.2 kJ