91.3k views
3 votes
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

1 Answer

0 votes

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

User Ilya Palkin
by
7.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.