Question

How much heat is absorbed when 52.3 g H2O(l) at 100 C and 101.3 kPa is converted to steam at 100 C? The molar heat of vaporization of water is 40.7 kJ/mol.

Answer 1

Answer:

118.25 kJ heat is absorbed when 52.3 g H2O(l) at 100 C and 101.3 kPa is converted to steam at 100 C.

Explanation:

Mass of water = 52.3 g

Moles of water = $\frac{52.3 g}{18 g/mol}=2.9055 mol$

$H_2O(l)\rightleftharpoons H_2O(g),\Delta H_{vap}=40.7 kJ/mol$

When 1 mole of water is vaporized it absorbs 40.7 kJ of energy.

So, when 2.9055 moles of water gets vaporized the heat absorbed by the water is:

$2.0955 mol\times 40.7kJ/mol=118.25 kJ$

118.25 kJ heat is absorbed when 52.3 g H2O(l) at 100 C and 101.3 kPa is converted to steam at 100 C.