Question

The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 9.00 g of steam condenses to liquid water at 100°c.

Answer 1

Answer:

20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.

Explanation:

Heat is being consumed during vaporization and heat is being released during condensation.

To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.

Molar mass of water = 18.02 g/mol

Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.

So, to vaporize 9.00 g of water, $(\frac{40.66}{18.02}\times 9.00)kJ$ of heat or 20.3 kJ of heat is being consumed

As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.