Question

How much heat is required to vaporize 1.5 KG of liquid water at its boiling point

Answer 1

The heat required to vaporize 1.5 kg of liquid water at its boiling point is 3,375,000 J or 3.375 MJ, using the heat of vaporization of water which is 2250 J per gram.

Step-by-step explanation:

The amount of heat required to vaporize a substance is given by its enthalpy of vaporization. For liquid water at its boiling point, the enthalpy of vaporization is about 40.7 kJ/mol. To determine the amount of heat required to vaporize 1.5 kg of liquid water at its boiling point, we use the heat of vaporization. The heat of vaporization of water is approximately 2250 J per gram. Since there are 1000 grams in a kilogram, for 1.5 kg of water we have 1.5 kg * 1000 g/kg = 1500 g of water. To vaporize this amount of water, the heat required is calculated as:

Heat required = mass (in grams) * heat of vaporization

Heat required = 1500 g * 2250 J/g

Thus, the heat required to vaporize 1.5 kg of liquid water at its boiling point is:

Heat required = 3,375,000 J or 3.375 MJ (since 1 MJ = 1,000,000 J).

Answer 2

When water boils it changes from liquid to a gas. In order to do this it is necessary to add energy (the energy 'frees' the water molecules from the liguid and gives them enough energy to exists separate from othe molecules in a gaseous state.) Note - Although you are adding heat (energy) the temperature does not change during the vapourisation. Both th eboiling water and the gaseous steam are at 100degC (under normal conditions) The energy is absorbed by the vapourisation process The amount of energy required to do this (for each Kg) is given by Hv The units of Hv are J/kg Joules PER kilogram PER means 'for each' so 1 kg of water takes 2.26x10^6 J to vapourise. You should see an easy sum to work out how much heat 1.5kg takes.....