Question

How much heat is required to convert $10{\text{ }}g$ of ice at $- 10^\circ C$ into steam at $100^\circ C$?

Answer 1

The molecular formula of water is${{\text{H}}_{\text{2}}}{\text{O}}$. The molecular mass of water is${\text{18g}}$. The water has intermolecular hydrogen bonding between the water molecules. Water can be seen in all the states of matter. The solid-state of water is called ice cubes. The gas form of water is called such water vapor. The liquid form of water is called such water. Hydrogen combines with oxygen in the presence of an electric current to form the product of water. Water is one of the good conductors of an electric city.
${\text{2}}{{\text{H}}_{\text{2}}}{\text{ + }}{{\text{O}}_{\text{2}}} \to {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}}$

Complete answer:
Water is made up of a hydrogen bond. There are two types of hydrogen bonds. They are intermolecular hydrogen bonds and intramolecular bonds. The water is made up of an intermolecular hydrogen bond at room temperature.
The latent heat of melting ice of $1{\text{ }}g$ water changes into steam is $333.6{\text{ }}J/g$.
The latent heat of melting ice of $10{\text{ }}g$ water changes into steam is $3336{\text{ }}J$.
According to the above discussion, we conclude the amount of heat is required to convert $10{\text{ }}g$of ice $- 10^\circ C$into steam at $100^\circ C$ is $3336{\text{ }}J$.

Note:
Water is classified into two types. This classification is based on the amounts of the mineral content. The two types of water are hard water and soft water. Hard water means the water contains high amounts of minerals ions. Soft water means the water contains low amounts of minerals ions. The hardness of water is divided into two types. There is the temporary and permanent hardness of the water.