Question

How much energy transferred when 1 gram of boiling water at $100^\circ C$ turns to water at 0 $^\circ C$ ?

Answer 1

Hint: The elements in nature are present in one of the three forms namely solid, liquid and gas. There comes a state during the change in the state that the temperature remains the same though there is a constant supply of heat. This type of energy is the latent energy for the process.

Complete step by step answer:
When the temperature of the water is raised to $100^\circ C$ it starts to boil. During the process of boiling the water which is being heated changes its state from liquid to gas, that is it changes from water to water vapour and the energy needed for this is called the heat of vaporization.
The mass of water is given in the question as 1 gram. So we need to calculate the energy needed for the conversion of water and water vapour at $100^\circ C$ to water at 0 $^\circ C$.
For solving the questions we need to know that the heat of vaporisation of water is 540 $cal/gram$
The heat needed is represented by Q, and it is given by the relation
$Q = m \times {L_v}$ , where ${L_v}$ is the latent heat of vaporisation.
So, $Q = 1gm \times 540$
$540cal/gm$
So the energy needed for 1 gram of water is 540 cal.
This is the heat needed for the conversion of boiling water at 100 $^\circ C$ to water at 100 $^\circ C$. To change the water to 0 $^\circ C$ the transfer of energy will be an additional 100 cal.
So the total energy that is being transferred in the above process will be represented as
$Q = {Q_1} + {Q_2}$
$540 + 100 = 640\,cal$.

Thus, the total energy transferred is 640 cal.

Note: During the entire cycle of the phase change of the element, there are many processes involved. The process which is involved can be termed as vaporisation. Condensation, evaporation, cooling and freezing. All these processes take place resulting in the change of phase.