Question

A thermally isolated vessel contains $100g$ of water at $0^\circ C$. When air above the wall is pumped out, some of the water freezes and some evaporates at $0^\circ C$ itself. Calculate the amount of the ice formed if no water is left in the vessel. Latent heat of vaporization of water at $0^\circ C$ is $2.5\times {{10}^{6}}Jk{{g}^{-1}}$ and latent heat of fusion of ice is $3.34\times {{10}^{5}}Jk{{g}^{-1}}$.

Answer 1

Hint: Use principle of calorimetry, which states that, in an isolated system the total heat lost by a hot body is equal to the total heat gained by a cold body. This law is based on the law of conservation of energy.
Latent heat is the amount of energy required in the form of heat to change the state of 1 gram of the substance.

Formula used:
$Q=mL$

Complete step by step answer:
As the system is isolated, we can use the law of conservation of energy or say law of calorimetry which says for an isolated system the sum of the exchanges of heat is zero.
Let us assume that mass ${{m}_{i}}$ of the water in kilograms converted to ice.
As total mass of the water contained by vessel is 100 g = 0.1 kg, mass converted to vapor is ${{m}_{v}}=0.1-{{m}_{i}}$
The amount of water which freezes releases energy in the form of heat which is used by the remaining portion of water to evaporate. At conversion point, the type of heat which comes into play is latent heat.
Latent heat is the amount of energy required in the form of heat to change the state of 1 gram of the substance. Therefore,
$Q=mL$
According to law of calorimetry,
Heat gained = Heat lost
This implies that,
${{m}_{v}}{{L}_{v}}={{m}_{i}}{{L}_{f}}$
Where ${{L}_{i}}$ and ${{L}_{f}}$ are latent heat of vaporization and latent heat of fusion respectively. It is given that latent heat of vaporization of water at $0^\circ C$ is $2.5\times {{10}^{6}}Jk{{g}^{-1}}$ and latent heat of fusion of ice is $3.34\times {{10}^{5}}Jk{{g}^{-1}}$. Substituting these values, we have
$\left( 0.1-{{m}_{i}} \right)\times \left( 2.5\times {{10}^{6}} \right)={{m}_{i}}\times 3.34\times {{10}^{5}}$
On rearranging and solving this equation, we get
${{m}_{i}}=0.0882kg$
That is, amount of water converted to ice is 88.2 g

Note:
We can also do this kind of problem by checking the amount of heat released and absorbed when converting into a different state. The heat released by freezing of water is absorbed by some amount of water to evaporate.
The latent heat of fusion of water is 79.5 calories/g and latent heat of vaporization for water is 540 calories/g.