Question

How much heat (in $ \text{kJ} $ ) is absorbed when $ 24.8~\text{g} {{\text{H}}_{2}}\text{O}(l) $ at $ 100^{\circ} \text{C} $ and $ 101.3 \text{kPa} $ is converted to steam at $ 100^{\circ} \text{C} ? $

Answer 1

Hint: The Heat of Vaporization is the amount of heat that needs to be absorbed at a constant temperature to vaporize a specific amount of liquid. The kinetic energy of the steam is demonstrated to be higher than the kinetic energy of the fluid if the solutions of vapour and liquid states are compared.

Formula Used: We will use the following formula to solve the question above
 $ q=m\times \Delta {{H}_{v}} $
Where
 $ q $ is the heat energy
 $ m $ is the mass of water
 $ \Delta {{H}_{v}} $ is the heat of vaporization.

Complete Step-by-Step Solution
According to the question, the reaction can be shown as follows:
 $ H_{2} O(l)+\Delta \rightarrow H_{2} O(g) $
Water is converted to its vapour form. This indicated that the water is already at its boiling temperature and the only thing we require to solve this question is the heat of vaporization of water
The heat of vaporization of water, $ \Delta {{H}_{v}} $ is $ 2257 J/g $
The mass of the water is given as $ 24.8 g $
Now, we will put all the known values into the formula stated above
 $ q=m\times \Delta {{H}_{v}} $
Upon substituting the values, we obtain
 $ q=24.8\times 2257 $
On solving, we obtain
 $ q=5973.6 J $
We can rewrite it as
 $ \therefore q=55.97 kJ $
Hence, the required heat is $ 55.97 kJ $ .

Additional Information
In other words, we can say that vaporization heat is the total quantity of heat required to convert a specific quantity of liquid into its vapor form without any major temperature increase.

Note
This is a very simple problem which is directly based on a single formula. Students should remember the value of heat of vaporization of water or else they would never be able to come to the conclusion of the given problem.