Question
Question: The latent heat of vaporization of water is \(2260\;{\text{J}}/{\text{g}}\). How many kilojoules per...
The latent heat of vaporization of water is 2260J/g. How many kilojoules per gram is this, and how many grams of water will be vaporized by the addition of 2.260×103J heat energy at 100∘C ?
Solution
In order to solve the first part of the question, we need to convert the given value in joules per gram into kilojoules per gram. It is a very simple conversion which can easily be done by unitary method. To solve the second part of the question, we must know the concept of latent heat of vaporization.
Formula Used:
We will use the following conversion factor to solve this question:
1000 J=1kJ
Complete step-by-step answer: For the first part of the question:
We know that 1000J is equal to 1kJ
Then 2260J/g will be equal to
2260gJ×1000J1kJ
On solving we get
2260J/g=2.26kJ/g
For the second part of the question:
In the question, it is stated that 2260J/g is the latent heat of vaporization of water.
We can also observe the second part of the question that the quantity provided to us is 2.260×103J
This is in turn equal to 2.260×103J=2260J, that is, the latent heat of vaporization of water.
So, by the definition of latent heat of vaporization, we can say that 2.260×103J or 2260J is the heat required to vaporize 1g of water at 100∘C.
Hence,
The answer to the first part is 2.26kJ/g
The answer to the second part is 1g of water
Note: Latent heat is defined as the heat or energy that during a phase change of a substance is absorbed or released. It could be from gas to liquid or from liquid to solid or vice versa. A heat property called enthalpy is related to latent heat. The heat consumed or discharged when matter disintegrates is the latent heat of vaporization, changing phase at a consistent temperature from fluid to gas stage.