The latent heat of vaporization 5gram of steam converted to liquid at 100°C is 11.3 KJ.
The latent heat of vaporization for a given substance tells you how much energy is required for one mole of that substance to undergo a phase transition or go from a liquid to a gas, at its boiling point.
Joules per gram, an alternative to the more popular kilojoules per mole, are used to express the latent heat of vaporization for water.
Therefore, we must determine how many kilojoules per gram are needed for a certain sample of water to transition from a liquid to a vapor at its boiling point.
As you know, the conversion factor that exists between Joules and kilojoules is 1 kJ = 10³ J
2260 J/g will be equivalent to
[tex]2260 \frac{J}{g} . \frac{1kJ}{1000J } = 2.26 kJ/g\\\\[/tex]
As we know,
2260 = 2.26 . 10³
which means that 2.26 .10³ = 2260J
This is the latent heat of vaporization 5gram of water= 2260J/g × 5g
= 11,300J
= 11.3 KJ
Therefore, the latent heat of vaporization 5gram of steam converted to liquid at 100°C is 11.3 KJ.
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