Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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.
Learn more about latent heat here:
https://brainly.com/question/12775811
#SPJ4
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.