The latent heat of vapourisation for 1g water is 536 Cal. It`s value in Joule/kg will be ?
Preeti , 8 Years ago
Grade 11
Thermal Physics> The latent heat of vapourisation for 1g w...
2 Answers
Adarsh
4.18 Joule = 1 calorie .
so, latent heat for 1g water = 536calorie = 4.18 x 536 = 2240.48 Joule.
2240.48 joule of heat required to convert 1g of water to vapour .
so, 2240.48 x 103 joule heat is required to convert 1 kg water into steam .
hence, latent heat of vapourisation = 2240.48 x 103J/kg = 2240.48KJ/kg
hope this will help you ..
Khimraj
Since 1cal = 4.184 J
so latent heat for 1g water in J = 536*4.184 = 2242.62 J
latent heat for 1kg water in J = 2242.62*103 J/Kg.
This is your require answer.
Hope it clears.
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