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5 votes
How much heat is absorbed when 1 kg of a liquid gets vaporized at its boiling point

asked
User Ankurrc
by
8.5k points

1 Answer

3 votes
22.6 x 10^5 J

‘Heat of vaporization’ is the amount of ‘heat’ required to convert '1g' of mass of a liquid to vapor. Normal “boiling point of water” is 100 C. The heat of vaporization at this temperature is 2260 J/g. It means 2260 J/g of heat is required to convert ‘1g of water’ to 1g of vapor at 100 C. So to convert 1 kg of mass of water or 1000g of water we require 22.6 x 10^5 J of heat.
answered
User Borexino
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8.2k points
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