Question

What is the value of latent heat of vapourisation of boiling water of 2 grams ? ​

Answer 1

Answer:

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Answer 2

Thus the latent heat of vaporization for 2 grams is 1130 x 10^4 KJ / g

Explanation:

• The amount of heat energy that is absorbed or released by a substance during the change in its physical state is called the latent heat of vaporization.
• Water absorbs 2260 KJ/kg of heat energy to vaporize. It means this is for 1 kg.  
• 1 Kg = 1000 grams
• 2260 x 1000 = 2260000 KJ/ g
• Now For 2 grams  
• 2260000 / 2 = 1130000 KJ / g
• Thus the latent heat of vaporization for 2 grams is 1130 x 10^4 KJ / g