Question

If 200 Cal heat energy is required 40 grams
of liquid to its gaseous state, then the
latent heat of vapourization of it is ........
1) 4 Cal/gm 2) 5 Cal/gm [2]
3) 8000 Cal/gm 4) 240 Cal/gm
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Answer 1

Given:

200 Cal heat energy is required=40

To Find:

Latent heat of vaporization

Solution:

Latent heat of vaporization is 200/40 calorie per gram=5 cal/g

                                                                           

∴ latent heat of vaporization is 5 cal/g

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Answer 2

Answer:

5cal per gram

Explanation:

latent latent heat of vaporization is equal to 200 calories by 40 grams then the answer is 200 divided by 40 is equal to 5