How much heat is required to vaporize 25g of water at 100*c

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals,

Q = mL,

Where, m is the mass of water present

L = specific latent heat of vaporization

Here, m = 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25*533 = 13,325 Calories

Or, Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.