Question

How much energy is required to convert 65 grams of water of 100℃ into steam ?

Answer 1

As the water is at a temperature of 100°C, the latent heat of vaporization comes into the picture.

Latent heat of vaporization(L) of water is 540cal/g/°C

Therefore the heat required to convert the whole 65 grams of water at 100°C into steam is 
                                             Q = mL
                                             Q = 65 x 540
                                                 = 35100 calories
                                                 = 35.1 kilocalories = 146.85 kilo joules