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