How much heat is required to change 100 g of water at 100°C into steam? (Latent heat of vaporization of water is 2.26 ｘ10⁶  Jkg⁻¹. Ans. (2.26 ｘ 10⁵ J)


Given:

Mass of ice  = m = 100 g = 0.1 kg
Temperature of water = T₁ = 100°C = (100 + 273) K = 373 K
Temperature of steam  = T₂ = 100°C = (100 + 273) K = 373 K
Latent heat of Vaporization of water = Hᵥ = 2.26 ｘ10⁶ J kg⁻¹

To Find:

Heat  = Q  = ?

Solution: 


We have the formula for Latent heat of Vaporization is

Q = m Hv

By putting values 

Q = 0.1 Kg ｘ 2.26 ｘ10⁶ J kg⁻¹

Q = 0.226 ｘ10⁶ J kg⁻¹

or

Q = 2.26 ｘ10⁵ J

Thus the heat required to change the water into steam according to the given conditions is 2.26 ｘ10⁵ J

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