Question

A thermally insulated vessel contains 150 g of water at 0 °C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0 °C itself. 

The mass of evaporated water will be closest to :

(Latent heat of vaporization of water = 2.10 × 106 J kg^-1 and Latent heat of Fusion of water = 3.36 × 105 J kg^-1)

(A) 150g 

(B) 20g 

(C) 130g 

(D) 35g

Answer 1

(B) 20g

m = 150 g = 0.15 kg.

The heat required to evaporate ‘m’ grams of water,

∆Q_required = mL_v ………. (1)

(015 – m) is the amount of mass that converts into ice

∴ ∆Q_released = (0.15 – m) L_f …………. (2)

Now, amount of heat required = amount of heat released

∴ From (1) and (2),

mL_v = (0.15 – m)L_f