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 x 10⁶ Jkg^–1 and Latent heat of fusion of water = 3.36 x 10⁵ J kg^–1 )

(1) 130 g  (2) 35 g   (3) 150 g  (4) 20 g

Answer 1

Correct option is (4) 20 g

Suppose ′m′ gram of water evaporates then, heat required

△Q_req​ = mL_v​

Mass that converts into ice = (150 − m)

So, heat released in this process

△Q_rel​ = (150 − m) L_f​

Now,

△Q_rel​ = △Q_req​

(150 − m)L_f​ = mL_v

m(L_f​ + L_v​) = 150L_f​

m = \(\frac{150 L_f}{L_f + L_v}\)

m = 20 g.

Answer 2

Correct option (4) 20 g

Explanation: