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)

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) 130 g (B) 35 g (C) 20 g (D) 150 g

Tardigrade Admin · Accepted Answer

Suppose'm' gram o f water evaporates then, heat required

Δ Q_{re q} = m L_{V}

Mass that converts into ice = (150 - m)
So, heat released in this process

Δ Q_{re l} = (150 - m) L_{f}

Now,

Δ Q_{re l} = Δ Q_{re q}

(150 - m) L_{f} = m L_{V}

m (L_{f} + L_{v}) = 150 L_{f}

m = \frac{150 L _{f}}{L _{f} + L _{v}}

m = 20 g

130 g

35 g

20 g

150 g

Suppose'm' gram o f water evaporates then, heat required ΔQreq​=mLV​ Mass that converts into ice = (150 - m) So, heat released in this process ΔQrel​=(150−m)Lf​ Now, ΔQrel​=ΔQreq​ (150−m)Lf​=mLV​ m(Lf​+Lv​)=150Lf​ m=Lf​+Lv​150Lf​​ m=20g