A thermally insulted vessel contains 150 g of water at \(0^{\circ} \mathrm{C}\). Then the air from the vessel is pumped out a adiabatically. A fraction of water turns into ice and the rest evaporates at \(0^{\circ} \mathrm{C}\) itself. The mass of evaporated water will be closes to: (Latent heat of vaporization of water \(=2.10 \times 10^{6} \mathrm{Jkg}^{-1}\) and Laten heat of Fusion of water \(=3.36 \times 10^{5} \mathrm{Jkg}^{-1}\) )

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Heat & Thermodynamics

A thermally insulted vessel contains 150 g of water at

0^{\circ} C

. Then the air from the vessel is pumped out a adiabatically. A fraction of water turns into ice and the rest evaporates at

0^{\circ} C

itself. The mass of evaporated water will be closes to: (Latent heat of vaporization of water

= 2.10 \times 1 0^{6} Jkg^{- 1}

and Laten heat of Fusion of water

= 3.36 \times 1 0^{5} Jkg^{- 1}

)

Answer: (D)

Pumping out the air lets some water evaporate at

0^{\circ}

C, and the latent heat for that evaporation can come only from the rest of the water, which therefore freezes. If a mass

m

evaporates it needs

m L_{v}

, while freezing the remaining

(150 - m)

g releases

(150 - m) L_{f}

. Balancing,

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

, so

m = \frac{150 L _{f}}{L _{f} + L _{v}} = \frac{150 \times 3.36 \times 1 0 ^{5}}{3.36 \times 1 0 ^{5} + 2.10 \times 1 0 ^{6}} \approx 20

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option