1. Tardigrade
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  3. Physics
  4. 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)

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Q. A thermally insulated vessel contains $150\,g$ of water at $0^{\circ}C$. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at $0^{\circ}C$ itself. The mass of evaporated water will be closest to :
(Latent heat of vaporization of water $= 2.10 \times 10^6 \; J \; kg^{-1}$ and Latent heat of Fusion of water = $3.36 \times 10^5 \; J \; kg^{-1}$)

JEE MainJEE Main 2019Thermodynamics

Solution:

Suppose'm' gram o f water evaporates then, heat required
$\Delta Q_{req} = mL_V$
Mass that converts into ice = (150 - m)
So, heat released in this process
$\Delta Q_{rel} = (150 - m) L_f$
Now,
$\Delta Q_{rel} = \Delta Q_{req}$
$(150 - m ) L_f =mL_V$
$m(L_f + L_v) = 150 L_f$
$ m = \frac{150 L_f}{L_f + L_v} $
$m = 20 \,g$