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A thermally insulted vessel contains $150\, g$ of water at $0\,^oC$. Then the air from the vessel is pumped out a adiabatically. A fraction of water turns into ice and the rest evaporates at $0\,^oC$ itself. The mass of evaporated water will be closes to ....... $g$ (Latent heat of vaporization of water $= 2.10 \times10^6\, Jkg^{-1}$ and Laten heat of Fusion of water $ = 3.36 \times10^5\,Jkg^{-1}$ )
$35$
$150$
$130$
$20$
Solution
Suppose $'m'$ gram of water evaporates then, heat required $\Delta {Q_{req}} = m{L_v}$
Mass that converts into ice $=(150-m)$
So, heat released in this process
$\Delta {Q_{rel}} = \left( {150 – m} \right){L_f}$
Now,
$\Delta {Q_{rel}} = \Delta {Q_{req}}$
$\left( {150 – m} \right){L_f} = m{L_v}.$
$M\left( {{L_f} – {L_v}} \right) = 150\,{L_f}$
$m = \frac{{150{L_f}}}{{{L_f} + {L_v}}}\,\,;\,\,m = 20\,g$
