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Question

A thermally insulated vessel contains $$150g$$ 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 $$= 3.36\times 10^{5} J\ kg^{-1})$$

A
$$20\ g$$
B
$$130\ g$$
C
$$35\ g$$
D
$$150\ g$$
Solution
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Correct option is C. $$20\ g$$
Suppose $$'m'$$ gram of water evaporates then, heat required
$$\triangle Q_{req} = mL_{v}$$
Mass that converts into ice $$= (150 - m)$$
So, heat released in this process
$$\triangle Q_{rel} = (150 - m) L_{f}$$
Now,
$$\triangle Q_{rel} = \triangle Q_{req}$$
$$(150 - m)L_{f} = mL_{V}$$
$$m(L_{f} + L_{v}) = 150 L_{f}$$
$$m = \dfrac {150L_{f}}{L_{f} + L_{v}}$$
$$m = 20\ g$$.

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