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 -1A- 130 gB- 35 gC- 20 gD- 150 g

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Standard XII

Physics

Specific Heat Capacity

A thermally i...

Question

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 $= 2.10 \times 10^{6} J k g^{- 1}$ and Latent heat of fusion of water $= 3.36 \times 10^{5} J k g^{- 1})$

130 g

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35 g

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20 g

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150 g

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Solution

The correct option is C $20 g$
Suppose $m$ grams of water evaporate. Then, heat required
$△ Q_{r e q} = m L_{v}$
Mass that converts into ice $= (150 - m)$
Heat released in this process
$△ Q_{r e l} = (150 - m) L_{f}$
Now,
$△ Q_{r e l} = △ Q_{r e q}$
$(150 - m) L_{f} = m L_{V}$
$m (L_{f} + L_{v}) = 150 L_{f}$
$m = \frac{150 L_{f}}{L_{f} + L_{v}}$
$m = 20 g$

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Q. A thermally insulated vessel contains

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The correct option is C 20 gSuppose m grams of water evaporate. Then, heat required △Qreq=mLv Mass that converts into ice =(150−m) Heat released in this process △Qrel=(150−m)Lf Now, △Qrel=△Qreq (150−m)Lf=mLV m(Lf+Lv)=150Lf m=150LfLf+Lv m=20 g