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Question

A closed thermally insulated vessel contains $100 g$ of water at $0^{o} C$ . If the an from this vessel in rapidly pumped out, intensive evaporation will produce cooling and as a result of this water will freeze. How much ice will be formed by this method? If latent heat of fusion is $80 c a l / g$ and of evaporation $560 c a l / g$

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Solution

Let $x$ gram of ice is formed by this method and $y$ gram of water gets evaporated.

The energy liberated by formation of ice is used for water to get evaporated.

So

$80 x = 560 y$

$y = \frac{80 x}{560}$

Also $x + y = 100$

$x + \frac{80 x}{560} = 100$

$640 x = 560 \times 100$

$x = \frac{560 \times 100}{640}$

$x = 87.5 g$

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Similar questions

C_{i c e} = 0.53 c a l / g -^{\circ} C

C_{w a t e r} = 1.0 c a l / g -^{\circ} C

(L_{f})_{w a t e r} = 80 c a l / g

(L_{v})_{w a t e r} = 529 c a l / g

100 g

0^{\circ} C

g

80 c a l / g

560 c a l / g

100 g

0^{o} C

0^{o} C

0^{o} C = \times 10^{6} J k g^{- 1}

= 3.36 \times 10^{5} J k g^{- 1}

0^{o} C

0^{o} C = 21 \times 10^{5} J / k g

= 3.36 \times 10^{5} J / k g .

150 g

0^{\circ} C

0^{\circ} C

= 2.10 \times 10^{6} J k g^{- 1}

= 3.36 \times 10^{5} J k g^{- 1})

150 g

0^{\circ} C

0^{\circ} C

= 2.10 \times 10^{6} J k g^{- 1}

= 3.36 \times 10^{5} J k g^{- 1})

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