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Take Cice =...

Question

Take $C_{i c e} = 0.53 c a l / g -^{\circ} C$ , $C_{w a t e r} = 1.0 c a l / g -^{\circ} C$ , $(L_{f})_{w a t e r} = 80 c a l / g$ and $(L_{v})_{w a t e r} = 529 c a l / g$ . A closely thermally. Insulated vessel contains $100 g$ of water at $0^{\circ} C$ . If the air from this vessel is 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 (in $g$ ) ? If latent heat of fusion is $80 c a l / g$ and of evaporation $560 c a l / g$ .

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Solution

Since the heat lost in fusion will convert water to steam and raise temperature to $100^{0} C$ .
If $m$ gram ice is formed, $m L_{f} = (100 - m) L_{v}$
Hence, $m (80) = (100 - m) (560 + 100 (1))$
$m = 89.2 g$

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100 g

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560 c a l / g

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

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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

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

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

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