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How should 1 kg of water at 5^@C be divi...

How should 1 kg of water at `5^@C` be divided into two parts so that if one part turned into ice at `0^@C`, it would release enough heat to vaporize the other part? Latent heat of steam`=540 cal//g` and latent heat of ice `=80 cal//g`.

Text Solution

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Let the mass be divided into x grams (for ice) and (1000`-x`) grams (for vapour).
Heat released by x grams of water `=x xx1xx5+x xx80`
Heat absorbed by `(1000-x)` grams of water
`=(1000-x)xx1xx95+(1000-x)xx540`
Assuming that the conversion of the other part takes place at `100^@C`.
`85x=95(1000-x)+540(1000-x)` or `x=882g`
Thus the mass is to be divided into 882 g for conversion into ice and 118 g for conversion into vapour.
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