5g of water at `30^@C` and 5 g of ice at `-29^@C` are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = `0.5 cal//g.^@C` and latent heat of ice `=80 cal//g`.
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In this case heat is given by water and taken by ice Heat available with water to cool `30^(@)C` to `0^(@)C` `=ms theta=5xx1xx30=150` cal Heat required by 5 gm ice to increase its temperature up to `0^(@)C` is `ms theta=5xx0.5xx20=50 `cal Out of 150 cal heat available, 50 cal is used for increasing temperature of ice from `-20^(@)C` to `0^(@)C`. The remaining heat 100 cal is used fro melting the ice. If mass of ice melted is m-gm then `mxx80=100impliesm=1.25gm,.` Tus 1.25 gm ice out of 5 gm melts and mixture of ice and water is at `0^(@)C`.
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