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5g of water at 30^@C and 5 g of ice at -...

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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The correct Answer is:
A
In this case heat is given by water and taken by ice
Heat available with water to cool from `30^@C` to `0^@C`
`=msDeltatheta = 5xx1xx30`
`=150cal`
Heat required by `5g` ice to increase its temperature upto `0^@C`
`msDeltatheta = 5xx 0.5 xx 20`
`= 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 for melting the ice.
If mass of ice melted is m g, then
`mxx 80 = 100`
`rArr m = 1.25g `
Thus, `1.25 g` ice out of 5 g melts and mixture of ice and water is at `0^@C.`
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