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In a container of negligible mass 140 g ...

In a container of negligible mass 140 g of ice initially at `-15^@C` is added to 200 g of water that has a temperature of `40^@C`. If no heat is lost to the surroundings, what is the final temperature of the system and masses of water and ice in mixture ?

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The correct Answer is:
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Let the mixture is water at `theta^@C("where "0^@Cltthetalt40^@C)`.
Heat given by water = Heat taken by ice
`:. (200)(1)(40-theta) = (140)(0.53)(15)`
`+(140)(80) + (140)(1)(theta - 0)`
Solving we get,
`theta = -12.7^@C`.
Since, `thetalt0^@C` and we have assumed the mixture
to be water whose temperature can't be less than
`0^@C`.
Hence, mixture temperature `theta = 0^@C`.
Heat given by water in reaching upto `0^@C` is,
`theta = (200)(1)(40 - 0) = 8000 cal`.
Let m mass of ice melts by this heat , then
`8000 = (140)(0.53)(15) + (m)(80)`
Solving we get m = 86g
`:.` Mass of water` = 200 + 86 = 286 g `
Mass of ice `= 140 -86 = 54g` .
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