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A calaorimeter of water equivalent 83.72...

A calaorimeter of water equivalent `83.72 Kg` contains `0.48 Kg` of water at `35^@ C`. How much mass of ice at `0^@C` should be added to decrease the temperature of the calorimeter to `20^@ C`.
`(S_(w) = 4186 J//Kg -K and L_(ice) = 335000 J//Kg)`.

Text Solution

Verified by Experts

Heat capacity of the calorimeter `= 83.72 J K^(-1)` From law of method of mixtures,
`{:("Heat", "lost","by","calorimeter",),(,,+,,),("Heat","lost","by","water",):}} = "Heat gained by the ice"`
`83.72xx 15+0.48xx4186 xx15 xx(335000+83720)`
`:. m = 0.07498 Kg`.
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