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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)`.

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