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450 g water at 40^(@)C was kept in a cal...

`450` g water at `40^(@)C` was kept in a calorimeter of water equivalent `50 g 25 g ice at 0^(@)C` is added and simultaneously `5` g steam at `100^(@)C` was passed to the calorimeter. The final temperature of the calorimeter. Will be
`(L_(ice)=80 cal // g.L_(steam) =540cal//g)`

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