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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)`

A

`0^(@)C`

B

`100^(@)C`

C

`40^(@)C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`5xx540+5xx1xx60=25xx25xx1xx40`
`2700+300implies2000+1000`
Hence, the final temperature `=40^(@)C`
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