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Ice at -20^(@)C is added to 50 g of wate...

`Ice at -20^(@)C` is added to 50 g of water at `40 ^(2)C` When the temperature of the mixture reaches `0^(@) C` it is found that 20 g of ice is still unmelted .The amount of ice added to (Specific heat of water `=4.2 j//g//^(@)C`
Specific heat of Ice `=2.1 J//g //^(@)C` M
Heat of fusion of water of `0^(@)C=334 J//g)`

Text Solution

Verified by Experts

Let m gram of ice of added.
From principal of calorimeter
heat gained (by ice) =heat lost (by water)
`:.20xx2.1xxm+(m-20)xx334`
`=50xx4.2xx40`
`376m=8400+6680`
`m=40.1`
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