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50g of ice at 0^(@)C is mixed with 200g ...

`50g` of ice at `0^(@)C` is mixed with `200g` of water at `0^(@)C.6` kcal heat is given to system [Ice +water]. Find the temperature (in `.^(@)C`) of the system.

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The correct Answer is:
8

`6000 = 50 xx 80 +250 xx 1 xx Delta theta`
`6000 - 400 = 250 xx Delta theta rArr Delta theta = (2000)/(250) = 8^(@)C`
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