100g ice at 0^(@)C is mixed with 100g wa...

`100g` ice at `0^(@)C` is mixed with `100g` water at `100^(@)C`. The resultant temperature of the mixture is

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The correct Answer is:

283

Let the final temperature of mixture be `theta`
Then `100x80+100(theta-0)=100xx1xx(100-theta)`
Solving, we get `theta=10^(@)C=283K`

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