300 grams of water at `25^@ C` is added to 100 grams of ice at `0^@ C.` The final temperature of the mixture is _____ `^@C`

Answer

`0^(@)C`

Explanation

Heat ilberated when `300g` water at `25^(@)C` goes to water at `0^(@)C`:
`Q = ms Delta theta`
`= (300) (1) (25) = 7500 cal`
From `Q = mL`, this much heat can melt mass of ice given by
`m = (Q)/(L) = (7500)/(80) = 93.75 g`
i.e. whole ice will not melt.
Hence, the mixture will be at `0^(@)C` Mass of water in mixture
`= 300 + 93.75 = 393.75 g` and Mass of ice is mixture `=100 - 93.75 = 6.25 g`