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

100g of ice at `0^(@)` is mixed with 100g of water at `100^(@)C`. What will be the final temperature of the mixture?

Text Solution

Verified by Experts

Heat required by ice to raise its temperature to `100^(@)C`,
`Q_(1) = m_(1)L_(1) + m_(1)c_(1)Deltatheta_(1)= 5xx80+ 5 xx 1xx100= 400+500 = 900"cal"`
Heat given by the steam when condensed `Q_(2)= m_(2)L_(2)=5xx536=2680"cal"`
As `Q_(2)gtQ_(1)`. This means that whole steam is not even condensed.
Hence temperature of mixture will remain at `100^(@)C`.
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