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  4. 540 g of ice of 0° C is mixed with 540 g of water at 80° C . The final temperature of the mixture is

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Q. 540 g of ice of $ 0{}^\circ C $ is mixed with 540 g of water at $ 80{}^\circ C $ . The final temperature of the mixture is

Rajasthan PMTRajasthan PMT 2010

Solution:

Heat taken by ice to melt at $ {{0}^{o}}C $ is $ {{Q}_{1}}=mL=540\times 80=43200\,\,cal $ Heat given by water to cool upto $ {{0}^{o}}C $ is $ {{Q}_{2}}=m\times s\times \Delta \theta $ $ =540\times 1\times (80-0) $ $ =43200\,\,cal $ Heat given by water = Heat absorbed by ice $ \therefore $ Three will be no change in the temperature of mixture. Hence, the temperature of mixture will be $ {{0}^{o}}C $ .