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  4. 5 g of ice of 0° C is dropped in a beaker containing 20 g of water at 40° C . The final temperature will be

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Q. 5 g of ice of $ 0{}^\circ C $ is dropped in a beaker containing 20 g of water at $ 40{}^\circ C $ . The final temperature will be

MGIMS WardhaMGIMS Wardha 2011

Solution:

Let final temperature be $ \theta $ . Now heat taken by ice $ ={{m}_{1}}L+m{{c}_{1}}\theta $ $ =5\times 80+5\times 1(\theta -0) $ $ =400+5\theta $ Heat given by water at $ 40{}^\circ C $ $ ={{m}_{2}}{{l}_{2}}{{\theta }_{2}}=20\times 1\times (40-\theta ) $ Heat given = Heat taken $ 800-20\theta =400+5\theta $ $ 25\theta =400 $ $ \theta =\frac{400}{25}={{16}^{o}}C $