Thank you for reporting, we will resolve it shortly
Q.
5 g of ice at $ 0{}^\circ C $ is dropped in a beaker containing 20 g of water at $ 40{}^\circ C $ . The final temperature will be:
JamiaJamia 2004
Solution:
Let final temperature be $ \theta $ Now heat taken by ice $ ={{m}_{1}}L+{{m}_{1}}{{c}_{1}}{{\theta }_{1}} $ $ =5\times 80+5\times 1(\theta -0) $ $ =400+5\theta $ ...(1) Heat given by water at 40°C $ ={{m}_{2}}{{l}_{2}}{{\theta }_{2}}=20\times 1\times ({{40}^{o}}-\theta ) $ .. .(2) Heat given = Heat taken $ 800-20\theta =400+5\theta $ Or $ 25\theta =400, $ Or $ \theta =\frac{400}{25} $ $ =16{}^\circ C $