A) \[32{}^\circ C\]
B) \[16{}^\circ C\]
C) \[8{}^\circ C\]
D) \[24{}^\circ C\]
Correct Answer: B
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 \[\theta =\frac{400}{25}\] \[=16{}^\circ C\]
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