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\[={{m}_{1}}L+m{{c}_{1}}\theta \] \[=5\times 80+5\times 1(\theta -0)\] \[=400+5\theta \] Heat given by water at\[40{}^\circ C\] \[={{m}_{2}}{{l}_{2}}{{\theta }_{2}}=20\times 1\times (40-\theta )\] Heat given = Heat taken \[800-20\theta =400+5\theta \] \[25\theta =400\] \[\theta =\frac{400}{25}={{16}^{o}}C\]
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