A) 3.75 minutes
B) 8.6 minutes
C) 9.6 minutes
D) 6.5 minutes
Correct Answer: B
Solution :
From Newton's law of cooling, \[t=\frac{1}{k}{{\log }_{e}}\left( \frac{{{\theta }_{2}}-{{\theta }_{0}}}{{{\theta }_{1}}-{{\theta }_{0}}} \right)\] From question and above equation, \[5=\frac{1}{k}{{\log }_{e}}\left( \frac{40-30}{80-30} \right)\]..........(1) and\[t=\frac{1}{k}{{\log }_{e}}\frac{(32-30)}{(62-30)}\]...........(2) Dividing equation (2) by (1), \[\frac{1}{5}=\frac{\frac{1}{k}lo{{g}_{e}}\frac{(32-30)}{(62-30)}}{\frac{1}{k}lo{{g}_{e}}\frac{(40-30)}{(80-30)}}\] On solving we get, time taken to cool down from \[62{}^\circ C\] to \[32{}^\circ C\], t = 8.6 minutes.
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