Question

A hot body, obeying Newton's law of cooling is cooled down from its peak value ${80}^{\circ }C$ to an ambient temperature of ${30}^{\circ }C$. It takes 5 minutes in cooling down from ${80}^{\circ }Cto{40}^{\circ }C$. How much time will it take to cool down from ${62}^{\circ }Cto{32}^{\circ }C$?
(Given $ln2=0.693,ln5=1.609$)

• A. 9.6 minutes
• B. 3.75 minutes
• C. 8.6 minutes
• D. 6.5 minutes

Answer 1

The correct option is C 8.6 minutes

Answer is option $\left(C\right)$
We know from Newton's law of cooling that .
$T=T_f+(T_i-T_f)e^{-kt}$
$40=30+(80-30)e^{-5k}$
-$5k=log\left(1\right)-log\left(5\right)$
or, $k=1.609/5=\mathrm{0.3218.}$
Now,
$32=30+(62-30)e^{-0.3218t}$
$4log2=0.3218t$
$t=4\times 0.693/0.3218=8.61min$