Question

A hot body, obeying Newton's law of cooling is cooling down from its peak value ${80}^{o}C$ to ${40}^{o}C$. How much time will it take to cool down from ${62}^{o}C$ to ${32}^{o}C$? (given $\ln 2=0.693,\ln 5=1.609$)

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

Answer 1

Answer: (A)

$8.6$ minutes

From Newton's law of cooling,

$t=\frac{1}{k}\log e\left(\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}\right)$, from this equation

$5=\frac{1}{k}\log e\left(\frac{40-30}{80-30}\right)$ ___(I)

$t=\frac{1}{k}\log e\frac{(32-30)}{(62-30)}$ ____(II)

Dividing equ. (II) by (I)

$\frac{t}{5}=\frac{\frac{1}{k}\log e\left(\frac{2}{32}\right)}{\frac{1}{k}\log e\left(\frac{10}{50}\right)}$

$\frac{t}{5}=\frac{\log e\left(\frac{1}{16}\right)}{\log e\left(\frac{1}{5}\right)}\Rightarrow t=\frac{1.20\times 5}{0.69}=8.6$ minutes