Question

The temperature of a body falls from ${40}^{o}C$ to ${36}^{o}C$ in $5$ minutes when placed in a surrounding of constant temperature ${16}^{o}C$. Then, what is the cooling constant (per minute)?

• A. $-(\frac{2}{55})$
• B. $-(\frac{1}{15})$
• C. $-\frac{lo{g}_e(\frac{5}{6})}{5}$
• D. $-\frac{lo{g}_e(\frac{6}{5})}{5}$

Answer 1

The correct option is C $-\frac{lo{g}_e(\frac{5}{6})}{5}$

Newton's law of cooling gives us the relation:
$\frac{d\theta }{dt}=-bA(\theta -{\theta }_0)$
Integrating this equation between the appropriate limits, we get:
$-\frac{1}{t}ln\left(\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}\right)=bA=$ Newton's constant
Substituting we get:
Newtons constant $=-\frac{1}{5}ln\left(\frac{36-16}{40-16}\right)=-\frac{1}{5}ln\left(\frac{5}{6}\right)$