Question

A body at ${80}^{o}C$ cools down to ${64}^{o}C$ in 5 minutes; and in 10 minutes it cools down to ${52}^{o}C$. What will be its temperature after $20$ minutes? What is the temperature of the environment?

Answer 1

Using approximate version of Newton's law of cooling

$\frac{{\theta }_1-{\theta }_2}{t}=\alpha (\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0)$

Now using condition 1:

$\frac{80-64}{5}=\alpha [\frac{80+64}{2}-{\theta }_0]$ (1)

Now using condition 2:

$\frac{64-52}{10}=\alpha [\frac{64+54}{2}-{\theta }_0]$ (2)

Divide 1 and 2

$\frac{16}{5}\times \frac{10}{12}=\frac{72-{\theta }_0}{58-{\theta }_0}$

$\frac{8}{3}=\frac{72-{\theta }_0}{58-{\theta }_0}$

$464-{8\theta }_{\theta }=216-3{\theta }_0$

$248={5\theta }_{\theta }$

${\theta }_{\theta }=\frac{248}{5}=49.6°C$

Temperature of environment =$49.6°C$