Question

A body cools from ${62}^{\circ }C$ to ${50}^{\circ }C$ in $10$ minutes and to ${42}^{\circ }C$ in the next $10$ minutes. Find the temperature of surroundings.

Answer 1

For the first ten minutes,
$\frac{dT}{dt}=-\left(\frac{{62}^{\circ }-{50}^{\circ }}{10}\right)=-{1.2}^{\circ }C/min$ and
$△T=\left(\frac{{62}^{\circ }+{50}^{\circ }}{2}\right)-T_0=(56-T_0{)}^{\circ }C$
$\Rightarrow -{1.2}^{\circ }C/min=-KA(56-T_0{)}^{\circ }C.....(1)$
Similarly for the next ten minutes
$\frac{dT}{dt}=\left[\frac{{42}^{\circ }-{50}^{\circ }}{2}\right]=-{0.8}^{\circ }C/min$ and
$△T=\left(\frac{{42}^{\circ }-{50}^{\circ }}{2}\right)-T_0=(46-T_0{)}^{\circ }C$
$\Rightarrow -{0.8}^{\circ }C/min=-KA(46-T_0{)}^{\circ }C.....(2)$
Dividing (1) by (2)
$\frac{3}{2}=\frac{{56}^{\circ }-T_0}{{46}^{\circ }-T_0}$
$\Rightarrow T_0={26}^{\circ }C$.