Explanation

(b) According to Newton’s law of cooling
$\frac{(\theta_{1} - \theta_{2})}{t} = K\left( \frac{\theta_{1} + \theta_{2}}{2} - \theta_{0} \right)$
$\therefore\ \ \ \frac{(62 - 50)}{10} = K\left( \frac{62 + 50}{2} - \theta_{0} \right)$
$\frac{12}{10} = K(56 - \theta_{0})$ …(i)
For further cooling
$\frac{(50 - 42)}{10} = K\left( \frac{50 + 42}{2} - \theta_{0} \right)$
$\frac{8}{10} = K(46 - \theta_{0})$ …(ii)
Dividing Eq (i) by Eq. (ii), we get,
$\frac{12}{8} = \frac{(56 - \theta_{0})}{(46 - \theta_{0})}$
$3\left( 46 - \theta_{0} \right) = 2(56 - \theta_{0})$
$138 - 3\theta_{0} = 112 - 2\theta_{0}
$ $\theta_{0} = 26℃$