b
(b) According to Newton's law of cooling
\(\frac{{{\theta _1} - {\theta _2}}}{t} = K\left[ {\frac{{{\theta _1} + {\theta _2}}}{2} - {\theta _0}} \right]\)
In the first case, \(\frac{{(60 - 50)}}{{10}} = K\,\left[ {\frac{{60 + 50}}{2} - {\theta _0}} \right]\)
\(1 = K\,(55 - \theta )\)….\((i)\)
In the second case, \(\frac{{(50 - 42)}}{{10}} = K\,\left[ {\frac{{50 + 42}}{2} - {\theta _0}} \right]\)
\(0.8 = k\,(46 - {\theta _0})\)….\((ii)\)
Dividing \((i)\) by \(  (ii)\) , we get \(\frac{1}{{0.8}} = \frac{{55 - \theta }}{{46 - \theta }}\)
or \(46 - {\theta _0} = 44 - 0.8\theta \) ==> \({\theta _0} = {10^o}C\)