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Q.
Hot water cools from $ 60{}^\circ C $ to $ 50{}^\circ C $ in the first $10 \,min$ and to $ 42{}^\circ C $ in the next $10\, min$. The temperature of the surroundings is
According to Newtons law of cooling
$ \frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right] $
In the first case $ \Rightarrow $ $ \frac{60-50}{10}=K\left[ \frac{60+50}{2}-{{\theta }_{0}} \right] $
$ \Rightarrow $ $ 1=K(55-\theta ) $ ...(i)
In the second case $ \Rightarrow $ $ \frac{50-42}{10}=K\left[ \frac{50+42}{2}-\theta \right] $
$ \Rightarrow $ $ 0.8=K[46-\theta ] $
...(ii) Dividing Eq. (i) by Eq. (ii), we get
$ \frac{1}{0.8}=\frac{55-\theta }{46-\theta } $ or $ 46-\theta =44-0.8\theta $
$ \theta =10{}^\circ C $