Hot water cools from ${60}^{\circ }C$ to ${50}^{\circ }C$ in the first 10 mins and to ${42}^{\circ }C$ in the next 10 mins. The temperature of surroundings

$Given,T_0={22}^{\circ }C,$ From Newton's law of cooling,  $\frac{T_i-T_f}{t}=(\frac{T_i+T_f}{2}-T_0)$

For case-I, Hot soup cools from 60°C to 50°C in 10 min.

$\frac{60-50}{10}=-\mathrm{K}\left[\frac{60+50}{2}-{\theta }_0\right]$

For case-II, Time taken for cooling from $50°\text{C}$ to $42°\text{C}$

$\frac{50-42}{10}=-\mathrm{K}\left[\frac{50+42}{2}-{\theta }_0\right]$

After solving we get,

$\begin{array}{l}\frac{1}{0.8}=\frac{55-{\theta }_0}{46-{\theta }_0}\\ \Rightarrow \frac{10}{8}=\frac{55-{\theta }_0}{46-{\theta }_0}\\ \Rightarrow 460-10{\theta }_0=440-8{\theta }_0\\ \Rightarrow 2{\theta }_0=20\\ \Rightarrow {\theta }_0={10}^{\circ }\mathrm{C}\end{array}$