Hot water cools from $60{}^{\circ }\mathrm{C}\text{ to }50{}^{\circ }\mathrm{C}$ in the first 10 minutes and to $42°\mathrm{C}$ in the next 10 minutes. The temperature (in ${}^{\circ }\mathrm{C}$ ) of the surroundings is:

$\begin{array}{l}\text{ By Newton's law of cooling }\\ \frac{{\mathrm{\theta }}_1-{\mathrm{\theta }}_2}{\text{t}}=-\text{K}[\frac{{\mathrm{\theta }}_1+{\mathrm{\theta }}_2}{2}-{\mathrm{\theta }}_0]{\text{ where θ}}_0\text{ is the temperature of surrounding. }\\ \text{Now, hot water cools from }60{}_{ }{}^{\circ }\text{C to }50°\text{C in }10\text{ minutes, }\\ \frac{60-50}{10}=-\text{K}[\frac{60+50}{2}-{\text{θ}}_0]\mathrm{......}\left(1\right)\text{ }\\ \text{Again, it cools from }50°\text{C to }42°\text{C in next }10\text{ minutes.,}\\ \text{}\frac{50-42}{10}=-\text{K}[\frac{50+42}{2}-{\text{θ}}_0]\mathrm{......}\left(2\right)\text{ }\\ \text{Dividing equations (i) by (ii) we get }\\ \frac{1}{0.8}=\frac{55-{\text{θ}}_0}{46-{\text{θ}}_0}\\ \Rightarrow \frac{10}{8}=\frac{55-{\text{θ}}_0}{46-{\text{θ}}_0}\text{}\\ \Rightarrow 460-10{\text{θ}}_0=440-8{\text{θ}}_0\text{}\\ \Rightarrow 2{\text{θ}}_0=20\text{}\\ \Rightarrow {\text{θ}}_0=10°\text{C}\end{array}$