Certain quantity of water cools from 70°C  to 60°C  in the first 5 minutes and to 54°C  in the next 5 minutes. The temperature of the surroundings is

$\text{ Let }{T}_s\text{ be the temperature of the surroundings. }$

According to Newton's law of cooling

$\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_s\right)$

for first 5 minutes,

$\begin{array}{c}{T}_1={70}^{\circ }\mathrm{C},T_2={60}^{\circ }\mathrm{C},t=5\text{ minutes }\\ \therefore \frac{70-60}{5}=K\left(\frac{70+60}{2}-T_s\right)\\ \frac{10}{5}=K\left(65-T_s\right)\end{array}$

$\begin{array}{c}{T}_1={60}^{\circ }\mathrm{C},T_2={54}^{\circ }\mathrm{C},t=5\text{ minutes }\\ \therefore \frac{60-54}{5}=K\left(\frac{60+54}{2}-T_s\right)\\ \frac{6}{5}=K\left(57-T_s\right)\end{array}$

Divide eqn. (i) by eqn. (ii), we get

$\frac{5}{3}=\frac{65-T_s}{57-T_s}$

$\begin{array}{l}285-5T_s=195-3T_s\\ 2T_s=90\text{ or }{T}_s={45}^{\circ }\mathrm{C}\end{array}$