In an experiment to verify Newton’s law of cooling, a graph is plotted between, the temperature difference ( $\mathrm{\Delta }$ T) of the water and surroundings and time as shown in figure. The initial temperature of water is taken as 80°C. The value of t₂ as mentioned in the graph will be

$\mathrm{\Delta T}={\mathrm{T}}_{\text{water }}-{\mathrm{T}}_{\text{surroundings }}$

Initially, at

$\mathrm{t}=0,\mathrm{\Delta T}={60}^{\circ }\mathrm{C},{\mathrm{T}}_{\mathrm{W}}={80}^{\circ }\mathrm{C}\text{ So, }{\mathrm{T}}_{\mathrm{s}}=80-60={20}^{\circ }\mathrm{C}$
By Newton’s law of cooling, $\frac{\mathrm{\Delta T}}{\mathrm{t}}=\mathrm{K}\left[\frac{{\mathrm{T}}_{\mathrm{i}}+{\mathrm{T}}_{\mathrm{f}}}{2}-{\mathrm{T}}_{\mathrm{s}}\right]$

Case I :

Body temperature changes from 80 to 60 in 6 min.

$\frac{20}{6}=\mathrm{K}\left[\left(\frac{80+60}{2}\right)-20\right]$

Case II :

Body temperature changes from 60 to 40 in t₂-6 min.

$\frac{20}{{\mathrm{t}}_2-6}=\mathrm{K}\left[\left(\frac{60+40}{2}\right)-20\right]$...(ii) 

From eqn. (i) and (ii)

 $\frac{{\mathrm{t}}_2-6}{6}=\frac{5}{3}\Rightarrow {\mathrm{t}}_2=16\min$