Question

In an experiment ot verify Newton's law of cooling, a graph is plotted between, the temperature difference $(\Delta T)$ of the water and surroundings and time as shown in figure. The initial temperature of water is taken as $80^{\circ }C$. The value of $t_2$ as mentioned in the graph will be ______

Answer 1

Answer: 16

$T-T_0\left(T_i-T_0\right)e^{-\frac{Bt}{ms}}$
$6\lambda =\ln 1.5$
$40=60e^{-\lambda (6)}\Rightarrow 6\lambda =\ln 1.5$
$20=60e^{-\lambda t_2}\Rightarrow t_2\lambda =\ln 3$
$\frac{t_2}{6}=\frac{\ln 3}{\ln 1.5}$
$\therefore t_2=16.25\min$
$So\approx 16$