In an experiment to verify Newton's law of cooling. a graph is...

According to the Newton's law of cooling, $- \dfrac{dT}{dt} = K\left(T - {T}_{0}\right)$.

Applying approximation, $- \dfrac{∆ T}{∆ t} = K\left(\dfrac{{T}_{1} + {T}_{2}}{2} - {T}_{0}\right)$

For first $6min$, ${T}_{1} = 80 ° C, {T}_{2} = \left(40 + 20\right) = 60 ° C, {T}_{0} = 20 ° C$

$⇒ - \dfrac{\left(60 - 80\right)}{6} = K\left(70 - 20\right) ⇒ K = - \dfrac{\left(60 - 80\right)}{6 \times \left(70 - 20\right)}$

$⇒ K = \dfrac{20}{6 \times 50} = \dfrac{1}{15}$

For $6\text{to}{t}_{2}$, ${T}_{1} = 60 ° C, {T}_{2} = \left(20 + 20\right) = 40 ° C, {T}_{0} = 20 ° C$

$- \dfrac{\left(60 - 80\right)}{\left({t}_{2} - 6\right)} = K\left(50 - 20\right) ⇒ {t}_{2} - 6 = 15 \times \dfrac{20}{30} ⇒ {t}_{2} = 16min$