. A cup of tea cools from 80 °C to 60 °C is one minute. The ambient temperature is 30 °C. In cooling from 60 °C to 50 °C. It will take :

From Newton's law of cooling

$\frac{\mathrm{dT}}{\mathrm{dt}}=-\mathrm{K}\left(\mathrm{T}-{\mathrm{T}}_0\right)$

$\frac{{\mathrm{T}}_1-{\mathrm{T}}_2}{\mathrm{t}}=\mathrm{K}\left(\frac{{\mathrm{T}}_1+{\mathrm{T}}_2}{2}-{\mathrm{T}}_0\right)$

$\frac{80-60}{60}=K\left(\frac{80+60}{2}-30\right)$

$\frac{1}{3}=K\times 40$    ….(1)

$\text{ and }\frac{60-50}{t}=K\left(\frac{60+50}{2}-30\right)$

$\frac{10}{t}=K\times 25\dots \left(2\right)$

From eqn (1) and (2)

$\frac{t}{30}=\frac{40}{25}\Rightarrow t=48\sec$