Question

A cup of tea cools from $80^{\circ} C$ to $60^{\circ} C$ in one min. The ambient temperature is $30^{\circ} C$. In cooling from $60^{\circ} C$ to $50^{\circ} C$, how much time (in s) will it take?

Answer 1

According to Newton's law of cooling
$\frac{\Delta \theta}{\Delta t}=k\left[q-q_{0}\right]$
$q_{0}=$ Surrounding's temperature
$\Rightarrow \frac{80-60}{60}=k\left[\frac{80+60}{2}-30\right]$ ...(i)
and $\frac{60-50}{t}=k\left[\frac{60+50}{2}-30\right]$ ...(ii)
$\Rightarrow t=48\, s$