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

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Q. A cup of tea cools from $80^{\circ} C$ to $60^{\circ} C$ in one minute. The ambient temperature is $30^{\circ} C$. In cooling from $60^{\circ} C$ to $50^{\circ} C$. It will take

Thermodynamics

Solution:

Newton's law of cooling
$\frac{\Delta \theta}{\Delta t}=k\left[\theta-\theta_0\right]$
$\theta_0=$ surrounding's temperature
$\Rightarrow \frac{80-60}{t}=k\left[\frac{80+60}{2}-30\right]$ ...(i)
and $\frac{60-50}{t}=k\left[\frac{60+50}{2}-30\right]$ ....(ii)
$\Rightarrow t=48\, \sec .$