Question

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

• A. $50sec$
• B. $90sec$
• C. $60sec$
• D. $48sec$

Answer 1

Answer: (D)

From Newton's law of cooling
$\frac{dT}{dt}=-K\left(T-T_0\right)$
$\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_0\right)$
$\frac{80-60}{60}=K\left(\frac{80+60}{2}-30\right)$
$\frac{1}{3}=K\times \mathrm{40...}\left(1\right)$
and $\frac{60-50}{t}=K\left(\frac{60+50}{2}-30\right)$
$\frac{10}{t}=K\times \mathrm{25...}\left(2\right)$
From eqn $\left(1\right)$ and $\left(2\right)$
$\frac{t}{30}=\frac{40}{25}\Rightarrow t=48sec$