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
$\frac{d T}{d t}=-K\left(T - T_{0}\right)$
$\frac{T_{1} - T_{2}}{t}=K\left(\frac{T_{1} + T_{2}}{2} - T_{0}\right)$
$\frac{80 - 60}{60}=K\left(\frac{80 + 60}{2} - 30\right)$
$\frac{1}{3}=K\times 40...\left(1\right)$
$$ and $\frac{60 - 50}{t}=K\left(\frac{60 + 50}{2} - 30\right)$
$\frac{10}{t}=K\times 25...\left(2\right)$
From eqn $\left(1\right)$ and $\left(2\right)$
$\frac{t}{30}=\frac{40}{25}\Rightarrow t=48sec$