Question

A cup of tea cools from ${80}^{\circ }\text{C}$ to ${60}^{\circ }\text{C}$ in $1\text{min}$. The ambient temperature is ${30}^{\circ }\text{C}$. How much time will it take to cool from ${60}^{\circ }\text{C}$ to ${50}^{\circ }\text{C}$?

• A. $30\text{s}$
• B. $60\text{s}$
• C. $90\text{s}$
• D. $48\text{s}$

Answer 1

The correct option is D $48\text{s}$

According to Newton's law of cooling
$\frac{{\theta }_1-{\theta }_2}{t}\propto [\frac{{\theta }_1+{\theta }_2}{2}-\theta ]$
For the first condition
$\frac{80-60}{60}\propto [\frac{80+60}{2}-30]$ ... (i)
and for the second condition
$\frac{60-50}{t}\propto [\frac{60+50}{2}-30]$ ... (ii)
Dividing (i) by (ii) :
$\frac{20}{60}\times \frac{t}{10}=\frac{40}{25}$
$\Rightarrow t=48\text{s}$