Question

The temperature of a liquid drops from $365\text{K}$ to $361\text{K}$ in $2\text{min}$. Find the time during which temperature of the liquid drops from $344\text{K}$ to $342\text{K}$.
[Take the room temperature as $293\text{K}$]

• A. $84\text{sec}$
• B. $72\text{sec}$
• C. $66\text{sec}$
• D. $60\text{sec}$

Answer 1

The correct option is A $84\text{sec}$

Newton's law of cooling says that,
$\frac{{\theta }_1-{\theta }_2}{t}\propto (\frac{{\theta }_1+{\theta }_2}{2}-{\theta }_0)$
From the data given in the question,
$\frac{365-361}{2}=K[\frac{365+361}{2}-293]$
$\Rightarrow K=\frac{1}{35}{\text{min}}^{-1}$
Again, for change in temperature from $344\text{K}$ to $342\text{K}$,
$\frac{344-342}{t}=\frac{1}{35}[\frac{344+342}{2}-293]$
$\Rightarrow \frac{2}{t}=\frac{10}{7}$
$\Rightarrow t=\frac{7}{5}\text{min}=\frac{7}{5}\times 60=84\text{sec}$
Thus, option (a) is the correct answer.