Question

A body cools from ${50}^{\circ }\text{C}$ to ${49.9}^{\circ }\text{C}$ in $5\text{sec}$. It cools from ${40}^{\circ }\text{C}$ to ${39.9}^{\circ }\text{C}$ in $t\text{sec}$. Assuming Newton's law of cooling to be valid and temperature of surroundings to be ${30}^{\circ }C$ , the value of $t$ will be

Answer 1

From Newton's law of cooling,
$\left(\frac{{\theta }_1-{\theta }_2}{t}\right)=k(\frac{{\theta }_1+{\theta }_2}{2}-\theta )\{\because \Delta \theta \text{is very less}\}$
When the body cools from ${50}^{\circ }\text{C}$ to ${49.9}^{\circ }\text{C}$ at surrounding temperature ${30}^{\circ }\text{C}$,
we have $\frac{0.1}{5}=k\left(\frac{39.9}{2}\right)......\left(a\right)$
When the body cools from ${40}^{\circ }\text{C}$ to ${39.9}^{\circ }\text{C}$ at surrounding temperature ${30}^{\circ }\text{C}$
$\frac{0.1}{t}=k\left(\frac{19.9}{2}\right).....\left(b\right)$
From $\left(a\right)\&\left(b\right)$, we get
$\frac{t}{5}=\frac{39.9}{19.9}\approx 2$
$\Rightarrow t=10\text{sec}$