Question

In a room where the temperature is ${30}^{\circ }C$ a body cools from ${61}^{\circ }C$ to ${59}^{\circ }C$ in $4$ minutes. The time taken by the body to cool from ${51}^{\circ }C$ to ${49}^{\circ }C$ will be about :

• A. $5$ minutes
• B. $8$ minutes
• C. $4$ minutes
• D. $6$ minutes

Answer 1

The correct option is D $6$ minutes

The average temperature of the liquid in the first case
${\theta }_1=\frac{61+59}{2}={60}^{\circ }C$
Temperature difference from surrounding
${\theta }_1-{\theta }_0=60-30={30}^{\circ }C$
The rate of fall of temperature is
$-\frac{d{\theta }_1}{dt}=\frac{{61}^{\circ }C-{59}^{\circ }C}{4}$
$=\frac{2}{4}={\frac{1}{2}}^{\circ }C/min$
From Newton's law of cooling
${\frac{1}{2}}^{\circ }C/min=K\left({30}^{\circ }\right)\Rightarrow K=\frac{1}{60}.....\left(i\right)$
In the second case, average temperature
${\theta }_2=\frac{51+49}{2}={50}^{\circ }C$
Temperature difference with surrounding
${\theta }_2-{\theta }_0={50}^{\circ }C-{30}^{\circ }C$
$={20}^{\circ }C$
If it takes a time $t$ to cool from ${51}^{\circ }C$ to ${49}^{\circ }C$
then $-\frac{d{\theta }_2}{dt}=\frac{51-49}{t}=\frac{2^{\circ }C}{t}$
From Newton's law of cooling
$\frac{2^{\circ }C}{t}=\frac{1}{60}\times 20$
$\Rightarrow t=6min$