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

Answer 1

The Correct Option is B

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
$-\frac{d \theta_{1}}{d t} =\frac{61^{\circ} C -59^{\circ} C }{4}$
$=\frac{2}{4}=\frac{1}{2}{ }^{\circ} C / min$
From Newton's law of cooling
$\frac{1}{2}^{\circ} C / min =K\left(30^{\circ}\right)$
$\Rightarrow K=\frac{1}{60}$ ...(i)
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 $-\frac{d \theta_{2}}{d t}=\frac{51-49}{t}=\frac{2^{\circ} C }{t}$
From Newton's law of cooling
$\frac{2^{\circ} C }{t}=\frac{1}{60} \times 20$
$t=6\, min$