Question

A body cools from $60^{\circ }C$ to $50^{\circ }C$ in $10$ minutes. If the room temperature is $25^{\circ }C$ and assuming Newton's cooling law holds good, the temperature of the body at the end of next $10$ minute is

• A. $45^{\circ }C$
• B. $42.85^{\circ }C$
• C. $40^{\circ }C$
• D. $38.5^{\circ }C$

Answer 1

Answer: (B)

Using the relation we have
$\frac{ms\left(60^{\circ }-50^{\circ }\right)}{10}$
$=K\left(\frac{60^{\circ }+50^{\circ }}{2}-25^{\circ }\right)$
$\frac{10ms}{10}=K\left(55^{\circ }-25^{\circ }\right)$
$\frac{10ms}{10}=30K$ ...(1)
Suppose the required temperature is $T$ Then
$\frac{ms\left(50^{\circ }-T\right)}{100}$
$=K\left(\frac{50^{\circ }+T}{2}-25^{\circ }\right)$
$=K\left(\frac{50^{\circ }+T-50^{\circ }}{2}\right)$
$\frac{m\left(50^{\circ }-T\right)}{10}=\frac{KT}{2}$ ...(2)
Dividing equation (2) by (1) we get
$\frac{50^{\circ }-T}{10}=\frac{\frac{T}{2}}{30^{\circ }}=\frac{T}{60^{\circ }}$
$50^{\circ }-T=\frac{T}{6}$ or $300^{\circ }-6T=T$
So, $T=300^{\circ }$
or $T=\frac{300^{\circ }}{7}=42.85^{\circ }C$