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