Question

A body cools from ${400}^{\circ }C$ to ${325}^{\circ }C$ in the room temperature of ${25}^{\circ }C$ in $20$ minutes. Its temperature after $1$ hour is

• A. ${216}^{\circ }C$
• B. ${215}^{\circ }C$
• C. ${218}^{\circ }C$
• D. ${217}^{\circ }C$

Answer 1

Answer: (D)

${217}^{\circ }C$

Let $T\left(t\right)$ be the temperature of the body in $t$ minutes.

According to Newton's rate of cooling,

$dT\left(t\right)=-k\left(T\right(t)-T)dt$ where $T$ is the surrounding temperature and k is a constant.

$\Rightarrow \frac{dT\left(t\right)}{T\left(t\right)-T}=-kdt$

Integrating on both sides,

$\ln \left(T\right(t)-T)=-kt+\ln C$

When $t=0\Rightarrow \left(T\right(0)-T)=(400-25{)}^{0}C={375}^{0}C=C$

When $t=20\Rightarrow \ln \left(T\right(20)-{25}^{0}C)=-20k+\ln {375}^{0}C$

$k=-\frac{\ln \frac{325-25}{375}}{20}=\frac{1}{20}\ln \frac{375}{300}=\frac{1}{20}\ln \frac{5}{4}$

So, $\frac{T\left(t\right)-{25}^{0}C}{{375}^{0}C}=(\frac{5}{4}{)}^{-\frac{1}{20}t}$

$\Rightarrow T\left(60\right)=(\frac{4}{5}{)}^3\times {375}^{0}C+{25}^{0}C=(4^3\times 3+25{)}^{0}C={217}^{0}C$

ie, Option D is the correct answer.