Question

The temperature of a body falls from ${40}^{\circ }C$ to ${36}^{\circ }C$ in $5$ minutes, when placed in a surrounding at a constant temperature of ${16}^{\circ }C.$ Find the time taken for the temperature of the body to become ${32}^{\circ }C$.

• A. $4.1\text{Min.}$
• B. $5.1\text{Min.}$
• C. $6.1\text{Min.}$
• D. $7.1\text{Min.}$

Answer 1

The correct option is C $6.1\text{Min.}$

We solve this problem by an approximation method that makes use of mean temperature. The approximation errors are small because temperature differences are not large.

The mean temperature of the body as it cools from ${40}^{\circ }C$ to ${36}^{\circ }C$ is $T_{m_1}=\frac{(40+36)}{2}={38}^{\circ }C.$

Newton's law of cooling can be written as
$\frac{\Delta T}{\Delta t}=-b(T_m-T_s)$

$\frac{(36-40)}{5}=-b(38-16),$

Which gives $b=\frac{0.8}{22}{\text{Min}}^{-1}$
Note that $\frac{\Delta T}{\Delta t}$ is negative.

Let the time taken for the temperature to become ${32}^{\circ }C$ be $t$.

During this interval, $T_{m_2}=\frac{(36+32)}{2}={34}^{\circ }C$

$\therefore \frac{(32-36)}{t}=-b(34-16)$

$\frac{-4}{t}=-\left(\frac{0.8}{22}\right)18$,

$\therefore t=6.1\text{Min.}$

Hence, $\left(C\right)$ is the correct answer.