Question

A liquid takes 5 minutes to cool from ${80}^{\circ }C$ to ${50}^{\circ }C$. How much time(in mins) will it take to cool from ${60}^{\circ }C$ to ${30}^{\circ }C$? The temperature of surrounding is ${20}^{\circ }C$. Use exact method.

Answer 1

The mathematical interpretation of Newton's law of cooling states that:
$\frac{dT}{dt}=-bA(T-T_0)$; where $T_{0}isroomtemperature$
Solving this differential equation, with basic integration and applying the limits:

$T$:- varying from $T_{1}to{T}_2$

• $t$:- varying from 0 to t. The time it takes to change from $T_{1}to{T}_2$

Thus we get:
$\frac{1}{t}ln\left(\frac{T_2-T_0}{T_1-T_0}\right)=-bA=constant$
Here, 't' is the time taken to for cooling from $T_1$ to $T_2$, $T_0$ is the room temperature.
Applying this here:
$\frac{1}{5}ln\left(\frac{30}{60}\right)=\frac{1}{t}ln\left(\frac{10}{40}\right)$

$\frac{1}{5}ln\left(\frac{1}{2}\right)=\frac{1}{t}ln\left(\frac{1}{4}\right)$

$\frac{-ln2}{5}=\frac{-2ln2}{t}$

$t=10minutes$