Question

A cup of coffee at temperature ${100}^{o}C$ is placed in a room whose temperature is ${15}^{o}C$ and it cools to ${60}^{o}C$ in $5$ minutes.
Find its temperature after a further interval of $5$ minutes

Answer 1

The initial temperature of the coffee $T_0={100}^{\circ }C$.

The temperature of the room ${15}^{\circ }C$.

Let $T$ be the temperature at time $t$ in minutes.

By Newton's law of cooling,

$\frac{dT}{dt}\alpha (T-15)$

$\therefore \frac{dT}{dt}=k(T-15)$

$\Rightarrow \frac{dT}{T-15}=kdt$

$\Rightarrow \int \frac{dT}{T-15}=\int kdt$

$\Rightarrow log(T-15)=kt+logc$

$\Rightarrow log\left(\frac{T-15}{c}\right)=kt$

$\Rightarrow T-15=c{e}^{kt}$

At $t=0,T=100$

We have $100-15=c{e}^0\Rightarrow 85=c$

Thus we get $T-15=85e^{kt}$.

When $t=5minutes,T=60$

$\Rightarrow 60-15=85e^{5k}$.

$\Rightarrow e^{5k}=\frac{45}{85}=\frac{9}{17}$

We have to find the temperature after $5$ minutes. That is temperature at $t=10minutes$

$\therefore T-15=85e^{10k}$

$\Rightarrow T-15=85(e^{5k}{)}^2$

$\Rightarrow T-15=85{\left(\frac{9}{17}\right)}^2[\because e^{5k}=\frac{9}{17}]$

$\Rightarrow T-15=85\left(\frac{81}{289}\right)$

$\Rightarrow T-15=23.82$

$\Rightarrow T=15+23.82$

$\therefore T={38.82}^{\circ }C$

Hence the temperature after $10$ minutes is ${38.82}^{\circ }C$