Question

The temperature of a body in a room is ${100}^{o}F$. After five minutes, the temperature of the body becomes ${50}^{o}F$. After another $5$ minutes, the temperature becomes ${40}^{o}F$. What is the temperature of surroundings?

Answer 1

Let at time $t$, temperature of a body is $T$. Let $S$ be the surrounding temperature, then according to Newton's cooling law
$\frac{dT}{dt}\propto (T-S)$
Since body looses temperature constant of variation is negative
$\therefore \frac{dT}{dt}=-k(T-S)$
Now integrate both sides
$\frac{dT}{T-S}=-kdt$
$\int \left(\frac{1}{T-S}\right)=-k\int dt$
$\therefore \log (T-S)=-kt+c....\left(1\right)$
Now $t=1$ and we have $T={100}^{o}C$
$\therefore \log (100-S)=c$
From equation (1)
$\log (T-S)=-kt+\log (100-S)$
$t=5\Rightarrow T-{50}^{o}F$
$\log (50-S)=-5t+\log (100-S)...\left(2\right)$
Also $t=100\Rightarrow T-{40}^{o}F$
$\log (40-S)=-10t+\log (100-S)...\left(3\right)$
From equation (2) and (3)
$\frac{1}{5}\log \left(\frac{50-S}{100-S}\right)=-k=\frac{1}{10}\log \left(\frac{40-S}{100-S}\right)$
$2\log \left(\frac{50-S}{100-S}\right)=\log \left(\frac{40-S}{100-S}\right)$
$\therefore {\left(\frac{50-S}{100-S}\right)}^2=\left(\frac{40-S}{100-S}\right)$
$\therefore {(50-S)}^2=(40-S)(100-S)$
$\therefore 2500-100S+S^2=4000-140S+S^2$
$\therefore 49S=1500=\frac{75}{2}$
$\therefore S={\left(37.5\right)}^{o}F$
$\therefore$ Temperature of room is ${\left(37.5\right)}^{o}F$