Question

Certain quantity of water cools from $85^{\circ} C$ to $55^{\circ} C$ in the first $10$ minutes and to $43 .{ }^{\circ} C$ in the next 10 minutes. The temperature of the surroundings is ______ ${ }^{\circ} C$.

Answer 1

By Newton's law of cooling,
$\frac{\theta_{1}-\theta_{2}}{\Delta t}=K\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{0}\right)$
For first $10$ minutes,
$ \frac{85-55}{10}=K\left(70-\theta_{0}\right)$
$\therefore 3=K \times\left(70-\theta_{0}\right)$ ..... (i)
Similarly, for next $10$ minutes,
$\frac{55-43}{10}= K \left(49-\theta_{0}\right)$
$1.2= K \times\left(49-\theta_{0}\right)$ ..... (ii)
Dividing equation (i) by equation (ii),
$ \frac{3}{1.2}=\frac{70-\theta_{0}}{49-\theta_{0}} $
$ 2.5\left(49-\theta_{0}\right)=70-\theta_{0} $
$ \therefore 122.5-2.5 \theta_{0}=70-\theta_{0} $
$ \therefore 1.5 \theta_{0}=52.5 $
$ \therefore \theta_{0}=\frac{52.5}{1.5}=35^{\circ} C $