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

Answer: 35

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),
$\frac{3}{1.2}=\frac{70-{\theta }_0}{49-{\theta }_0}$
$2.5\left(49-{\theta }_0\right)=70-{\theta }_0$
$\therefore 122.5-2.5{\theta }_0=70-{\theta }_0$
$\therefore 1.5{\theta }_0=52.5$
$\therefore {\theta }_0=\frac{52.5}{1.5}=35^{\circ }C$