Certain quantity of water cools from $70^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in the first

Question

Certain quantity of water cools from $70^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in the first 5 minutes and to $54^{\circ} \mathrm{C}$ in the next 5 minutes. The temperature of the surroundings is\

Professor PC Thomas Classes · Answer

(a) : Let $T_{s}$ be the temperature of the surroundings. According to Newton's law of cooling $$ \frac{T_{1}-T_{2}}{t}=K\left(\frac{T_{1}+T_{2}}{2}-T_{s}ight) $$ For first 5 minutes,\ $T_{1}=70^{\circ} \mathrm{C}, T_{2}=60^{\circ} \mathrm{C}, t=5$ minutes\ $	herefore \quad \frac{70-60}{5}=K\left(\frac{70+60}{2}-T_{s}ight)=K\left(65-T_{s}ight)$\ For next 5 minutes,\ $T_{1}=60^{\circ} \mathrm{C}, T_{2}=54^{\circ} \mathrm{C}, t=5$ minutes\ $	herefore \quad \frac{60-54}{5}=K\left(\frac{60+54}{2}-T_{s}ight)$ $$ \frac{6}{5}=K\left(57-T_{s}ight) 	ag{ii} $$ Divide eqn. (i) by eqn. (ii), we get $$ \begin{aligned} & \frac{5}{3}=\frac{65-T_{s}}{57-T_{s}} \ & 285-5 T_{s}=195-3 T_{s} \ & 2 T_{s}=90 	ext { or } T_{s}=45^{\circ} \mathrm{C} \end{aligned} $$