Question

A body cools down from ${40}^{\circ }\text{C}$ to ${35}^{\circ }\text{C}$ in $10$ minutes and to ${30}^{\circ }\text{C}$ in another $15$ minutes. Find the temperature of the surroundings.

• A. ${20}^{\circ }\text{C}$
• B. ${22.5}^{\circ }\text{C}$
• C. ${25}^{\circ }\text{C}$
• D. ${27}^{\circ }\text{C}$

Answer 1

The correct option is B ${22.5}^{\circ }\text{C}$

Let $T_0$ be the temperature of surroundings.
Now, for cooling from ${40}^{\circ }\text{C}$ to ${35}^{\circ }\text{C}$ in 10 minutes,
$\frac{dT}{dt}=-K(T_{avg}-T_0)$
$=-K(\frac{T_1+T_2}{2}-T_0)\left[\begin{array}{c}{T}_1={35}^{\circ }\text{C}\\ T_2={40}^{\circ }\text{C}\end{array}\right]$
$\Rightarrow \frac{5}{10\times 60}=-k[37.5-T_0]---\left(1\right)$

From ${35}^{\circ }\text{C}$ to ${30}^{\circ }\text{C}$ in 15 minutes
$\frac{dT}{dt}=-k(\frac{T_1+T_2}{2}-T_0)\left[\begin{array}{c}{T}_1={30}^{\circ }\text{C}\\ T_2={35}^{\circ }\text{C}\end{array}\right]$
$\Rightarrow \frac{5}{15\times 60}=-k(32.5-T_0)---\left(2\right)$
From eq ($1$) and eq($2$),
$\frac{5}{10\times 60}\times \frac{15\times 60}{5}=\frac{-k(37.5-T_0)}{-k(32.5-T_0)}$
$\Rightarrow \frac{3}{2}=\frac{37.5-T_0}{32.5-T_0}$
$\Rightarrow 3(32.5-T_0)=2(37.5-T_0)$
$\Rightarrow T_0={22.5}^{\circ }\text{C}$