Question

A body cools from ${80}^{\circ }C$ to ${70}^{\circ }C$ in 5 minutes and to ${62}^{\circ }C$ in the next 5 minutes. Calculate temperature of the surrounding.

Answer 1

Hint:Newton’s law of cooling says that the rate of change of the temperature with respect to time is equal to the difference of the temperature of an object and the temperature of the surrounding. The change in temperature decreases as the difference in temperature of the surrounding and the object decreases.

Formula used:The formula of change in temperature with time is given by,
$\Rightarrow {\displaystyle \frac{dT}{dt}}=-k\left(T-T_o\right)$
Where average temperature is T, the time is t, the constant is k and the temperature of the surrounding is$T_o$.

Complete step by step solution:
It is given in the problem that a body cools from ${80}^{\circ }C$ to ${70}^{\circ }C$ in 5 minutes and to ${62}^{\circ }C$ in the next 5 minutes and we need to calculate the temperature of the surrounding.
For first decrease in temperature. The average temperature is equal to,
$\Rightarrow T={\displaystyle \frac{80+70}{2}}$
$\Rightarrow T={\displaystyle \frac{150}{2}}$
$\Rightarrow T={75}^{\circ }C$
The formula of change in temperature with time is given by,
$\Rightarrow {\displaystyle \frac{dT}{dt}}=-k\left(T-T_o\right)$
Where average temperature is T, the time is t, the constant is k and the temperature of the surrounding is$T_o$.
The body cools from ${80}^{\circ }C$ to ${70}^{\circ }C$ in 5 minutes. According to Newton’s law of cooling.
$\Rightarrow {\displaystyle \frac{80-70}{5}}=-k\left(75-T_o\right)$
$\Rightarrow {\displaystyle \frac{10}{5}}=-k\left(75-T_o\right)$
$\Rightarrow 2=-k\left(75-T_o\right)$………eq. (1)
For the decrease temperature from ${70}^{\circ }C$ to ${62}^{\circ }C$ in 5 minutes.
The average temperature is equal to,
$\Rightarrow T={\displaystyle \frac{70+62}{2}}$
$\Rightarrow T={66}^{\circ }C$
The body cools from ${70}^{\circ }C$ to ${62}^{\circ }C$ in 5 minutes. According to Newton’s law of cooling.
$\Rightarrow {\displaystyle \frac{dT}{dt}}=-k\left(T-T_o\right)$
$\Rightarrow {\displaystyle \frac{70-62}{5}}=-k\left(66-T_o\right)$
$\Rightarrow {\displaystyle \frac{8}{5}}=-k\left(66-T_o\right)$………eq. (2)
Dividing the equation (2) with equation (1) we get,
$\Rightarrow {\displaystyle \frac{8}{10}}={\displaystyle \frac{66-T_o}{75-T_o}}$
$\Rightarrow 8\left(75-T_o\right)=10\left(66-T_o\right)$
$\Rightarrow 600-8T_o=660-10T_o$
$\Rightarrow 2T_o=60$
$\Rightarrow T_o={30}^{\circ }C$.

The surrounding temperature is equal to$$T_o={30}^{\circ }C$$.

Note:It is advisable for students to understand and remember the formula of Newton's law of cooling as it is very helpful in solving problems like these. The average temperature is taken of the initial and final temperature of the object.