Question

A body cools in $7$ minutes from ${60}^{\circ }C$ to ${40}^{\circ }C$. What will be its temperature after next $7$ minutes, if the temperature of the surroundings is ${10}^{\circ }C$?

• A. ${20}^{\circ }C$
• B. ${24}^{\circ }C$
• C. ${28}^{\circ }C$
• D. ${35}^{\circ }C$

Answer 1

The correct option is C ${28}^{\circ }C$

Given that,
Temperature of surrounding, ${\theta }_o={10}^{\circ }$
${\theta }_1={60}^{\circ }$
${\theta }_2={40}^{\circ }$
Time taken to change from ${\theta }_1$ to ${\theta }_2=7\text{min}$

According to Newton's law of cooling $\frac{d\theta }{dt}=-K(\theta -{\theta }_0)$
${\int }_0^{t}dt=\frac{1}{K}{\int }_{{\theta }_1}^{{\theta }_2}-\frac{d\theta }{(\theta -{\theta }_0)}$
$t=\frac{1}{K}{\log }_e\left[\frac{{\theta }_1-{\theta }_{\circ }}{{\theta }_2-{\theta }_{\circ }}\right]$
When temperature changes from ${60}^{\circ }$ to ${40}^{\circ }$
$\Rightarrow 7=\frac{1}{K}{\log }_e\left[\frac{60-10}{40-10}\right]$ ...(i)
Let the the temperature after next $7$ minutes is $\theta$
$\Rightarrow 7=\frac{1}{K}{\log }_e\left[\frac{40-10}{\theta -10}\right]$ ...(ii)
From equation (i) and (ii)
$\log \frac{50}{30}=\log \frac{30}{\theta -10}$
$\frac{5}{3}=\frac{30}{(\theta -10)}\Rightarrow \theta ={28}^{\circ }C$
Hence, option, (c) correct