Question

A body cools from ${80}^{o}C$ to ${50}^{o}C$ in 5 minutes. Calculate the time it takes to cool from ${60}^{o}C$ to ${30}^{o}C$. The temperature of the surroundings is ${20}^{o}C$.

Answer 1

According to Newton’s law of cooling, we have:

$(-dT/dt)=K(T-T_0)$

$dT/K(T-T_0)=-Kdt$ .... (i)
Where,
Temperature of the body = T
Temperature of the surroundings $T_0={20}^{o}C$
K is a constant
Temperature of the body falls from ${80}^{o}C$ to ${50}^{o}C$ in time, t = 5 min = 300 s
Integrating equation (i), we get:

${\int }_{50}^{80}\frac{dT}{K(T-T_0)}=-{\int }_0^{300}Kdt$

$lo{g}_e[T-T_0{]}_{50}^{80}=-K[t{]}_0^{300}$

$\frac{2.3026}{K}lo{g}_{10}\frac{80-20}{50-20}=-300$

$\frac{2.3026}{K}lo{g}_{10}2=-300$

$\frac{-2.3026}{300}lo{g}_{10}2=K$ ... (ii)

The temperature of the body falls from ${60}^{0}C$ to ${30}^{o}C$ in time =t'

Hence, we get:

$\frac{2.3026}{K}lo{g}_{10}\frac{60-20}{30-20}=t^{\prime }$

$\frac{-2.3026}{t^{\prime }}lo{g}_{10}4=K$ ... (iii)

Equating (ii) and (iii), we get:

$\frac{-2.3026}{t^{\prime }}lo{g}_{10}4=\frac{-2.3026}{300}lo{g}_{10}2$

$\therefore t^{\prime }=300\times 2=600s=10min$

Therefore, the time taken to cool the body from ${60}^{o}C$ to ${30}^{o}C$ is 10 minutes