A body cools from $70 °\mathrm{C}$ to $40 °\mathrm{C}$ in $5$ minutes. Calculate the time it takes to cool from $60 °\mathrm{C}$ to $30 °\mathrm{C}$. The temperature of the surroundings is $20 °\mathrm{C}$.

According to Newton's law of cooling, Rate of cooling is directly proportional to the temperature difference between body and surrounding.

$$$\frac{dT}{dt}=K\left(T-T_0\right)=K\left(\frac{T_i+T_f}{2}-T_0\right)$

Where, $T=\frac{T_i+T_f}{2}$ is average temperature of body.

The substitute the value as per first condition

$\Rightarrow \frac{70-40}{5}=K \left(\frac{{\displaystyle 70+40}}{{\displaystyle 2}}-20\right)$

$\therefore K=\frac{6}{35}$
As per the second condition,

$\frac{60-30}{t}=\frac{6}{35}\left(\frac{{\displaystyle 60+30}}{{\displaystyle 2}}-20\right)$

The time required is given to cool the body from $60°$ to $30°$ is given by

$\Rightarrow t=\frac{30\times 35}{25\times 6}=7 \min$