Question

For a liquid allowed to cool down in a room whose temperature is ${20}^{\circ }C$, it takes $5min$ to cool down from $50$ to ${46}^{\circ }C$. Find out the temperature of the liquid after the next $5min$.

Answer 1

$T_0=20°C$

$t=5min$

$T_1=50°C$

$T_2=46°C$

By Newtons law of cooling,

$\frac{MC}{K}=\frac{\left(\frac{T_1+T_2}{2}\right)-T_0}{\frac{T_1-T_2}{t}}$

substituting the values ,we get $\frac{MC}{K}=35$

Now again after next 5 mins,

$T_2=?$

$T=46°C$

$t=5min$

$T_0=20°C$

$35=\frac{(\frac{46+T_2}{2}-20)}{\left(\frac{46-T_2}{5}\right)}$

$7(46-T_2)=\frac{46+T_2}{2}-20$

$322-7T_2=3+\frac{T_2}{2}$

$\frac{T_2}{2}+7T_2=322-3=319$

$7.5T_2=319$

$T_2=42.5°C$

So , in the next 5 minutes the temperature of the liquid will drop to $42.5°C$