Question

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

Answer 1

Here,

Initial temperature $\left(T_i\right)={80}^{\circ }$C

Final temperature $\left(T_f\right)={50}^{\circ }$C

Temperature of the surrounding $\left(T_0\right)={20}^{\circ }$C

$t=5$ min

According to Newton's law of cooling,

Rate of cooling $\frac{dT}{dt}=K[\frac{(T_i+T_f)}{2}-T_o]$

$\frac{(T_f-T_i)}{t}=K[\frac{(80+50)}{2}-20]$

$\frac{80-50}{5}=K[65-20]$

$6=K\times 45$

$K=\frac{6}{45}=\frac{2}{15}$

In second condition,

initial temperature $=T_i={60}^{\circ }$C

Final temperature $T_f={30}^{\circ }$C

Time taken for cooling is $t$

According to Newton's law of cooling

$\frac{(60-30)}{t}=\frac{2}{15}[\frac{(60+30)}{2}-20]$

$\frac{30}{t}=\frac{2}{15}\times 25$

$\frac{30}{t}=\frac{50}{15}=\frac{10}{3}$

$\therefore t=9$ min