Answer to Question #104563 in Differential Equations for Suraj Singh

Question

According to Newton’s law of cooling, the rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is 290°C and the substance cools from 370°C to 330°C in 10 minutes, find when the temperature will be 295°C.

Answer 1

Newton's law of cooling is $Q/t \propto (T_s-T_f)$

Solving for it, we get;

${T\left( t \right) = {T_S} }+{ \left( {{T_0} – {T_S}} \right){e^{ – kt}}}$

where; t is the time ;

$T_o=$ initial temperature of substance

$T_s=$ temperature of surrounding fluid

In this case

$$T_s=290°C; T_o=370°C$$

For t=10 minutes and $T=330°C$ ---(given)

To find: t for $T=295°C$

$330=290+80e^{-10k}$

$\implies k =ln2/10$

$295=290+80e^{-kt}$

$\implies e^{kt}=16 \implies t =4ln2/k$

$\implies t=40 min.$ (Answer)