Answer to Question #309160 in Differential Equations for Anny

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 year if the temperature of the year is 298 k a and the substance cools from 478 k to 428 k in 20 minutes. find when the temperature will be 405 k

Answer 1

Solution;

According to Newton's law of cooling;

$-\frac{dT}{dt}\alpha(T-T_s)$

$\frac{dT}{dt}=-k(T-T_s)$

Integrating;

$\int\frac{dT}{T-T_s}=-k\int dt$

$ln(T-T_s)=-kt$

$T-T_s=e^{-kt}+C$

$T(t)=T_s+Ce^{-kt}$

$T_s=298k$

Boundary conditions;

$t=0,T=478k$

$t=20,T=428$

$478=298+Ce^0$

$C=478-298=180$

Also;

$\frac{478}{428}=\frac{Ce^o}{Ce^{-20k}}$

$e^{20k}=1.116$

$k=0.00552$

Therefore;

$T(t)=298+180e^{-0.00552t}$

At 405,t=?;

$405=298+180e^{-0.00552t}$

$107=180e^{-0.00552t}$

$ln(\frac{107}{180})=-0.0055t$

$t=94.23minutes$