Answer to Question #179491 in Differential Equations for Joy Lubigo

Question

The temperature of air is 30°C, and the substance cools from 100°C to 70°C in 15 minutes. Find the temperature after 20 min.  

Answer 1

By Newton’s Cooling Law:

$\lambda\frac{dT}{dt}=30-T$

$t=\intop\frac{\lambda}{30-T}dT=\lambda ln(T-30)+c$

We have: $T=100\degree C$ when $t=0$

$c=\lambda ln70$

$t=\lambda(ln70-ln(T-30))=\lambda ln(\frac{70}{T-30})$

We have: $T=70\degree C$ when $t=15\ min$

$\lambda=\frac{15}{ln(70/40)}=26.8$

$T=70e^{-t/\lambda}+30=70e^{-t/26.8}+30$

For $t=20\ min$ :

$T=70e^{-20/26.8}+30=63.19\degree C$