Question #179491

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.  


1
Expert's answer
2021-04-15T07:27:10-0400

By Newton’s Cooling Law:

λdTdt=30T\lambda\frac{dT}{dt}=30-T

t=λ30TdT=λln(T30)+ct=\intop\frac{\lambda}{30-T}dT=\lambda ln(T-30)+c


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

c=λln70c=\lambda ln70

t=λ(ln70ln(T30))=λln(70T30)t=\lambda(ln70-ln(T-30))=\lambda ln(\frac{70}{T-30})


We have: T=70°CT=70\degree C when t=15 mint=15\ min

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


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


For t=20 mint=20\ min :


T=70e20/26.8+30=63.19°CT=70e^{-20/26.8}+30=63.19\degree C


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