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.
By Newton’s Cooling Law:
λdTdt=30−T\lambda\frac{dT}{dt}=30-TλdtdT=30−T
t=∫λ30−TdT=λln(T−30)+ct=\intop\frac{\lambda}{30-T}dT=\lambda ln(T-30)+ct=∫30−TλdT=λln(T−30)+c
We have: T=100°CT=100\degree CT=100°C when t=0t=0t=0
c=λln70c=\lambda ln70c=λln70
t=λ(ln70−ln(T−30))=λln(70T−30)t=\lambda(ln70-ln(T-30))=\lambda ln(\frac{70}{T-30})t=λ(ln70−ln(T−30))=λln(T−3070)
We have: T=70°CT=70\degree CT=70°C when t=15 mint=15\ mint=15 min
λ=15ln(70/40)=26.8\lambda=\frac{15}{ln(70/40)}=26.8λ=ln(70/40)15=26.8
T=70e−t/λ+30=70e−t/26.8+30T=70e^{-t/\lambda}+30=70e^{-t/26.8}+30T=70e−t/λ+30=70e−t/26.8+30
For t=20 mint=20\ mint=20 min :
T=70e−20/26.8+30=63.19°CT=70e^{-20/26.8}+30=63.19\degree CT=70e−20/26.8+30=63.19°C
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