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Sagot :
The roast was 82.2degrees Celsius when it was taken out of the hardwood smoker
It would of taken 80 mins for it to cool from 180 to 140
As every 20 mins it would decrease it's temperature by 10degrees C so it would be
80mins for it to reach 140 from 180
It would of taken 80 mins for it to cool from 180 to 140
As every 20 mins it would decrease it's temperature by 10degrees C so it would be
80mins for it to reach 140 from 180
By Newton's law of cooling
[tex]T-S=Ce^{-kt}[/tex]
S=75
at t=0
[tex]180-75=C=105[/tex]
at t=20
[tex]170-75=105e^{-20k}=95 \\ e^{-20k}= \frac{95}{105} [/tex]
[tex]k=- \frac{2.303}{20} log_{10}( \frac{95}{105}=0.00500507463889254879012536550336[/tex]
at T=140
[tex]140-75=65=105e^{-kt}[/tex]
[tex]t= -\frac{2.303}{0.00500507463889254879012536550336}log_{10}( \frac{65}{105}==[/tex]
[tex]t=96 minutes[/tex]
[tex]T-S=Ce^{-kt}[/tex]
S=75
at t=0
[tex]180-75=C=105[/tex]
at t=20
[tex]170-75=105e^{-20k}=95 \\ e^{-20k}= \frac{95}{105} [/tex]
[tex]k=- \frac{2.303}{20} log_{10}( \frac{95}{105}=0.00500507463889254879012536550336[/tex]
at T=140
[tex]140-75=65=105e^{-kt}[/tex]
[tex]t= -\frac{2.303}{0.00500507463889254879012536550336}log_{10}( \frac{65}{105}==[/tex]
[tex]t=96 minutes[/tex]
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