IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Ask your questions and receive comprehensive and trustworthy answers from our experienced community of professionals.
A pork roast was taken out of a hardwood smoker when its internal temperature had reached 180°F and it was allowed to rest in a 75°F house for 20 minutes after which its internal temperature had dropped to 170°F. Assuming that the temperature of the roast follows Newton’s Law of Cooling,
a) Express the temperature of the roast, T as a function of time t.b) Find the amount of time that has passed when the roast would have dropped to 140°F had it not been carved and eaten.
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
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]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
