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.

Answers

Answer 1

Answer: 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

Answer 2

Answer: By Newton's law of cooling
$T-S=Ce^(-kt)$
S=75
at t=0
$180-75=C=105$
at t=20
$170-75=105e^(-20k)=95 \n e^(-20k)= (95)/(105)$
$k=- (2.303)/(20) log_(10)( (95)/(105)=0.00500507463889254879012536550336$
at T=140
$140-75=65=105e^(-kt)$
$t= -(2.303)/(0.00500507463889254879012536550336)log_(10)( (65)/(105)==$
$t=96 minutes$

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Answers

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Answers

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