GenevaKrawchuk818
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# 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.

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