Question

this exercise uses newtons law of cooling.
a roasted turkey is taken from an oven when its temperature has reached 185°f and is placed on a table in a room where the temperature is 72°f.
find the following values for newtons law of cooling (t(t) = t_0 + d_0e^{-kt}), where time is measured in minutes and the temperature is measured in degrees fahrenheit, if the temperature of the turkey is 145°f
after half an hour.
t_0 = 72
d_0 = 113
k=
enter an exact answer with no rounding.
(a) what is its temperature (in °f) after 45 min? (round your answer to the nearest whole number.)
131
°f
(b) after how many hours will the turkey cool to 100°f? (round your answer to one decimal place.)
1.6
hr
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Explanation:

Step1: Find the value of \(k\)

We know that \(T(t)=T_{s}+D_{0}e^{-kt}\). When \(t = 30\) (half - hour), \(T(30)=145\), \(T_{s}=72\), \(D_{0}=113\).
Substitute into the formula: \(145 = 72+113e^{-30k}\).
First, subtract 72 from both sides: \(145 - 72=113e^{-30k}\), so \(73 = 113e^{-30k}\).
Then, divide both sides by 113: \(\frac{73}{113}=e^{-30k}\).
Take the natural logarithm of both sides: \(\ln(\frac{73}{113})=- 30k\).
Solve for \(k\): \(k=-\frac{1}{30}\ln(\frac{73}{113})=\frac{1}{30}\ln(\frac{113}{73})\approx\frac{1}{30}\times0.447\approx0.0149\).

Step2: (a) Find the temperature after \(t = 45\) minutes

We have \(T(t)=72 + 113e^{-0.0149t}\).
Substitute \(t = 45\) into the formula: \(T(45)=72+113e^{-0.0149\times45}\).
First, calculate \(-0.0149\times45=-0.6705\).
Then, \(e^{-0.6705}\approx0.511\).
\(T(45)=72 + 113\times0.511=72+57.743\approx130\).

Step3: (b) Find the time \(t\) when \(T(t)=100\)

Set \(T(t)=100\), so \(100 = 72+113e^{-0.0149t}\).
Subtract 72 from both sides: \(28 = 113e^{-0.0149t}\).
Divide both sides by 113: \(\frac{28}{113}=e^{-0.0149t}\).
Take the natural logarithm of both sides: \(\ln(\frac{28}{113})=-0.0149t\).
Solve for \(t\): \(t=-\frac{1}{0.0149}\ln(\frac{28}{113})=\frac{1}{0.0149}\ln(\frac{113}{28})\).
\(\ln(\frac{113}{28})\approx1.44\), \(t=\frac{1.44}{0.0149}\approx96.64\) minutes.
Convert to hours: \(t=\frac{96.64}{60}\approx1.6\) hours.

Answer:

(a) \(130^{\circ}F\)
(b) \(1.6\) hours