Question

a pizza baked at 425°f is removed from the oven at 4:00 pm into a room that is a constant 72°f. after 5 minutes, the pizza is at 300°f.
(a) at what time can you begin eating the pizza if you want its temperature to be 130°f?
(b) using a graphing utility, graph the relation between temperature and time.
(c) using intersect, determine the time that needs to elapse before the pizza is 170°f.
(d) trace the function for large values of time. what do you notice about y, the temperature?

(a) the temperature of the pizza is 130°f at about 4:21 pm.
(b) choose the correct graph below. the graphing window is 0,130,10 by 0,425,85.
(c) using intersect, approximately □ minutes need to elapse before the pizza is 170°f.
(type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Find the cooling - law formula

Newton's Law of Cooling is \(T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}\), where \(T_{a}\) is the ambient temperature, \(T_{0}\) is the initial temperature, \(t\) is the time in minutes. Here, \(T_{a} = 72^{\circ}F\), \(T_{0}=425^{\circ}F\). When \(t = 5\), \(T(5)=300^{\circ}F\).
Substitute into the formula: \(300=72+(425 - 72)e^{-5k}\).
First, simplify: \(300 - 72=353e^{-5k}\), so \(228 = 353e^{-5k}\). Then \(e^{-5k}=\frac{228}{353}\). Take the natural logarithm of both sides: \(-5k=\ln(\frac{228}{353})\), and \(k=-\frac{1}{5}\ln(\frac{228}{353})\approx0.089\).
The formula becomes \(T(t)=72+(425 - 72)e^{-0.089t}=72 + 353e^{-0.089t}\).

Step2: Solve for \(t\) when \(T(t)=170\)

Set \(T(t)=170\), so \(170=72+353e^{-0.089t}\).
Subtract 72 from both sides: \(170 - 72=353e^{-0.089t}\), \(98 = 353e^{-0.089t}\).
Then \(e^{-0.089t}=\frac{98}{353}\). Take the natural logarithm of both sides: \(-0.089t=\ln(\frac{98}{353})\).
Solve for \(t\): \(t=\frac{\ln(\frac{98}{353})}{- 0.089}\approx14.4\)

Answer:

\(14.4\)