Question

a pizza baked at 425°f is removed from the oven at 10:00 pm into a room that is a constant 71°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 10:21 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: Recall Newton's Law of Cooling

Newton's Law of Cooling formula is \(T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}\), where \(T(t)\) is the temperature at time \(t\), \(T_{a}\) is the ambient temperature, \(T_{0}\) is the initial temperature. Here, \(T_{a} = 71^{\circ}F\), \(T_{0}=425^{\circ}F\). When \(t = 5\), \(T(5)=300^{\circ}F\).
Substitute into the formula: \(300=71+(425 - 71)e^{-5k}\).
First, simplify the equation: \(300-71 = 354e^{-5k}\), so \(229=354e^{-5k}\). Then \(e^{-5k}=\frac{229}{354}\). Take the natural logarithm of both sides: \(-5k=\ln(\frac{229}{354})\), and \(k=-\frac{1}{5}\ln(\frac{229}{354})\approx0.087\).

Step2: Find \(t\) when \(T(t) = 170\)

Substitute \(T(t)=170\), \(T_{a} = 71\), \(T_{0}=425\), and \(k\approx0.087\) into \(T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}\).
We get \(170=71+(425 - 71)e^{-0.087t}\).
Simplify: \(170 - 71=354e^{-0.087t}\), so \(99 = 354e^{-0.087t}\). Then \(e^{-0.087t}=\frac{99}{354}\).
Take the natural logarithm: \(- 0.087t=\ln(\frac{99}{354})\). Solve for \(t\): \(t=\frac{\ln(\frac{99}{354})}{- 0.087}\approx14.5\)

Answer:

\(14.5\)