Question

a pizza baked at 425°f is removed from the oven at 8:00 pm into a room that is a constant 70°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 8:20 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 formula

Newton's Law of Cooling is given by \(T(t)=T_{a}+(T_{0} - T_{a})e^{-kt}\), where \(T(t)\) is the temperature of the object at time \(t\), \(T_{a}\) is the ambient temperature, \(T_{0}\) is the initial temperature of the object, and \(k\) is the cooling constant.

We know that \(T_{0}=425^{\circ}F\), \(T_{a} = 70^{\circ}F\), and when \(t = 5\), \(T(5)=300^{\circ}F\).

Substitute these values into the formula: \(300=70+(425 - 70)e^{-5k}\)

First, simplify the equation: \(300-70=(425 - 70)e^{-5k}\), so \(230 = 355e^{-5k}\)

Then, solve for \(k\): \(e^{-5k}=\frac{230}{355}\approx0.6479\)

Take the natural logarithm of both sides: \(-5k=\ln(0.6479)\)

\(k=-\frac{\ln(0.6479)}{5}\approx-\frac{- 0.434}{5}=0.0868\)

The formula becomes \(T(t)=70 + 355e^{-0.0868t}\)

Step2: Find \(t\) when \(T(t)=170^{\circ}F\)

Substitute \(T(t) = 170\) into the formula: \(170=70+355e^{-0.0868t}\)

Simplify: \(170 - 70=355e^{-0.0868t}\), so \(100 = 355e^{-0.0868t}\)

\(e^{-0.0868t}=\frac{100}{355}\approx0.2817\)

Take the natural logarithm of both sides: \(-0.0868t=\ln(0.2817)\)

\(t=\frac{\ln(0.2817)}{- 0.0868}\approx\frac{-1.266}{-0.0868}\approx14.6\)

Answer:

\(14.6\)