Question

a bottle of sun tea initially has a temperature of 84°f. it is left to cool in a refrigerator that has a temperature of 40°f. after 10 minutes the temperature of the sun tea is 58°f. complete parts a through c.
a. use newton’s law of cooling, ( t = c + (t_0 - c)e^{kt} ), to find a model for the temperature of the sun tea, ( t ), after ( t ) minutes.
( t = 40 + (44)e^{-0.0894t} )
(simplify your answer. use integers or decimals for any numbers in the equation. round to four decimal places as needed.)
b. what is the temperature of the sun tea after 15 minutes?
( square )°f
(round to nearest degree as needed.)

Explanation:

Step1: Identify values for Newton's Law

Newton's Law of Cooling is \( T = C + (T_0 - C)e^{kt} \). Here, \( C = 40 \) (refrigerator temp), \( T_0 = 84 \) (initial tea temp), and at \( t = 10 \), \( T = 58 \). First, find \( k \) (though part a seems to have a typo, but part b: use \( T = 40 + 44e^{-0.0894t} \) as given? Wait, let's proceed with part b: \( t = 15 \).

Step2: Substitute \( t = 15 \) into the formula

Given \( T = 40 + 44e^{-0.0894 \times 15} \). Calculate the exponent: \( -0.0894 \times 15 = -1.341 \). Then \( e^{-1.341} \approx e^{-1.341} \approx 0.259 \) (more accurately, \( e^{-1.341} \approx e^{-1.341} \approx 0.259 \), but let's compute precisely: \( e^{-1.341} \approx 0.259 \)? Wait, \( e^{-1.341} \): \( e^{-1} = 0.3679 \), \( e^{-1.3} \approx 0.2725 \), \( e^{-1.34} \approx e^{-1.3 - 0.04} = e^{-1.3} \times e^{-0.04} \approx 0.2725 \times 0.9608 \approx 0.2619 \), \( e^{-1.341} \approx 0.2619 \times e^{-0.001} \approx 0.2619 \times 0.999 \approx 0.2616 \). Then \( 44 \times 0.2616 \approx 11.5104 \). Then \( T = 40 + 11.5104 \approx 51.5104 \), round to nearest degree: 52? Wait, wait, maybe the formula is \( T = 40 + 44e^{-0.0894t} \). Let's recalculate \( e^{-0.0894 \times 15} \): \( 0.0894 \times 15 = 1.341 \), \( e^{-1.341} \approx e^{-1.341} \approx 0.259 \)? Wait, no, let's use calculator: \( e^{-1.341} \approx 0.259 \)? Wait, actually, \( \ln(0.259) \approx -1.34 \), so yes. Wait, but let's do it step by step. \( e^{-1.341} \): using calculator, \( e^{-1.341} \approx 0.259 \). Then \( 44 \times 0.259 \approx 11.396 \). Then \( 40 + 11.396 = 51.396 \), round to nearest degree: 51? Wait, maybe my approximation is wrong. Let's use more precise calculation: \( e^{-1.341} \). Let's compute \( 1.341 \), \( e^{1.341} \): \( 1.341 \), \( e^{1} = 2.71828 \), \( e^{0.3} = 1.34986 \), \( e^{0.04} = 1.04081 \), \( e^{0.001} = 1.0010005 \). So \( e^{1.341} = e^{1 + 0.3 + 0.04 + 0.001} = e^1 \times e^{0.3} \times e^{0.04} \times e^{0.001} \approx 2.71828 \times 1.34986 \times 1.04081 \times 1.0010005 \). First, \( 2.71828 \times 1.34986 \approx 3.679 \), then \( 3.679 \times 1.04081 \approx 3.830 \), then \( 3.830 \times 1.0010005 \approx 3.834 \). So \( e^{-1.341} = 1 / 3.834 \approx 0.2608 \). Then \( 44 \times 0.2608 \approx 11.4752 \). Then \( T = 40 + 11.4752 \approx 51.4752 \), which rounds to 51 (nearest degree) or 52? Wait, 51.4752 is closer to 51. But maybe the formula was \( T = 40 + 44e^{-0.0894t} \), let's check with \( t = 10 \): \( T = 40 + 44e^{-0.894} \). \( e^{-0.894} \approx e^{-0.9} \approx 0.4066 \), \( 44 \times 0.4066 \approx 17.89 \), \( 40 + 17.89 = 57.89 \), which is close to 58 (given). So that's correct. Then for \( t = 15 \), \( e^{-0.0894 \times 15} = e^{-1.341} \approx 0.2608 \), \( 44 \times 0.2608 \approx 11.475 \), \( 40 + 11.475 = 51.475 \), round to nearest degree: 51. Wait, but maybe I made a mistake. Wait, let's use calculator for \( e^{-0.0894 \times 15} \): \( 0.0894 \times 15 = 1.341 \), \( e^{-1.341} \approx 0.2607 \). Then \( 44 \times 0.2607 = 11.4708 \), \( 40 + 11.4708 = 51.4708 \), which is approximately 51 when rounded to the nearest degree.

Answer:

51