b. Suppose you pour a cup of tea that is at \( 150^{\circ} \mathrm{F} \) and leave it in a room that is at \( 60^{\circ} \mathrm{F} \). After 15 minutes, the tea is at \( 100^{\circ} \mathrm{F} \). How long before the tea is at \( 80^{\circ} \mathrm{F} \) Show your work.

To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula is given by: \[ \frac{dT}{dt} = -k(T - T_a) \] where: - \( T \) is the temperature of the object (the tea), - \( T_a \) is the ambient temperature (the room temperature), - \( k \) is a positive constant, - \( t \) is time. We can also express the solution to this differential equation as: \[ T(t) = T_a + (T_0 - T_a)e^{-kt} \] where: - \( T_0 \) is the initial temperature of the object. ### Step 1: Set up the equation In this case: - \( T_a = 60^{\circ} \mathrm{F} \) - \( T_0 = 150^{\circ} \mathrm{F} \) After 15 minutes, the temperature of the tea is \( 100^{\circ} \mathrm{F} \). We can use this information to find the value of \( k \). ### Step 2: Use the information to find \( k \) At \( t = 15 \) minutes, we have: \[ 100 = 60 + (150 - 60)e^{-15k} \] Simplifying this: \[ 100 = 60 + 90e^{-15k} \] \[ 40 = 90e^{-15k} \] \[ e^{-15k} = \frac{40}{90} = \frac{4}{9} \] Taking the natural logarithm of both sides: \[ -15k = \ln\left(\frac{4}{9}\right) \] \[ k = -\frac{1}{15} \ln\left(\frac{4}{9}\right) \] ### Step 3: Find the time when the temperature is \( 80^{\circ} \mathrm{F} \) Now we want to find the time \( t \) when \( T(t) = 80^{\circ} \mathrm{F} \): \[ 80 = 60 + (150 - 60)e^{-kt} \] This simplifies to: \[ 80 = 60 + 90e^{-kt} \] \[ 20 = 90e^{-kt} \] \[ e^{-kt} = \frac{20}{90} = \frac{2}{9} \] Taking the natural logarithm of both sides: \[ -kt = \ln\left(\frac{2}{9}\right) \] \[ t = -\frac{1}{k} \ln\left(\frac{2}{9}\right) \] Substituting \( k \): \[ t = -\frac{15}{\ln\left(\frac{4}{9}\right)} \ln\left(\frac{2}{9}\right) \] ### Step 4: Calculate \( t \) Now we need to compute \( \ln\left(\frac{4}{9}\right) \) and \( \ln\left(\frac{2}{9}\right) \): 1. Calculate \( \ln\left(\frac{4}{9}\right) \): \[ \ln\left(\frac{4}{9}\right) = \ln(4) - \ln(9) = 2\ln(2) - 2\ln(3) = 2(\ln(2) - \ln(3)) \] 2. Calculate \( \ln\left(\frac{2}{9}\right) \): \[ \ln\left(\frac{2}{9}\right) = \ln(2) - \ln(9) = \ln(2) - 2\ln(3) \] Now substituting these values into the equation for \( t \): \[ t = -\frac{15}{2(\ln(2) - \ln(3))} \cdot (\ln(2) - 2\ln(3)) \] This simplifies to: \[ t = -\frac{15}{2} \cdot \frac{\ln(2) - 2\ln(3)}{\ln(2) - \ln(3)} \] ### Step 5: Numerical Calculation Using approximate values: - \( \ln(2) \approx 0.693 \) - \( \ln(3) \approx 1.099 \) Calculating: \[ \ln(2) - \ln(3) \approx 0.