Ice cream removed trom a freezer has a temperature of \( -45^{\circ} \mathrm{F} \). Ten minutes after being placed on a table in a room with surrounding temperat decimal places, in Newton's Law of Cooling equation \( T(10)=76-121 e^{-10 k} \). (Enter a number.) \( k=\square \) Need Help? Readli

To solve for \( k \) in the equation given by Newton's Law of Cooling, we start with the equation: \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \] where: - \( T(t) \) is the temperature of the object at time \( t \), - \( T_s \) is the surrounding temperature, - \( T_0 \) is the initial temperature of the object, - \( k \) is the cooling constant, - \( t \) is the time in minutes. From your problem, we have: - The initial temperature of the ice cream, \( T_0 = -45^{\circ} \mathrm{F} \). - The surrounding temperature, \( T_s = 76^{\circ} \mathrm{F} \). - The temperature after 10 minutes, \( T(10) = 76 - 121 e^{-10k} \). First, we need to find \( T(10) \) using the initial conditions. 1. Calculate \( T(0) \): \[ T(0) = T_s + (T_0 - T_s) e^{-0} = T_s + (T_0 - T_s) = 76 + (-45 - 76) = 76 - 121 = -45 \] 2. Now, we need to find \( T(10) \): \[ T(10) = T_s + (T_0 - T_s) e^{-10k} \] Substituting the values: \[ T(10) = 76 + (-121) e^{-10k} = 76 - 121 e^{-10k} \] 3. We need to find \( k \). To do this, we need a specific temperature after 10 minutes. If we assume that the temperature of the ice cream after 10 minutes is known (let's say it is \( T(10) = T_f \)), we can set up the equation: \[ T_f = 76 - 121 e^{-10k} \] 4. Rearranging gives: \[ 121 e^{-10k} = 76 - T_f \] \[ e^{-10k} = \frac{76 - T_f}{121} \] 5. Taking the natural logarithm of both sides: \[ -10k = \ln\left(\frac{76 - T_f}{121}\right) \] \[ k = -\frac{1}{10} \ln\left(\frac{76 - T_f}{121}\right) \] To find \( k \), you need the value of \( T_f \) (the temperature of the ice cream after 10 minutes). If you provide that value, I can help you calculate \( k \).