Problem
A hot bowl of soup is served at a dinner party. It starts to cool according to Newton's Law of Cooling. Its temperature $T$ (measured in degrees Fahrenheit) after $t$ minutes is given by \[ T(t)=66+208 e^{-0.06 t} \] What is the initial temperature of the food? degrees Fahrenheit. What is the temperature after 13 minutes? degrees Fahrenheit. How long will it take from the time the food is served until the temperature is 95 degrees Fahrenheit? minutes.
Solution
Step 1 ：The initial temperature of the food is given by the function \(T(t)\) at \(t=0\). We can substitute \(t=0\) into the function to find the initial temperature.
Step 2 ：Substitute \(t=0\) into the function \(T(t)=66+208 e^{-0.06 t}\) to get the initial temperature.
Step 3 ：\(T(0)=66+208 e^{-0.06 \times 0}\)
Step 4 ：\(T(0)=66+208 \times 1\)
Step 5 ：\(T(0)=66+208\)
Step 6 ：\(T(0)=274\)
Step 7 ：Final Answer: The initial temperature of the food is \(\boxed{274}\) degrees Fahrenheit.
