Question

A hot bowl of soup cools according to Newtons law of cooling. Its temperature (in degrees Fahrenheit) at time t is given by T(t)= 68+144e^-0.04t, where t is given in minutes.
Answer these questions
What was the initial temperature of the soup?


What was the temperature of the soup after 15 minutes?


How long after serving is the soup 125 degrees Fahrenheit?

Answer 1

still need help?

Answer 2

the initial temperature of the soup is what you get when you replace \(t\) by \(0\)

Answer 3

\[ T(t)= 68+144e^{-0.04t}\]
\[ T(0)= 68+144e^0\]

Answer 4

you good from there?

Answer 5

Yeah so then 68+144= 212