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

212°F
147°F
23.2 min

1) Initial temperature for t=0
T(0)=68+144e^(0)=68+144=212°F
2) After 15 min
T(15)=68+144e^(-0.04*15)=68+144e^(-0.6)=147°F
3) 125°F -> t=?
125=68+144e^(-0.04t)
144e^(-0.04t)=125-68=57
e^(-0.04t)=57/144=0.4 you can apply ln on both sides:
-0.04t=ln(0.4) solving you get t=23.3 min