At 10:00 AM, an object is removed from a furnace and placed in an environment with a constant temperature of 68 degrees. Its core temperature is 1600 degrees. At 11:00 AM, its core temperature is 1090 degrees. Find its core temperature at 5:00 PM on the same day.
The setup is y=Ce^(kt)+68
Then C=1532 when you plug in t=0
Then you find k. ...
Core temperature is found using Newton's Law of Cooling and differential equations.