# Newton's law of cooling relating to differential equations

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.

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#### Solution Preview

The setup is y=Ce^(kt)+68

Then C=1532 when you plug in t=0

Then you find k. ...

#### Solution Summary

Core temperature is found using Newton's Law of Cooling and differential equations.

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