Answer:
A:) T(t) = 305(230/305)^(t/10) +70
B:) T(20) ≈ 243.4 °F
Explanation:
You have a pie removed from a 375° oven to a room temperature of 70° that has cooled to 300° after 10 minutes. You want to know its temperature after 20 minutes.
Law of Cooling
Newton's law of cooling tells you the rate of change of an object's temperature is proportional to the difference between the object's temperature and that of its surroundings. The differential equation describing this has an exponential function as its solution. That exponential function can be written as ...
temperature = (initial temperature difference)·(cooling factor)^(time/(cooling time)) + (final temperature)
Application
Here, the initial temperature difference between the pie and room temperature is ...
initial temperature difference = 375° -70° = 305°
The final temperature is the room temperature:
final temperature = 70°
The cooling factor is the multiplier applied to the temperature difference in the time period "cooling time". Here, the pie is at 300°, a temperature difference of 300° -70° = 230° after a cooling time of 10 minutes. The cooling factor is ...
cooling factor = 230/305
A:) Equation
These values let us write the equation as ...
T(t) = 305(230/305)^(t/10) +70
B:) 20 minutes
The temperature after 20 minutes is found using this equation:
T(20) = 305(230/305)^(20/10) +70 = 305(46/61)^2 +70 = 243 27/61
T(20) ≈ 243.4 . . . . degrees F
__
Additional comment
The cooling factor can be expressed as a value that lets the exponent be t. Often, it is written as (e^k), where ...
k = 1/10·ln(230/305) ≈ -0.0282232
Then the equation is ...
T(t) = 305·e^(-0.0282232t) +70
It could also be written as ...
T(t) = 305(0.972171^t) +70
where 0.972171 = (46/61)^(1/10).