Answer:
Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the temperature of its surroundings. In this case, the rate of change of the temperature of the coffee can be given by:
dT/dt = -k(T - Ts)
where T is the temperature of the coffee, Ts is the temperature of the surroundings (in this case, the freezer), k is a constant of proportionality, and dT/dt represents the rate of change of the temperature with respect to time.
We can use the given information to find the value of k:
125F - 0F = (125F - Ts)e^(-k*5min)
125/(-5) = (125-Ts)e^(-5k)
-25/125 = (125-Ts)e^(-5k)
-1/5 = (125-Ts)e^(-5k)
-1/(5(125-Ts)) = e^(-5k)
ln(-1/(5(125-Ts))) = -5k
k = ln(-1/(5(125-Ts)))/(-5)
Note that the negative sign in front of the fraction is due to the fact that the coffee is cooling down.
Now, we can use this value of k to find the temperature of the coffee after 15 minutes:
83F - 0F = (83F - Ts)e^(-k*5min)
83/(-5) = (83-Ts)e^(-5k)
-83/5 = (83-Ts)e^(-5k)
-1/(5(83-Ts)) = e^(-5k)
ln(-1/(5(83-Ts))) = -5k
k = ln(-1/(5(83-Ts)))/(-5)
T(15min) - 0F = (T(15min) - Ts)e^(-k15min)
T(15min) - 0F = (T(15min) - 0F)e^(-k15min)
T(15min) = (T(15min) - 0F)e^(-k*15min)
Substituting the value of k that we just found, we get:
T(15min) = (T(15min) - 0F)e^(3*ln(-1/(5(83-Ts))))
T(15min) = (T(15min) - 0F)(-1/(5(83-Ts)))^3
T(15min) = (T(15min) - 0F)(-1/1953125)
Simplifying, we get:
T(15min) = -1/1953125 T(15min)
Multiplying both sides by -1953125, we get:
T(15min) = 0F
Therefore, the temperature of the coffee after 15 minutes in the freezer is 0F.