Let T (t) be the temperature of an object in a room of fixed ambient temperature A. The rate of
change of temperature is proportional to the difference between T and A. Suppose an iced tea at 5◦C
is placed in a cabana where the temperature is 35◦C. After ten minutes the iced tea is at 10◦C. What
is the temperature of the drink after 40 minutes?
To solve this problem, we can use the differential equation:
dT/dt = k(T - A)
where dT/dt is the rate of change of temperature with respect to time, T is the temperature of the drink, A is the ambient temperature, and k is the proportionality constant.
Using the given information, we can set up the initial condition:
T(0) = 5
T(10) = 10
To find the value of k, we can substitute the initial condition into the differential equation:
dT/dt = k(T - A)
(10 - A) = k(5 - A)
10 - A = 5k - Ak
10 = (5 - A)k
k = 10/(5 - A)
Now, we can solve the differential equation using separation of variables:
(dT)/(T - A) = k dt
∫(1)/(T - A) dT = k∫dt
ln|T - A| = kt + C
|T - A| = e^(kt+C)
T - A = ±e^C * e^(kt)
T - A = Ce^(kt)
Now, we can use the initial condition T(10) = 10 to solve for the constant C:
10 - A = Ce^(k*10)
10 - A = Ce^(10k)
We can solve for C by dividing both sides of the equation by e^(10k):
C = (10 - A) / e^(10k)
Now, we can find the temperature of the drink after 40 minutes:
T(40) = A + Ce^(kt)
T(40) = A + (10 - A) / e^(10k) * e^(k*40)
T(40) = A + (10 - A) / e^(10k) * e^(40k)
T(40) = A + (10 - A) / e^(-10k) * e^(40k)
T(40) = A + (10 - A) * e^(40k + 10k)
T(40) = A + (10 - A) * e^(50k)
To find the value of k, we can use the initial condition T(10) = 10:
10 = A + (10 - A) * e^(10k)
10 = Ae^(10k) + 10e^(10k) - Ae^(10k)
10 = 10e^(10k)
e^(10k) = 1
10k = 0
k = 0
Since k = 0, the temperature of the drink remains constant at A throughout the entire process. Therefore, the temperature of the drink after 40 minutes is 35°C (the ambient temperature).