Question

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?

Answer 1

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).