Question

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the carton, T, as a function of time follows Newton's cooling law:
dT / dt = k (T - a),
where a is the temperature of the surroundings and k is a constant. We assume the temperature
the surroundings are a = 20 (given in ◦C), and that T (0) = 4 (given in ◦C).
a) Show that T (t) = 20 - 16e ^ kt (given in ◦C).
b) After 5 minutes the temperature of the carton is 8◦C. What is the temperature after 15 minutes?

Answer 1

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$


$k=-0.2877$

After 15 minutes, the temperature will be,


$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$


$T=19.786$

After 15 minutes, the temperature is approximately 20°C