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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?

Respuesta :

kudzordzifrancis
We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

[tex]\frac{dT}{dt}=k(T-a)[/tex]

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

[tex]\frac{dT}{T-a}=kdt[/tex]

Integrating both sides of the differential equation, we obtain;

[tex]ln(T-a)=kt +c[/tex]

This natural logarithmic equation can be rewritten as;

[tex]T-a=e^{kt +c}[/tex]

Applying the laws of exponents, we obtain,

[tex]T-a=e^{kt}\times e^{c}[/tex]

[tex]T-a=e^{c}e^{kt}[/tex]

We were given the initial conditions,

[tex]T(0)=4[/tex]

Let us apply this condition to obtain;

[tex]4-20=e^{c}e^{k(0)}[/tex]

[tex]-16=e^{c}[/tex]

Now our equation, becomes

[tex]T-a=-16e^{kt}[/tex]

or

[tex]T=a-16e^{kt}[/tex]

When we substitute a=20,
we obtain,

[tex]T=20-16e^{kt}[/tex]

b) We were also given that,

[tex]T(5)=8[/tex]

Let us apply this condition again to find k.

[tex]8=20-16e^{5k}[/tex]

This implied

[tex]-12=-16e^{5k}[/tex]

[tex]\frac{-12}{-16}=e^{5k}[/tex]

[tex]\frac{3}{4}=e^{5k}[/tex]

We take logarithm to base e of both sides,

[tex]ln(\frac{3}{4})=5k[/tex]

This implies that,

[tex]\frac{ln(\frac{3}{4})}{5}=k[/tex]


[tex]k=-0.2877[/tex]

After 15 minutes, the temperature will be,


[tex]T=20-16e^{-0.2877\times 15}[/tex]

[tex]T=20-0.21376[/tex]


[tex]T=19.786[/tex]

After 15 minutes, the temperature is approximately 20°C


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