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Irina18 [472]
3 years ago
9

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

arton, 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?
Mathematics
1 answer:
Kisachek [45]3 years ago
6 0
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


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